2nd PUC Statistics Question Bank Chapter 5 Theoretical Distribution

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Karnataka 2nd PUC Statistics Question Bank Chapter 5 Theoretical Distribution

2nd PUC Statistics Bernoulli Distribution Exercise Problems

Question 1.
Define a Bernoulli variate
Answer:
If ‘X’ is a discrete random variable with probability mass function p(x) = p^xq^{1 – x}, where x = 0, 1; 0 < p < 1; q = 1 – p, then ‘X’ is a Bernoulli variate

Question 2.
Define a Bernoulli distribution.
Answer:
If ‘X’ is a discrete random variable with probability mass function p(x) = p^xq^{1 – x},where x = 0, 1; 0 < p < 1; q = 1 – p,then the distribution of ‘X’ is named as Bernoulli distribution

Question 3.
Write down the probability mass
Answer:
Function of a Bernoulli distribution.
Probability mass function is
p(x) = p^x q^{1 – x}, When x = 0, 1; 0 < p < 1; q = 1 – p

Question 4.
Write the range and parameter of a Bernoulli distribution ?
Answer:
Parameter is p and range x = 0,1

Question 5.
Give an example for Bernoulli variate.
Answer:
No. of female children when a baby is born.

Question 6.
Write the relationship between mean and variance of a Bernoulli distribution?
Answer:
mean > variance.

Question 7.
What are the mean and variance of a Bernoulli distribution?
Answer:
Mean = p; Variance = pq

Question 8.
What is meant by Bernoulli trial? Give an example?
Answer:
A trial is said to be a Bernoulli trial of it satisfies the following conditions:

• Every trial results either in a success or failure
• The probability of success ‘p’ remains same for every trial
  Example: The tossing of a fair coin once can be considered as a Bernoulli trial.

Question 9.
If X₁, X₂ …….. X_n are independently identically distributed Bernoulli variate with parameter p, what is the distribution of X = X₁ + X₂ + ……. + X_n ?
Answer:
If X₁, X₂ …… X_n are independently and identically distributed Bernoulli variate with common parameter p. Then their sum X = X₁ + X₂ + …… + X is a Binomial variate with parameters n and p.

Question 10.
Write down the Bernoulli distribution with parameter p = 0.23
Answer:

p.m.f: p(x) = (0.23)^x (0.77)^{1 – x} = 0,1

Question 11.
If p = \(\frac { 1 }{ 4 }\) = 0.25 for a Bernoulli distribution, write down the p.m.f and find its variance
Answer:
Given p = \(\frac { 1 }{ 4 }\) then q = 1 – \(\frac { 1 }{ 4 }\) = \(\frac { 3 }{ 4 }\)
The p.m.f is p(x) = p^xq^1-x
p(x) = \(\left(\frac{1}{4}\right)^{x} \cdot\left(\frac{3}{4}\right)^{1-x}\)
Variance = p.q = \(\frac{1}{4} \times \frac{3}{4}=\frac{3}{16}\)

Question 12.
Find the mean and variance of a Bernoulli distribution with parameter p = 0.43
Answer:
p = 0.43
q = 1 – 0.43 = 0.57
Mean p = 0.43
Variance = pq = (0.43)(0.57) = 0.2451

Question 13.
Write down the Bernoulli distribution with parameter p = \(\frac { 2 }{ 5 }\)
Answer:

Question 14.
For a Bernoulli distribution, if p = 0.27 write the PMF. Also find S.D
Answer:
p = 0.27 q = 1 – p ⇒ 1 – 0.27 = 0.73
PMF is p(x) = (0.27)^x (0.73)^{1 – x}, x = 0,1
SD = √Var = √pq
= \(\sqrt{0.27 \times 0.73}\)
⇒ \(\sqrt{0.1971}\) = 0.4439.

Question 15.
Find the mean and variance of a Bernoulli distribution with parameter p = \(\frac { 2 }{ 3 }\)
Answer:
p = \(\frac{2}{3}\) q = \(1-\frac{2}{3}=\frac{1}{3}\)
Mean = p = \(\frac{2}{3}\) variance = pq = \(\frac{2}{3} \times \frac{1}{3}=\frac{2}{9}\)

Question 16.
Define Binomial variate If x is a random variable with parameters ‘n’ and ‘p’ with
Answer:
PMF ⇒ p(x) = ⁿC_x, p^xq^{n – x}, x = 0,1, 2, …………. n, 0 < p < 1,
q = 1 – p
Then X is a Binomial variate.

Question 17.
Define Binomial distribution.
Answer:
If X is a discrete random variable with parameters ‘n’ and ‘p’ with
PMF ⇒ p(x) = ⁿC₀ p^xq^{n – x} x = 0, 1, 2, ………..  n, 0 < p < 1,
q = 1 – P
then the distribution of X is named as Binomial distribution

Question 18.
Write down the p.m.f of a Binomial distribution
Answer:
p.m.f – p(x) = ⁿC_x p^xq^{n – x} where x = 0,1, 2, …………. n
0 < p < 1, q = 1 – p

Question 19.
Write down the range of a binomial distribution?
Answer:
Range of x = 0,1, 2 …. n

Question 20.
Write down the parameters of a binomial distribution?
Answer:
Parameters are ‘n’ and ‘p’

Question 21.
Give two examples of Binomial variate.
Answer:

1. Number of heads obtained in 4 tosses of a coin
2. Number of defective articles in a random sample of 6 articles drawn from a manufactured lot.

Question 22.
What are the mean, variance and S.D. of a Binomial distribution?
Answer:
Mean = np
Variance = npq
S.D. = \(\sqrt{\mathrm{npq}}\)

Question 23.
Write the relationship between mean and variance of Binomial distribution?
Answer:
Mean > variance (or) np > npq.

Question 24.
Write down two features of a Binomial distribution.
Answer:

1. In a Binomial distribution n and p are the parameters
2. In a Binomial distribution x = 0,1,2, n is the range
3. The mean and mode is equal if np is an integer

Question 25.
Mention the recurrence relation for successive probabilities in a binomial distribution with parameters ‘n’ and ‘p’
Answer:
Recurrence relation
\(p(x)=\frac{n-x+1}{x} \frac{p}{q} p(x-1)\)

Question 26.
Mention the recurrence relation for successive frequencies in a Binomial distribution with parameters ‘n’ and ‘p’
Answer:
\(\mathrm{T}_{\mathrm{x}}=\frac{\mathrm{n}-\mathrm{x}+1}{\mathrm{x}}. \frac{\mathrm{p}}{\mathrm{q}} \cdot \mathrm{T}_{\mathrm{x}-1}\)

Question 27.
Mention the conditions under which Binomial distribution tends to Poisson distribution.
Answer:
The Binomial distribution turns to Poisson distribution when n → ∞,p → 0,np = λ is a constant.

One of the phrases used to quantify a solution’s concentration is “formula of normality” in chemistry.

Question 28.
Mention the conditions under which Binomial distribution tends to normal distributions.
Answer:
The Binomial distribution turns to Normal distribution when n is large and p and q are almost equal.

Question 29.
For what value of ‘P’, is Binomial distribution symmetrical ?
Answer:
The Binomial distribution is symmetrical if p = q = \(\frac { 1 }{ 2 }\)

Question 30.
The mean and variance of a Binomial distribution are 4 and 5 respectively. Comment on this statement and give reason to your comment.
Answer:
Mean = np = 4
Variance = npq = 5
The statement is incorrect as for a Binomial distribution mean is greater than variance

Question 31.
In a Binomial distributions, if n = 6 and p = \(\frac { 1 }{ 3 }\) find mean, mode and variance.
Answer:
n = 6, p = \(\frac { 1 }{ 3 }\) ,q = 1 – p ⇒ 1 — \(\frac { 1 }{ 3 }\) ⇒ \(\frac { 2 }{ 3 }\)
Mean = np = 6 × \(\frac { 1 }{ 3 }\) = 2
Mode = Mean = 2 (Since np = 2 is an integer, for a BD mean = mode)
Variance = npq = \(6 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{3}\)

Question 32.
In a Binomial distribution with 5 trials, the mean is 3, find ‘p’ and S.D
Answer:
n = 5, mean = np = 3
5p = 3, p = \(\frac { 3 }{ 5 }\) = 0.6
q = 1 — \(\frac { 3 }{ 5 }\) = 0.4
SD = \(\sqrt{\mathrm{npq}}=\sqrt{(5)(0.6)(0.4)}=\sqrt{1.2}=1.0954\)

Question 33.
If a Binomial distribution has mean 3 and variance 2, find the parameters
Answer:
Mean = np = 3 .
Variance = npq = 2

q = \(\frac { 2 }{ 3 }\)
p = 1 – q = 1 — \(\frac { 2 }{ 3 }\) ⇒ \(\frac { 1 }{ 3 }\)
Given
Mean = np = 3
n\(\left(\frac{1}{3}\right)\) = 3 n = 3 × 3 = 9
Parameters n = 9, p = \(\frac { 1 }{ 3 }\)

Question 34.
The mean and S.D of a Binomial distribution are 8 & 2 respectively. Find the parameters
Answer;
Mean = np = 8
SD = \(\sqrt{\mathrm{npq}}\) = 2

∴ npq = 4

q = \(\frac { 1 }{ 2 }\), p = 1 – q = 1 – \(\frac { 1 }{ 2 }\) ⇒\(\frac { 1 }{ 2 }\)
np = 8
n × \(\frac { 1 }{ 2 }\) = 8
n = 8 × 2 = 16
∴ Parameters are n = 16, p = \(\frac { 1 }{ 2 }\)

Question 35.
In a college 35% students are girls. Find the probability that (i) Two are girls (ii) At least one is a girl in a random sample of 5 students
Answer:
Given n = 5
p = \(\frac { 35 }{ 100 }\) = 0.35, q = 1 – p
⇒ 1 – 0.35
⇒ 0.65
Let X denote no. of girls in a random sample of 5 students then X ~ B(n = 5, p = 0.35)
The pmf is, p(x) = ⁿC_x . p^x . q^{5 – x}
where x = 0, 1, 2, 3 ……………… n, 0 < p < 1
p(x) = ⁵C_x . (0.35)^x . (0.65)^{5 – x}
where x = 0,1 ……….. 5
(i) P (Two are girls) = P(X = 2)
= ⁵C₂ (0.35)²(0.65)^5-2
= 10(0.1225)(0.2746)
= 0.3364

(ii) P(at least one is a girl)
= P(X ≥ 1) = 1 – P(X< 1)
= 1 – P(X = 0)
= 1 – ⁵C₀ (0.35)⁰(0.65)^{5 – 0}
= 1 – 0.1160 = 0.884

Question 36.
If the chance that a vessel (ship) arrives safely at a port is \(\frac { 9 }{ 10 }\), find the chance that out of 6 vessels which are expected to ar rive at the port, at least 5 will arrive safely
Answer:
Given n = 6
P = \(\frac { 9 }{ 10 }\), q = 1 – P
⇒ 1 – \(\frac{9}{10}=\frac{1}{10}\)
Let X denote no. of ships arriving safely out of 6 ships Then X is a Binomial variate with parameters n = 6
and p = \(\frac { 9 }{ 10 }\)
The p.m.f is
p(x) = \(^{6} C_{x}\left(\frac{9}{10}\right)^{x} \times\left(\frac{1}{10}\right)^{6-x}\)
where x = 0,1, ……. 6
(i) P(atleast 5 will arrive safely) = P(X ≥ 5)
= P(X = 5) + P(X = 6)

[6.(0.9)⁵ (0.1)¹] + [1(0.9)⁶ (0.1)⁰]
⇒ [6.(0.59049) (0.1) + 1(0.5314)]
⇒ 0.3542 + 0.5314 ⇒ 0.8856

Question 37.
Assuming that birth to male child and birth to female child to be equiprobable, find the probability that a family with three children has 2 or more male children
Answer:
n = 3, p = 0.5, q = 1 – p = 1 – 0.5 = 0.5
X denote number of male children in a family of 3 children Then ‘X’ is a binomial variate with parameters n = 3 and p = 0.5
p.m.f is
p(x) = ³C_x(0.5)^x(0.5)^{3 – x} where x = 0,1, …… 3
P(2 or more male children)
= P(X ≥ 2) = P(X = 2) + P(X = 3)
= [³c₂(0.5)²(0.5)^{3 – 2}] + [³C₃(0.5)³(0.5)^{3 – 3}]
= (3 × 0.25 × 0.5) + (1 × 0.125 × 1)
= (0.375)+ (0.125) = 0.5

Question 38.
The probability of an arrow hitting a tree is \(\frac { 1 }{ 3 }\). if 4 arrows are aimed at the tree, find the probability that
(i) 3 arrows miss the tree
(ii) At least two arrows hit the tree
Answer:
Given n = 4,p = \(\frac { 1 }{ 3 }\), q = 1 – p ⇒ 1 – \(\frac { 1 }{ 3 }\) = \(\frac { 2 }{ 3 }\)
Let X denote no. of arrows hitting the tree
Then, X is a Binomial variate with parameters n = 4
and –
The p.m.f is

(i) P(3 arrows miss the tree)
= P(1 arrow hits the three) = P(X = 1)
= ⁴c₁\(\left(\frac{1}{3}\right)^{1}\left(\frac{2}{3}\right)^{4-1}\)
= \(4 \times \frac{1}{3} \times \frac{8}{27}=\frac{32}{81}=0.3951\)

(ii) P(at least 2 arrows hit the tree)
= P(X ≥ 2) = 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1)]

Question 39.
In a certain university, the chance that a j professor suffers from dust allergy is \(\frac { 1 }{ 5 }\). Find the chance that out of 8 professors in a university 2 will not suffer from the allergy at least 2 will suffer from the allergy
Answer:
Given n = 8,p = \(\frac { 1 }{ 5 }\), q – 1-p ⇒ 1 – \(\frac { 1 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Let X denote no. of professor suffering from dust allergy out of 8 professors
Then X, is a Binomial variate with parameter n = 8
and P = \(\frac { 1 }{ 5 }\)
p.m.f is

(i) P(2 will not suffer from the allergy)
⇒ P(6 will suffer from the allergy) = P(X = 6)
p(x) = ⁸c₆\(\left(\frac{1}{5}\right)^{6}\left(\frac{4}{5}\right)^{8-6}\)
= \(28 \times \frac{1}{15625} \times \frac{16}{25}=\frac{44}{390625}=0.001147\)

(ii) P(At least 2 will suffer from the allergy)
= P(X ≥ 2) = 1 – P(X < 2)
= 1-[P(X = 0) + P(X = 1)]

Question 40.
In a city 60% people are non-vegetarians. In a random sample of 6 persons find the probability that (i) 3 are vegetarian, (ii) At least one is vegetarian
Answer:
Given p = 0.40, q = 1 – 0.4 = 0.60, n = 6
Let X denote number of vegetarians from a sample of 6 persons
Then X is Binomial variate with parameters, n = 6, p = 0.40
The p.m.f p(x) = ⁶C_x (0.40)^x (0.60)^{6 – x}
where x = 0,1, 2, ……….. 6
(i) P(3 are vegetarians) = P(X = 3) = ⁶C₃ (0.40)³ (0.60)^{6 – 3}
= 20.(0.064) (0.216) ⇒ 0.27648

(ii) P(At least one is vegetarian)
= P(X ≥ 1) = 1 – P (X = 0)
= 1 – ⁶C₀ (0.40)⁰ (0.60)^{6 – 0}
= 1 – 0.046656 = 0.9533

Question 41.
If on an average one ship in every 10 is sunk, find the chance that out of 5 ships atleast 4 will arrive safely
Answer:
Given n = 5 q = \(\frac { 1 }{ 10 }\), p = 1 – \(\frac { 1 }{ 10 }\) = \(\frac { 9 }{ 10 }\)
Let X denote No. of ships arriving safely out of 10 ships
Then X is a Binomial variate with parameters n = 5 & p = \(\frac { 9 }{ 10 }\)
The pmf is

(i) P(at least 4 will arrive safely)
= P(X ≥ 4) = P[X = 4) + P(X = 5)

= [5 × 0.6561 × 0.1] + [1 × 0.59049 × 1]
= 0.32805 + 0.59049 = 0.9186

Question 42.
In a village \(\frac { 1 }{ 3 }\) of the people are literates. If 100 investigators meet 5 persons each to see if they are literates, then how many investigators would you expect to report that 2 or more were literates
Answer:
Given n = 5, p = \(\frac { 1 }{ 3 }\),q = 1 – \(\frac { 1 }{ 3 }\) = \(\frac { 2 }{ 3 }\)
Let X denote number of persons who are literates among 5 persons
Then X is Binomial variate with parameters n = 5 & p = \(\frac { 1 }{ 3 }\)
p.m.f is

(i) P(2 or more literates) = P(X ≥ 2) = 1 – P(X < 2)
= 1- [P(X = 0) + P(X = 1)]

= 1 – (0.1316 + 0.3293) = 1 – 0.4609 = 0.539
∴ Out of 100 investigators who reported that 2 or more were literates
= 100 × 0.54 = 54

Question 43.
The incidence of an occupational disease in an industry is such that the workers have 20%, chance of suffering from it. What is the probability that out of 6 workers 4 or contract the disease.
Answer:
Let X denote number of workers contracting the disease out of 6 workers.
Then X is a Binomial variate with parameters
n = 6 and p = 20% ⇒ 0.2, q = 1 – p ⇒ 1 – 0.2 = 0.8
p.m.f is
p(x) = ⁶C_x(0.20)^x (0.80)^{6 – x} where x = 0,1, ……… 6
P(4 or more contract the disease)
= P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)
= [⁶C₄(0.20)⁴ (0.80)^{6 – 4}] + [⁶C₅(0.20)⁵ (0.80)^{6 – 5}] + [⁶C₆(0.20)⁶ (0.80)^{6 – 6}]
= [15(0.0016) (0.64)] + [6 × 0.00032 × 0.8] + [1 × 0.00064 × 1]
= 0.01536 + 0.001536 + 0.000064
= 0.01696

Question 44.
In a certain school 40% of the students have opted for first language Kannada. As suming 20 teachers take a sample of 4 students each, how many teachers report that 2 or 3 students opted for first language Kannada
Answer:
Let X denote number of students opted Kannada from a sample of 4 students
X is a Binomial variate with parameters
n = 4, p = 0.40 q = 1 – p ⇒ 1 – 0.4 ⇒ 0.6
p.m.f
p(x) = ⁴C_x(0.40)^x (0.6)^{4 – x} where x = 0,1, …………. 4
P (2 or 3 students opting for first language Kannada)
= P(X = 2) or P(X = 3) = P(X = 2) + P(X = 3)
= [⁴C₂(0.4)²(0.6)^{4 – 2}] + [⁴C₃(0.4)³(0.6)^{4 – 3}]
= (6 × 0.16 × 0.36) + (4 × 0.064 × 0.6)
= 0.3456 + 0.1536 = 0.4992 = 0.5
Out of 20 teachers, number of teachers who report that 2 or 3 students opted for first language Kannada is N.p(x) = 20 × 0.5 = 10

Question 45.
5 unbiased coins are tossed 64 times. Calculate the expected frequencies for the number of heads obtained
Answer;
Let X denote no. of Heads obtained in a toss of five unbiased coins
Given n = 5,p = \(\frac { 1 }{ 2 }\), q = \(\frac { 1 }{ 2 }\) , N = 64
The p.m.f is P(x) = ⁵C_x\(\left(\frac{1}{2}\right)^{x}\left(\frac{1}{2}\right)^{5-x}\)
where x = 0,1 ………… 5

Question 46.
7 fair coins are tossed 384 times. Find the expected frequencies for the number of heads obtained
Answer:
Let X denote Total number of heads obtained in a throw of 7 fair coins
X is Binomial variate with parameters
n = 7,p = \(\frac { 1 }{ 2 }\),q = \(\frac { 1 }{ 2 }\), N = 384
The p.m.f is P(X) = ⁷C_X\(\left(\frac{1}{2}\right)^{x}\left(\frac{1}{2}\right)^{7-x}\)
where x = 0,1 ……….. 7

Question 47.
six unbiased coins and tossed 128 times. Find the theoretical frequencies for the number of tails obtained
Answer:
Let x denote Number of tails obtained in a throw of 6 coins X is a Binomial variable with parameter
Given n = 6 P = \(\frac { 1 }{ 2 }\) q = \(\frac { 1 }{ 2 }\) , N = 128
The p.m.f is p(x) = ⁶c_x\(\left(\frac{1}{2}\right)^{x}\left(\frac{1}{2}\right)^{6-x}\)
Where x = 0, 1 ………… 6

Question 48.
Four unbiased coins are tossed 256 times. Find the theoretical frequencies for the number of heads obtained.
Answer:
Let X = The number of heads obtained in a toss of 4 coins
Then n = 4, p = \(\frac { 1 }{ 2 }\),q = \(\frac { 1 }{ 2 }\)
The p.m.f is P(x) = ⁴C_x\(\left(\frac{1}{2}\right)^{x}\left(\frac{1}{2}\right)^{4-x}\) ⇒ x = 0,1,2,3,4

Question 49.
Fit a Binomial distribution for the following data and obtain the theoretical frequencies

Answer:
Given n = 4, p is to be estimated
Let X denote The number of defective balloons

Here, np = mean = x̄ = \(\frac { 159 }{ 74 }\) = 2.14
n = 4, 4p = 2.14 ⇒ p = \(\frac { 2.14 }{ 4 }\) ⇒ 0.54, q = 0.46
The p.m.f is p(x) = ⁴C_x(0.54)^x(0.46)^{4 – x} x = 0,1,2,4…
The frequency function is
T_x = N.p(x) = 74 × ⁴C_x (0.54)^x (0.46)^{4 – x}
The frequencies are, when x = 0,
T₀ = 74 × ⁴C₀ × (0.54)⁰ (0.46)^{4 – 0} = 3.3078 ≅ 3
Remaining theoretical frequencies are calculated by using recurrence relation
Recurrence relation = T_X = \(\frac{n-x+1}{x} \cdot \frac{p}{q} \cdot T_{x}-1\)
T₁ = \(\frac{4-1+1}{1} \times \frac{0.54}{0.46} \times 3.3078\) = 4 × 1.1739 × 3.3078 = 15.52 = 16
T₂ = \(\frac{4-2+1}{2} \times \frac{0.54}{0.46} \times 15.52\) = 1.5 × 1.1739 × 15.52 = 27.32 = 27
T₃ = \(\frac{4-3+1}{3} \times \frac{0.54}{0.46} \times 27.32\) = 0.667 × 1.1739 × 27.32 = 21.5 = 22
T₄ = \(\frac{4-4+1}{4} \times \frac{0.54}{0.46} \times 21.5\) = 0.25 × 1.1739 × 21.4 = 6.30 = 6
Thus the observed and theoretical frequencies are tabulated

Question 50.
For the following data, fit a Binomial distribution and obtain the expected frequencies

Answer:
Given n = 5 and p is to be estimated

Here, np = x̄ = \(\frac { 600 }{ 200 }\) = 3
np = 3, n = 5, 5p = 3, p = \(\frac { 3 }{ 5 }\), q = \(\frac { 2 }{ 5 }\)
p = \(\frac { 3 }{ 5 }\) = 0.6,q = \(\frac { 2 }{ 5 }\) = 0.4
The p.m.f is
P(x) = ⁵C_X (0.6)^x (0.4)^{5 – x} ⇒ x = 0,1,2,3,4,5
The frequency function is
T_x = N × (P(x)) = 200 × ⁵C_X(0.6)^{X(0.4)5 – X}
The frequencies are
when x = 0,
T₀ = 200 × ⁵C₀(0.6)⁰ × (0.4)^{5 – 0} = 200 × 0.01024 = 2.048 = 2
Remaining theoretical frequencies are calculated by using Recurrence relation
T_x = \(\frac{n-x+1}{x} \cdot \frac{p}{q} \cdot T_{x-1}\)
T₁ = \(\frac{5-1+1}{1} \times \frac{0.6}{0.4} \times 2.048\) = 5 × 1.5 × 2.048 = 15.31 = 15
T₂ = \(\frac{5-2+1}{2} \times \frac{0.6}{0.4} \times 15.31\) = 2 × 1.5 × 15.31 = 46.08 = 46
T₃ = \(\frac{5-3+1}{3} \times \frac{0.6}{0.4} \times 46.08\) = 1 × 1.5 × 46.08 = 69.12 = 69
T₄ = \(\frac{5-4+1}{4} \times \frac{0.6}{0.4} \times 69.12\) = 0.5 × 1.5 × 69.12 = 51.84 = 52
T₅ = \(\frac{5-5+1}{5} \times \frac{0.6}{0.4} \times 51.84\) = 0.2 × 1.5 × 51.84 = 15.552 = 16
Thus the observed and theoretical frequencies are tabulated

Question 51.
Fit a Binomial distribution for the following data and obtain the theoretical frequencies

Answer;
Given n = 7 and p is to be estimated

Here np = x̄ = \(\frac{\sum f x}{N}=\frac{433}{128}=3.38 \)
np = 3.38,7p = 3.38, p = \(\frac { 3.38 }{ 7 }\) = 0.48,
q = 1 – 0.48 = 0.52
X ~ B(n = 7, p = 0.48)
The p.m.f is
P(x) = ⁷C_X (0.48)^x (0.52)^{7 – X} ⇒ x = 0,1,2,3,4,5,6,7
The frequency function is
T_x = N.P(x) = 128 ⁷C_X(0.48)^X(0.52)^{7 – X}
when x = 0,
T₀ = 128. ⁷C₀(0.48)⁰ (0.52)^{7 – 0} = 1.31 = 1
By using recurrence relation = T_x = \(\frac{n-x+1}{x} \cdot \frac{p}{q} \cdot T_{x}-1\)
when x = 1,
T₁ = \(\frac{7-1+1}{1} \times \frac{0.48}{0.52} \times 1.31\) = 7 × 0.923 × 1.31 = 8.46 = 8
T₂ = \(\frac{7-2+1}{2} \times \frac{0.48}{0.52} \times 8.46\) = 3 × 0.923 × 8.46 = 23.43 = 23
T₃ = \(\frac{7-3+1}{3} \times \frac{0.48}{0.52} \times 23.43\) = 1.67 × 0.923 × 23.42 = 36.11 = 36
T₄ = \(\frac{7-4+1}{4} \times \frac{0.48}{0.52} \times 36.11\) = 1 × 0.923 × 36.11 = 33.32 = 33
T₅ = \(\frac{7-5+1}{5} \times \frac{0.48}{0.52} \times 33.32\) = 0.6× 0.923 × 33.32 = 8.45 = 18
T₆ = \(\frac{7-6+1}{6} \times \frac{0.48}{0.52} \times 18.45\) = 0.33 × 0.923 × 18.45 = 5.62 = 6
T₇ = \(\frac{7-7+1}{7} \times \frac{0.48}{0.52} \times 5.62\) = 0.14 × 0.923 × 5.62 = 0.74 = 1

Question 52.
Fit a Binomial distribution for the following data and obtain the theoretical frequencies

Answer:
Given n = 4 and p is to be estimated.

Here, x̄ = \(\frac{\Sigma f x}{N}\) x̄ = \(\frac{192}{103}\) = 1.86
np = 1.86
4p = 1.86
p = \(\frac { 1.86 }{ 4 }\) = 0.47
q = 1 – p = 1 – 0.47 = 0.53
Then X ~ B(n = 4, p = 0.47)
The p.m.f is
P(x) = ⁴C_X (0.47)^x (0.53)^{4 – X}, x = 0,1,2 6
The frequency function is
T_x = N . P(x) = 103 ⁴C_X (0.47)^x (0.53)^{4 – x}
The frequencies one is using Recurrence Relation
When x = 0,
T₀ = 103.⁴C₀(0.47)⁰ × (0.53)^{4 – 0} = 103 × 0.0789 = 8.13 = 8
The remaining theoretical frequencies are obtained using recurrence relation
When x = 1,
T₁ = \(\frac{4-1+1}{1} \times \frac{0.47}{0.53} \times 8.13\) = 4 × 0.8867 × 8.13 = 28.83 = 29
When x = 2,
T₂= \(\frac{4-2+1}{2} \times \frac{0.47}{0.53} \times 28.83\) = 1.5 × 0.8867 × 28.83 = 38.35 = 38
When x = 3,
T₃ = \(\frac{4-3+1}{3} \times \frac{0.47}{0.53} \times 38.35\) = 0.67 × 0.8867 × 38.35 = 22.78 = 23
When x = 4
T₄ = \(\frac{4-4+1}{4} \times \frac{0.47}{0.53} \times 22.78\) = 0.25 × 0.8867 × 22.78 = 5.04 = 5

Question 53.
Fit a Binomial distribution for the following data and obtain the expected frequencies

Answer:
Given n = 6 and p is to be estimated

Here x̄ = \(\frac{\sum f_{x}}{N}\) = \(\frac{723}{256}\) = 2.82
np = x̄ = 2.82
6p = 2.82
p = \(\frac{2.82}{6}\) = 0.47
q = 1 – p = 1 – 0.47 = 0.53
Then X ~ B (n = 6, p = 0.47)
The p.m.f is
P(x) = ⁶C_x (0.47)^x (0.53)^{6 – x} x = 0,1,2 ………….. 6
The frequency function is
T_x = N . P(x) = 256 ⁶C_X (0.47)^x (0.53)^{6 – x}
The frequencies are
when x = 0,
T₀ = (256) ⁶C₀(0.47)⁰ × (0.53)^{6 – 0} = 5.67 = 6
The remaining theoretical frequencies are found using the recurrence relation.
Recurrence Relation = T_x = \(\frac{n-x+1}{x} \cdot \frac{p}{q} \cdot T_{x}-1\)
T₁ = \(\frac{6-1+1}{1} \times \frac{0.47}{0.53} \times 5.67\) = 6 × 0.8867 × 5.67 = 30.16 = 30
T₂ = \(\frac{6-2+1}{2} \times \frac{0.47}{0.53} \times 30.16\) = 2.5 × 0.8867 × 0.16 = 66.8 = 67
T₃ = \(\frac{6-3+1}{3} \times \frac{0.47}{0.53} \times 66.8\) = 1.33 × 0.8867 × 66.8 = 78.7 = 79
T₄ = \(\frac{6-4+1}{4} \times \frac{0.47}{0.53} \times 78.7\) = 0.75 × 0.8867 × 7g7 = 52 3o = 52
T₅ = \(\frac{6-5+1}{5} \times \frac{0.47}{0.53} \times 52.30\) = 0.4 × 0.8867 × 52.3 = 18.54 = 19
T₆ = \(\frac{6-6+1}{6} \times \frac{0.47}{0.53} \times 18.54\) = 0.167 × 0.8867 × 18.54 = 2.7 = 3
Thus theoretical frequencies are

2nd PUC Statistics Poisson Distribution Exercise Problems

Question 1.
Define a Poisson variate?
Answer:
If X is a discrete random variable with probability mass function
p(x) = \(\frac{e^{-\lambda} \cdot \lambda^{x}}{x !}\) where x = 0, 1, 2,…. ∞; λ > 0
Then X is a Poisson variate

Question 2.
Define Poisson distribution
Answer:
If X is a discrete random variable with p.m.f
p(x) = \(\frac{e^{-\lambda} \cdot \lambda^{x}}{x !}\) where x = 0, 1, 2,…. ∞; λ > 0
Then the distribution of X can be called as Poisson distribution

Question 3.
Write down the p.m.f of a Poisson distribution
Answer:
p.m.f = P(x) = \(\frac{e^{-\lambda} \cdot \lambda^{x}}{x !}\) x = 0, 1, 2,…. ∞; λ > 0

Question 4.
Write the range and parameter of a Poisson distribution
Ansewr:
Range ⇒ x = 0,1,2, ……… ∞; λ > 0
Parameter = λ

Question 5.
Give two examples for Poisson variate
Answer:

1. Number of births occurring in a city in a day
2. Numbret of accidents in a town in a day

Question 6.
What are the mean, variance and SD of a Poisson distribution?
Answer:
Mean = λ ; Variance = λ
SD = √Variance = √λ

Question 7.
Write the relationship between mean and variance of a Poisson distribution
Answer:
The relationship between mean and variance of a Poisson distribution is
Mean = variance

Question 8.
Mention two features of Poisson distribution
Answer:

1. Mean = Variance = λ
2. The range of Poisson distribution is x = 0, 1, 2, ………. ∞

Question 9.
Mention the recurrence formula for successive probabilities in a Poisson distribution with parameter λ
Answer:
Recurrence relation P(x) = \(\frac{\lambda}{x}\).P(x – 1)

Question 10.
Mention the recurrence formula for successive frequencies in a Poisson distribution with parameter λ
Answer:
Recurrence formula = T_x = \(\frac{\lambda}{x}\)T_{x – 1}

Question 11.
Under what condition Poisson distribution tends to Normal distribution?
Answer:
When λ is large, then Poisson distribution tends to Normal distribution.

Question 12.
In a Poisson distribution if standard deviation is 3, find its mean
Answer:
Variance = √λ = 3
Mean = λ = (√λ)² = (3)² = 9

Question 13.
In a Poisson distribution if mean 4. Find its SD
Answer:
Mean = λ = 4
SD = √λ = √4 = 2

Question 14.
Find p(x = 0) in a Poisson distribution with mean 5
Answer:
p(x = 0) = \(\frac{\mathrm{e}^{-5} 5^{0}}{0 !}\) = 0.0067

Question 15.
In a Poisson distribution, if P(X = 0) = 0.0408, find λ
Answer:
P(X = 0) = 0.0408
p(x) = \(\frac{e^{-\lambda} \cdot \lambda^{0}}{0 !}\) = 0.0408
p(x) = \(\frac{\mathrm{e}^{-\lambda} \cdot 1}{1}\) = 0.0408
(From the table) the value is 3.2
∴ λ = 3.2

Question 16.
In a Poisson distribution, if P(X = 0) = 0.1225. Write down the p.m.f
Answer:
Given P(X = 0) = 0.1225
p(x) = \(\frac{e^{-\lambda} \cdot \lambda^{0}}{0 !}\) = 0.0125
p(x) = \(\frac{\mathrm{e}^{-\lambda} \cdot 1}{1}\) = 0.0125
e^-λ 0.1225
(From the table) the value is 2.1
λ = 2.1 ∴ p(x) = \(\frac{\mathrm{e}^{-21}(2.1)^{\mathrm{x}}}{\mathrm{x} !}\)

Question 17.
In a Poisson distribution P(X = 2) = P(X = 3). Find the P(X = 4)
Answer:
P(X = 2) = P(X = 3)

Question 18.
In a Poisson distribution, the second probability term is half of the first probability term. Find the SD
Answer:
Second term = P(X = 1)

Question 19.
The first two frequency terms of a Poisson distribution are 150 and 180 respectively. Find the next frequency term
Answer:
First two terms = T₀ & T₁

Question 20.
The second and third frequency terms of a Poisson distribution are 100 and 80 respectively. Find the next frequency term
Answer:
Given 2^nd = T₁ = 100
3^rd = T₂ = 80

Using recurrence relation, T₃ = \(\frac{\lambda}{3}\) . T₂ = \(\frac{1.6}{3}\) × 80 = 42.67
Values of e^-λ taken from statistical tables

Question 21.
On an average a box contains 2 defective items. Find the probability that a randomly selected box has (i) No defective item (ii) At least two defective items
Answer:
Given λ – 2
X = Number of defective items in a box λ = 2
Then X ~ p (λ = 2)
The p.m.f is
p(x) = \(\frac{e^{-2} 2^{x}}{x !}\) where x = 0,1, 2 …………… ∞

(i) P (No defective item) = P(X = 0) (From the table e^-2 = 0.1353)
\(\frac{\mathrm{e}^{-2} \cdot 2^{0}}{0 !}=\frac{0.1353 \times 1}{1}=0.1353\)

(ii) P (at least 2 defective items) = P(X ≥ 2) = 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1)]
= 1 – \(\left[\frac{e^{-2} \cdot 2^{0}}{0 !}+\frac{e^{-2} \cdot 2^{1}}{1 !}\right]\) = 1 – 0.4059 = 0.5941

Question 23.
It has been found that on an average 4 patients visit a particular doctor during one hour. What is the probability that during a particular hour, (i) Doctor is free (ii) More than 3 patients visit the doctor
Answer:
Given λ = 4
X = Total number of patients visits a doctor during one hour. X = 4
then x ~ P (λ = 4)
The p.m.f is
p(x) = \(\frac{e^{-4} \cdot 4^{x}}{x !}\) ; x = 0,1,2 ……. ∞
(i) P(doctor is free) = p(x = 0) = \(\frac{\mathrm{e}^{-4} \cdot 4^{0}}{0 !}\) = 0.0183.
From the table e^-4 = 0.0183

(ii) P(>3 patients visit) = P(X > 3) = 1 – P(X ≤ 3)
1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= 1- [(0.0183) + (0.0183)4 + (0.0183)8 + (0.0183)10.67]
⇒ 1 – [0.4331] ⇒ 0.5669

Question 24
A car hire agency has two cars. On an average there is a demand for one car during a particular hour. What is the probability that (i) Both the cars are free (ii) Some demand is refused.
Answer:
Given λ = 1
X : no. of cars in demand
Then X is a Poisson variable with λ = 1
The p.m.f is
P(x) = \(\frac{e^{-\lambda} \cdot \lambda^{x}}{x !}\) ,x = 0,1,2, ……. ∞
(i) P(both the cars are free) = P(x = 0) = \(\frac{\mathrm{e}^{-1} \cdot 1^{0}}{0 !}\) = 0.3679
(From the table e^-1 = 0.3679)
(ii) P(Some demand is refused) = P(Both the cars are busy)
\(\Rightarrow \frac{e^{-1} \cdot 1^{2}}{2 !} \Rightarrow \frac{0.3679}{2}=0.0802\)

Question 25.
In a text book, on an average 0.3 mistakes per page is found. If there are 500 pages in that text book. In how many pages will there be (i) Three mistakes (ii) At the most two mistakes
Answer:
Given λ = 0.3, N = 500
Let X = number of mistakes in a page
Then X ~ p(λ = 0.3)
PMF = P(x) = \(\frac{e^{-\lambda} \cdot \lambda^{x}}{x !}\) ,x = 0,1,2 ….. ∞

(i) P(page has three mistakes).
\(\frac{\mathrm{e}^{-13}(1.3)^{\mathrm{x}}}{\mathrm{x} !}\)
(From the table e^-0.3 = 0.7408)
Out of 500 pages, number of pages having 3 mistakes is 500 × 0.0033 = 1.6668 = 2 pages

(ii) P(at the most two mistakes) = P(X < 2)
= P(X = 0) + P(X = 1) + P(X = 2)

= 0.7408 + 0.22224 + 0.033336
= 0.996376
Out of 500 pages, number of pages having at most 2 mistakes
500 × 0.996 = 498.18 = 498

Question 26.
The number of accidents in a year attributed to taxi drivers in a city follows Poisson distribution with mean 2.5, out of 1000 taxi drivers, find approximately the number of drivers with (i) One accident in a year (ii) More than three accidents in a year
Answer:
Given λ = 2.5
Let X : Number of accidents in a year atributed to take drivers
Then X ~ p(λ = 2.5)

(i) P(one accident in a year) = P(X = 1).

(From the table e^-2.5 = 0.0821) p(x ≥ 3) = 1 – P(x < 3)

(ii) P(more than 3 accidents in a year)
= 1 – P(X = 0) + P(X = 2) + P (X = 3)

= 1 – [0.0821 + 0.2052 + \(\frac{0.0821 \times(2.5)^{2}}{2 !}+\frac{0.0821 \times(2.5)^{3}}{3 !}\)]
= 1 – (0.0821 + 0.2052 + 0.25656 + 0.2138)
= 1 -0.7576 = 0.2423
Out of thousand taxi drives, number of drivers with more than 3 accidents in a year is 1000 × 0.2422 = 242 approx

Question 27.
On an average, one in every 50 valves manufactures by a firm is substandard if valves are supplied in packets of 100 each; in how many of a lot of 2000 packets would you expect substandard valves
Answer:
Given n = 100,p = \(\frac{1}{50}\)
X : Total number of substandard valves in a packet of 100 each
Since p is very small and n is large, X is treated as Poisson variate with parameter
λ = np = 100 × \(\frac{1}{50}\) = 2
P(Number of standards valves]
= 1 – P (no one are substandard Valves)
= 1 – P(X = 0)
= 1 – \(\frac{e^{-2} \cdot 2^{0}}{0 !}\) = 1 – 0.1353 = 0.8647
Out of 2000 packets, the number of substandard valves
= 2000 × 0.8647 = 1729 approx

Question 28.
Out of experience, it is known that 1% of the screws manufactured by a firm are defective. Screws are supplied in packets of 100 each. What is the probability that a randomly selected packet has 2 defective screws? Among 3000 packets in how many packets would you expect defective screws?
Answer:
Given n = 100, p = \(\frac { 1 }{ 100 }\)
X : The total number of defective screws in a packets of 100 each
Since p is small and n is large, X is treated as Poisson variate with parameter λ = np
np = 100 × \(\frac { 1 }{ 100 }\) = 1 – λ
PMF = p(x) = \(\frac{\mathrm{e}^{-1} \cdot 1^{\mathrm{x}}}{\mathrm{x} !}\), x = 0,1,2 ……. ∞
(i) p (2 defective screws)
P(x = 2) = \(\frac{e^{-1} \cdot 1^{2}}{2 !}=\frac{0.3679 \times 1}{2}\) = 0.1839

(ii) P (defective screws) =1 – P (No defective screws)
1 – P(x = 0) = 1 – \(\frac{\mathrm{e}^{-1} \cdot(1)^{0}}{0 !}\) ⇒ 1 – 0.3679 ⇒ 0.6321
Out of 3000 packets, number of packets with defective screws is 3000 × 0.6321 = 1896 approx.

Question 29.
In a certain factory turning out optical lenses there is a small chance of \(\frac{1}{500}\) for any one lens to be defective. The lenses are supplied in packets of 50 each. Use Poisson distribution to calculate the approximate number of packets containing one defective lens in a consignment of 20, 000 packets
Answer:
Given
n = 50, p = \(\frac{1}{500}\) = 0.002
X : The total number of defective lens in a consignment
Since p is very small and n is large, X is treated as Poisson variate with parameter.
λ = np
λ = 50(0.002) = 0.1
p.m.f P(x) = \(\frac{\mathrm{e}^{-0.1}(0.1)^{x}}{\mathrm{x} !}\); x = 0,1,2 ….. α
P (packets containing one defective lens) = P(X = 1).
= \(\frac{\mathrm{e}^{-0.1}(0.1)^{1}}{1 !}\) = o 09048 (From the table e^-0.1 = 0.9048)
Out of 20,000 packets, the number of packets containing one defective lens
= 20,000 × 0.09048 = 1809.6 = 1810

Question 30.
If 98% of electric bulbs manufactured by a company are known to be no defectives. What is the probability that a sample of 150 electric bulb taken from the production process of that company would contain (i) exactly one defective bulb (ii) more than two defective bulbs
Answer:
X : Total number of defective bulbs in a sample of 150 electric bulbs.
Given n = 150 p = \(\frac { 2 }{ 100 }\) = 0.02
Since p is small and n is large, X is treated as Poisson variate with parameters.
λ = np = 150 × 0.02 = 3
p.m.f = p(x) = \(\frac{e^{-3} 3^{x}}{x !}\), x = 0,1, 2 ……. ∞

(i) P(exactly one defective bulb) = P(X = 1).
⇒ \(\frac{\mathrm{e}^{-3} \times 3^{1}}{1 !}\) = 0.0498 × 3 = 0.1494
(From the table e^-3 = 0.0498)

(ii) P(more than two defective bulbs) = P(X > 2)
⇒ 1 – P(X > 2)
⇒ 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

= 1 – (0.0498 + 0.1494 + 0.2241)
= 1 – 0.4 = 0.5767

Question 31.
Fit a Poisson distribution for the following data and obtain the theoretical frequencies

Answer:
X = the number of TV sets sold

Here, Mean = λ = x̄ = \(\frac{\Sigma f x}{N}=\frac{285}{150}\) = 1.9
The p.m.f

(from the table e^-19 = 0.1496)
The frequency function T_x = N . P(x)
When x = 0, T₀ = 150(0.1496) = 22.44 = 22
Remaining frequencies are calculated by using recurrence relation

Question 32.
Fit a Poisson distribution and obtain the expected frequencies.

Answer:
X : Denote the number of deaths per day

Here λ = x̄ = \(\frac{\Sigma f x}{N}=\frac{630}{395}\) = 1.6
The p.m.f

(From the table e^-1.6 = 0.2019)
The frequency function T_x = N. p(x)
when x = 0, T₀ = 395(0.2019) = 79.75
Remaining frequencies are calculated by using recurrence relation
T_x = \(\frac{\lambda}{x}\)T_{x – 1}

Question 33.
Fit a Poisson distribution to the following data and hence find the theoretical frequencies

Answer:

Here λ = \(\frac{\Sigma \mathrm{fx}}{\mathrm{N}}=\frac{110}{100}\) = 1.1
The p.m.f

(From the table e^-1.1 = 0.3329)
The frequency function T_x = N. p(x)
when x = 0, T₀ = 100(0.3329) = 33.29 = 33
Remaining frequencies are calculated by using recurrence relation

2nd PUC Statistics Hyper – Geometric Distribution Exercise Problems

Question 1.
Define a Hyper-geometric variate?
Answer:
If X is random discrete variable with p.m.f

where x = 0,1, 2 …. Min (a, n);
a,b and n > 0
Then X is Hyper-geometric variate.

Question 2.
Define a hyper geometric distributions.
Answer:
If X is a discrete r.v with pmt

then the distribution of X is called hyper geometric distribution

Question 3.
Write down the p.m.f of a Hyper geometric distribution
Answer:
p.m.f

where x = 0,1, 2,…. Min (a. n);
a,b and n > 0

Question 4.
Write the range and parameters of a Hyper geometric distribution
Answer:
Range for x = 0,1, 2 …. Min (a, n)
Parameters = a, b and n

Question 5.
Give two examples for Hyper geometric variate?
Answer:

1. Number of non-vegetarians in a sample of 10 selected from a class in which there are 20 vegetarians and 25 non-vegetarians.
2. Number of red balloons in a selection of 5 balloons when there are 6 red and 5 blue balloons in a bag.

Question 6.
What are the mean, variance and SD of a hyper geometric distribution?
Answer:

Question 7.
Mention two features of a Hyper geometric distribution.
Answer:

1. In Hyper-geometric distribution, a, b and n are the three parameters
2. x = 0,1,2, ……Min (a,n) is the Range of hyper geometric distribution

Question 8.
Under what conditions hyper geometric distribution tends to binomial distribution
Answer:
Hypergeometric distribution tends distribution when (a + b) → ∞ = 2b and \(\frac{a}{a+b}=p\) With parameters a, b

Question 9.
State the conditions which characterise the Hypergeometric distribution.
Answer:

1. The result of each draw is classified into one of the two categories.
2. The probability of success changes on each draw.

Question 10.
If a = 5 b = 15 and n = 3 then find the variance of hyper-geometric distribution
Answer:

Question 11.
Write down the p.m.f and mean of a Hyper geometric distribution whose parameters area = 4, b = 7 and n = 5.
Answer:
(i) Given a = 4b = 7, n = 5

Question 12.
For a Hyper-geometric distribution the parameters are a = 7 b = 5 and n = 8 what is the range of the Hyper-geometric distribution and also SD
Answer:
(i) Range = x = 0, 1, 2,…. min(7, 8) = 0,1, 2, 3, 4, 5, 6, 7

Question 13.
In a college, there are 2100 students. Among them 900 are girls. A computer training centre in the city offers free computer training to 5 randomly selected students of the colleges. Find the mean and SD of the number of girls selected for computer training
Answer:
Given a = 900, n = 5, b = 2100 – 900 = 1200

Question 14.
There are 20 fruits in a basket, out of which 8 are mangoes and rest are oranges. A girl picks 5 fruits at random from the basket. Find the prob ability that she gets 3 mangoes.
Answer:
n = 5, a = 8, b = 12
X : number of mangoes picked among 5 fruits a b.

Question 15.
A student preparing for an examination studies only 20 out of 25 sections prescribed. If the teacher selects 10 sections at random what is the probability that the student will have studied 9 of these sections
Answer:
Given a = 20; b = 5; n = 10
X: number of sections studied in a selection of 10 sections
p.m.f

From the table ²⁰C₉ = 1,67,960, ⁵C₉ = 5,
²⁵C₉ = \(\frac{25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 !}{15 ! \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\)
After simplifying the equation, we get
5 × 23 × 22 × 19 × 2 × 17 × 2 = 32,68,760

Question 16.
There are 50 lecturers in a college out of them 23 belong to the science faculty. The college management builds 5 quarters and allots them to 5 randomly selected lecturers. Find the probability that all the quarters are allotted to science lecturers.
Answer:
Given a = 23, b = 27 n = 5
X : number of quarters allotted to science lectures among 5 randomly lecturers.

After simplifying the equation, we get
23 × 11 × 7 × 19 = 33,649 .
5 × 49 × 4 × 47 × 46 = 21,18,760

Question 17.
In a Hyper-geometric distribution if a = 6 b = 9 and n = 4 find P(X = 2)
Answer:
Given n = 4, a = 6, b = 9

(From the table ⁶C₂ = 15, ⁹C₂ = 36, ¹⁵C₄= 1365)

2nd PUC Statistics Normal Distribution Exercise Problems

Question 1.
Define a Normal variate?
Answer:
If ‘X’ is a continuous random variable with probability density function

where -∞ < X < ∞, σ > 0 then X is a Normal variate

Question 2.
Define a Normal distribution.
Answer:
If X is a continuous random variable with probability density function

where -∞ < X < ∞, σ > 0 and the distribution of X is Normal distribution.

Question 3.
Write down the p.d.f of a normal distribution.
Answer:

Question 4.
Write the range and parameters of a Normal distribution.
Answer:
Range for x = (-∞, ∞); Range for µ = (- ∞, ∞),
Parameters are µ and σ

Question 5.
Give two examples for Normal variate.
Answer:

1. Life time of electric bulbs manufactured by a firm.
2. Heights of students of a college

Question 6.
What are the mean, variance and SD of a Normal distribution?
Answer:
Mean = µ, Variance = σ², S.D. = σ

Question 7.
What are the values of co-efficient of skewness (β₁) and coefficient of kurtosis (β₂) for a Normal distribution?
Answer:
Co-efficient of skewness (β₁) = 0;
Co-efficient of kurtosis (β₂) = 3

Question 8.
What is the probability that a Normal variate takes a value greater than its mean?
Answer:
0.5

Question 9.
What is the total area under the Normal curve?
Answer:
One

Question 10.
Give the value of a Normal variate for which probability is maximum. Write the maximum probability?
Answer:
The maximum probability occurs at x = µ and is given by
[p(x)]_max = \(\frac{1}{\sigma \sqrt{2 \pi}}\)

Question 11.
Define a Standard normal variate?
Answer:
If Z is a normal variate with mean (μ) = 0 and SD (σ) = 1 then Z is named as Standard normal variate.

Question 12.
Define a Standard normal distribution.
Answer:
If Z is a normal variate with mean (μ) = 0 and SD (σ) = 1 then the distribution of Z is a Standard normal distribution.

Question 13.
Write down the p.d.f of a standard normal distribution.
Answer:
f(z) = \(\frac{1}{\sqrt{2 \pi}} e^{\frac{z^{2}}{2}}\) , where -∞ < z <+∞

Question 14.
What are the mean, variance, SD of a standard distribution?
Answer:
Mean = (μ) = 0 variance = 1 SD = (σ) = 1

Question 15.
Name the distribution for which variance and SD are equal.
Answer:
Standard normal distribution.

Question 16.
Write down the area property of Normal distribution.
Answer:
The total area under the curve in unity. Area property of Normal distribution is
P[μ – σ < x < μ + σ] = 0.6826 = 68.26%
P[μ – 2σ < x < μ + 2σ] = 0.9544 = 95.44%
P(μ – 3σ < x < μ + 3σ] = 0.9973 = 99.73%

Question 17.
State five properties of a normal distributions
Answer:

1. The normal curve is bell shaped.
2. Normal curve is symmetrical about the mean (B₁ = 0), Here mean = median = mode = μ
3. Normal distribution is unimodal
4. The distribution is mesokurtie (β₂ = 3)
5. For a normal distribution SD = σ, Q.D = \(\frac{2}{3}\)σ MD = \(\frac{4}{5}\)σ

Question 18.
If Z is a SNV’ indicate the value of P(Z < 0)
Answer:
0.5

Question 19.
If X is a Normal variate with mean (μ) and SD (σ) give the value of P(X < μ)
Answer:

Question 20.
Write the p.d.f. of normal distribution with mean 3 and S.D. 2.
Answer:
Here μ = 3, σ = 2; f(x) = \(=\frac{1}{2 \sqrt{2 \pi}} e^{\frac{(x-3)^{2}}{2}}\) ; -∞ < x < ∞

Question 21.
Consider the following PDF of a normal variate
of f(x) = \(\frac{1}{3 \sqrt{2 \pi}} e^{-1 / 2\left(\frac{x-6}{3}\right)^{2}}\) ; -∞ < x < ∞
Find the mean and variance of X.
Answer:
Mean μ = 6, SD = 3, variance = σ² = 3² = 9

Question 22.
The distribution of a variable X is given by the law
f(x) = \(=\frac{1}{5 \sqrt{2 \pi}} e^{-1 / 2\left(\frac{x-100}{5}\right)^{2}}\) ; -∞ < x < ∞
Find the mean and SD
Answer:
Mean = μ = 100 SD = σ = 5.

Question 23.
Write the p.d.f of a Normal distribution with mean 55 and variance 4
Answer:
μ = 55, σ = √var = √4 = 2
p.d.f f(x) = \(\frac{1}{2 \sqrt{2 \pi}} e^{-\frac{1}{2}\left[\frac{x-55}{2}\right]^{2}}\) ; -∞ < x < ∞

Question 24.
Write down the area under the normal curve in a neighbourhood of μ
Answer:
P[μ – σ < x < μ + σ] = 0.6826.

Question 25.
In a Normal distribution, given P(-0.8 < Z < 0.8) = 0.5762 find P(0 < Z < 0.8)
Answer:
P(-0.8 < Z< 0.8)
= Area from -0.8 to 0 + [area from 0 to 0.8] = 0.5762
= Area from 0 to 0.8 = \(\frac { 1 }{ 2 }\) × 0.5762 = 0.2881
And hence [0 < Z < 0.8] = [area from 0 to 0.8] = 0.2881

Question 26.
If the variance of a normal distribution is 9 cms2, then find Q.D
Answer:
Given var = σ² = 9
σ = √var = √9 = 3
Quartile deviation =\(\frac{2}{3} \sigma=\frac{2}{3} \times 3=2\)

Question 27.
Find the QD and MD of a Normal distribution with mean 30 and SD 6.
Answer:
Given Mean μ = 30, σ = 6
QD = \(\frac{2}{3} \sigma=\frac{2}{3} \times 6=4\)
MD = \(\frac{4}{5} \sigma=\frac{4}{5} \times 6=4.8\)

Question 28.
If SD is 2 for a Normal distribution find MD
Answer:
Given σ = 2, MD = \(\frac{4}{5} \sigma\)
MD = \(\frac{4}{5}\) × 2 ⇒ 1.6

Question 29.
If Q₁ = 30 and Q₃ = 70 find the mode of the Normal distribution
Answer:
Q₁ = 30; Q₃ = 70
Mode = \(\frac{Q_{3}+Q_{1}}{2}=\frac{30+70}{2}=50\)

Question 30.
In a Normal distribution if the values of first and third quartiles are 25 and 55 respectively find the mean
Answer:
Given Q₁ = 25, Q₃ = 55
Mean = \(\frac{Q_{3}+Q_{1}}{2}=\frac{55+25}{2}=40\)

Question 31.
If the lower and upper quartiles of a normal distribution are 11 and 39 respectively, find the median.
Answer:
Median = \(\frac{Q_{3}+Q_{1}}{2}=\frac{39+11}{2}=25\)

Question 32.
In a Normal distribution Q₁ = 20 and Q₃ = 50 then find mean QD and SD
Answer:
Q₁ = 20, Q₃ = 50
(i) Mean = \(\frac{Q_{3}+Q_{1}}{2}=\frac{20+50}{2}=35\)
(ii) QD = \(\frac{Q_{3}-Q_{1}}{2}=\frac{50-20}{2}=15\)
(iii) QD = \(\frac{2}{3} \sigma \Rightarrow 15=\frac{2}{3} \sigma\)
45 = 2σ ; σ = \(\frac{45}{2}\) = 22.5 = 22.5
Area under the Normal curve from Z to Infinity from statistical tables.

Question 33.
In a Normal distribution mean and variance are 50 and 16 respectively. Find the and Q₁ and Q₃
Answer:
Given µ = 50, σ = 16 SD = √var = √l6 = 4
Q₁ = µ – 0.6745(σ) ⇒ 50 – 0.6745(4) = 47.302
Q₃ = µ + 0.6745(σ) ⇒ 50 + 0.6745(4) = 52.69

Question 34.
Mean and SD of a distribution are 12 and 5 respectively. Find the lower and upper quartile
Answer:
Given µ = 12; σ = 5
Q₁ = µ – 0.6745 (σ) = 12 – 0.6745(5)
⇒ 12 – 3.3725 = 8.6275
Q₃ = µ + 0.6745 (σ) = 12 + 0.6745 (5)
⇒ 12 + 3.3725 = 15.3725

Question 35.
If Z is a SNV and P(Z > k) = 0.05 find the value of k
Answer:
P(Z > k) = 0.05
[area from k to ∞] = 0.05
k = 1.65 (from the table)

Question 36.
If Z is a SNV and P(Z > k) = 0.1 find the value of k
Answer:
P(Z > k) = 0.1
Area from k to ∞ = 0.1
k = 1.28 (from the table)

Question 37.
Heights of 360 children are normally distributed with mean 120 cms and variance 4 cms2. Find the expected number of children having heights (i) greater than 118 cms
(ii) between 116 and 119 cms (iii) less than 117 cms
Answer:
Given N = 360 µ = 120, variance = 4; σ = 2
X : height of children
X ~ N (µ = 120, σ = 2)
z = \(\frac{x-\mu}{\sigma}=\frac{x-120}{2}\) is N(0, 1)

(i) P(X > 118)
= \(P\left(\frac{x-120}{2}>\frac{118-120}{2}\right)\)
= p(z > -1)
= (area from -1 to ∞) = 0.8413
Total number of children having height greater than 118 is
8413 × 360 = 302.86 ⇒ 303

(ii) P(116< x< 119)
\(=P\left(\frac{116-120}{2}<\frac{x-\mu}{\sigma}<\frac{119-120}{2}\right)\)
= P(-2 < Z < -0.5)
= (area from -2 to ∞) – (area from -0.5 to ∞ )
= 0.9772 – 0.6915 = 0.2857
Total number of children having height between 116
and 119 is 0.2857 × 360 = 102.85 = 103

(iii) P(X < 117)
\(P\left(\frac{x-120}{2}<\frac{117-120}{2}\right)\)
P(Z < -1.5)
= (area from – ∞ to ∞) – (area from -1.5 to ∞)
= 1 – 0.9332 = 0.0668
∴ Total number of children having height less than 117 is
0.0668 × 360 = 24.

The Percentage Difference Calculator will find the percent difference between two positive numbers greater than 0.

Question 38.
The mean IQ of a large number of children of age 14 is 95 and the SD is 6. Assuming the distribution as normal. Find the percentage of the children having IQ under 85. (ii) what proportion of the children has IQ above 90
Answer:
X : IQ of children
∴ X ~ N (µ = 95; σ = 6)
µ = 95; σ = 6
z = \(\frac{x-\mu}{\sigma}=\frac{x-95}{6}\) is N(0,1)

(i) P(X < 85)
\(\mathrm{p}\left(\frac{\mathrm{x}-\mu}{6}<\frac{85-95}{6}\right)\)
p(Z < -1.667)
=(area from – ∞ to ∞) – (area from -1.667 to ∞)
= 1 – 0.9525 = 0.0475 = 0.0475 × 100 = 4.75%
4.75% of the children has IQ under 85 (ii) P(IQ > 90)

(ii) P(X > 90)
\(P\left(\frac{x-95}{6}>\frac{90-95}{6}\right)\)
= P(Z > -0.83) = area from -0.83 to ∞ = 0.7967

Question 39.
If x ~ N (100, 32), find the probability value of the variate which lies between (i) 97 and 106 (ii) 100 and 103.
Answer:
Given µ = 100, σ = 3
X ~ N (100, 32)

(i) P(97 < X < 106)
\(P\left(\frac{97-100}{3}<\frac{x-100}{3}<\frac{106-100}{3}\right)\)
P (-1 < Z < 2)
= (area from -1 to ∞) – (area from 2 to ∞)
= 0.8413 – 0.0228 = 0.8185

(ii) P(100 < X < 103)
\(P\left(\frac{100-100}{3}<\frac{x-100}{3}<\frac{103-100}{3}\right)\)
p(0 < z < 1)
= (area from 0 to ∞) – (area from 1 to ∞)
= 0.5 – 0.1587 = 0.3413.

Question 40.
The weekly wages of workers are normally j distributed with mean ₹ 3000 and SD ₹ 500. Find the probability of workers whose weekly wages will be (i) more than 3400 (ii) between ₹ 2500 and ₹ 3500 (iii) less than ₹ 3200
Answer:
Given µ = 3000, σ = 500
X = weekly wages of workers
X ~N (3000, 500);
z = \(\frac{x-3000}{500}\) is N(0, 1) (i) p(X > 3000)

(ii) P(2500 < Z < 3500)

P(-1 < Z < 1)
= (area from -1 to ∞) – (area from 1 to ∞)
= 0.8143 – 0.1587 = 0.6826

(iii) p(X < 3200)
\(\mathrm{P}\left(\frac{\mathrm{X}-3000}{500}<\frac{3200-3000}{500}\right)\)
p(Z < 0.4)m = (area from -∞ to ∞) – (area from 0.4 to ∞)
= 1 – 0.3446 = 0.6554

Question 41.
Bengaluru corporation authorities have in stalled 2000 electric lamps in MG road. The lamps have an average life of 1000 burning hours with a SD of 200 hours. If life of lamps follow normal distribution, then
(i) What number of lamps might be expected to fail in the first 800 burning hours?
(ii) After what period of burning hours would be expected
(a) 10% of the lamps would have failed
(b) 10% of the lamps would be still burning
Answer:
Given µ = 1000 σ = 200
Let X : life of electric bulbs
X ~ N (µ = 1000, σ = 200)
Z = \(\frac{x-1000}{200}\) ~ N(0,1)

(i) Probability that lamps fall in the first 800 burning hours is
P(X<800) = \(\mathrm{P}\left(\frac{\mathrm{X}-1000}{200}<\frac{800-1000}{200}\right)\)
= P(Z < – 1)
= (area from – ∞ to ∞) – area from (-1 to ∞)
= 1 – 0.8413 = 0.1587
∴ Number of electric lamps expected to fail in the first 800 burning hours is (Given N = 2000)
P[X < 8000] = 2000 × 0.1587 = 317.4 ⇒ 317

(ii) (a) Period of burning hours after which 10% of the lamps would have failed (90% are still burning) P(X > k) = 90% = 0.90

P(-Z₁ < Z < 0] = 0.4
From tables Z₁ = -1.28
\(\frac{k-1000}{200}=-1.28\)
k = -(1.28) 200 + 1000 = -256 + 1000 = 744
After 744 hours 10% of lamps would have failed

(b) Period of burning hours after which 10% of the lamps are still burning is P(X > k₂) = 0.10
\(P\left(\frac{X-1000}{200}>\frac{k_{2}-1000}{200}\right)=0.1\)
P(Z > Z₂) = 0.1
Area from 0 to Z₂ = 0.4
From tables Z₂ = 1.28
\(\frac{k_{2}-1000}{200}=1.28\) = 1.28
k₂ = (1.28) 200 + 1000 = 1256
After 1256 hours 10% of the lamps are still burning.

Question 42.
The weights of 1000 Rajasthani Youths are normally distributed with mean 60 kgs and SD 8 kgs. Find the number of Rajasthani Youths with weight.
(i) less than 65 kgs
(ii) between 66 kgs and 70 kgs
(iii) more than 68 kgs
Answer:
Given µ = 60 σ = 8
X : weight of Rajasthani youths
X ~ N (60, 8)
z = \(\frac{X-\mu}{\sigma}=\frac{X-60}{8} \sim N(0,1)\)

(i) Pfweight less than 65 kgs) = P(X < 65)
⇒ \(\mathrm{P}\left(\frac{\mathrm{X}-60}{8}<\frac{65-60}{8}\right)\)
= P(Z < 0.625)
= (area from -∞ to ∞) – (area from 0.625 to ∞)
= 1 – 0.2643 = 0.7357
∴ Out of 1000 Rajasthani youths the number of youth with weight less than 65 kgs is 0.7357 × 1000 = 735.7
⇒ 736

(ii) p(66 < X < 70)
\(\mathrm{p}\left(\frac{66-60}{8}<\frac{\mathrm{X}-60}{8}<\frac{70-60}{8}\right)\)
⇒ P(0.75 < Z < 1.25)
= (area from 0.75 to ∞) – (area from 1.25 to ∞)
= 0.2266 – 0.1056 = 0.121
∴ The number of youth with weight between 66 and 70 among 1000 youth is 0.121 × 1000 = 121

(iii) P(more than 68 kgs) X > 68; \(P\left(\frac{X-60}{8}>\frac{68-60}{8}\right)\)
P(Z > 1) = area from 1 to ∞ = 0.1587
∴ Out of 1000 youths, the number of youths with weight more than 68 kgs is 0.1587 × 1000 = 158.7 ⇒ 159

Question 43.
If X is a Normal variate with mean (µ) and SD (σ) find the probability that X takes a value in the 1 σ neighbourhood of µ.
Answer:
P(µ – σ < x < µ + σ)

= Area fro (-1) to 1
= Area from (-1) to ∞ – Area from 1 to ∞
= 0.8413 – 0.1587 = 0.6826

Question 44.
If X is a Normal variate with mean (µ) and SD (σ) find the probability that x takes a value in the 3 σ neighbourhood of µ.
Answer:
P(µ – 3σ < x < µ + 3σ)

P ( 3 < Z < 3)
= (area from -3 to ∞) – (area from 3 to ∞)
= 0.9987 – 0.0013 = 0.9974

2nd PUC Statistics Chi-Square Distribution Exercise Problems

Question 1.
Define a Chi-square distribution.
Answer:
A continuous probability distribution with PDF

where 0 ≤ χ² < ∞ and n > 0 is said to be Chi-square distribution.

Question 2.
Write down the p.d.f of a Chi-square distribution.
Answer:
p.d.f of χ² is

Question 3.
Write the range and parameters of a Chi-square distribution?
Answer:
Range is (0, ∞)
Parameter : n is the parameter of χ² distribution

Question 150.
Define degrees of freedom.
Answer:
The number of independent observations are called degrees of freedom.

Question 4.
What are the mean, variance and SD of a Chi- square distribution?
Answer:
Mean = n; variance = 2n; S.D. = √2n

Question 5.
Mention two features of a Chi-square distribution.
Answer:

1. Parameter of χ² distribution is n
2. (0, ∞) is The range of χ² distribution

Question 6.
If Z is a SNV then name the distribution Z² and find its variance.
Answer:
If Z is a SNV, Z² is chi-square distribution.
Variance = 2n = 2(1) = 2.

Question 7.
If Z₁ and Z₂ are two independent SNVs, then name the
distribution of Z₁² + Z₂² and find its mean
Answer:
If Z₁, Z₂ are two independent SNVs, then Z₁² + Z₂² is chi-square distribution with 2 d.f and its mean = n = 2.

Question 8.
For a Chi-square variate with 9df, find mean and SD.
Answer:
df = n = 9
Mean = n = 9, SD = √2n = √2 × 9 = 4.2426

Question 9.
What are the mean and variance of a Chi-square variates with 11 df?
Answer:
d.f. = n = 11
Mean = n = 11
Variance = 2n = 2 × 11 = 22

Question 10.
For a Chi-square variate χ² with 10 df P (0 < χ² < 9.34) = 0.5 find median and mode
Answer:
Given df = 10
P(0 < χ² < 9.34) = 0.52 ⇒ Median = 9.34
Mode = n – 2 = 10 – 2 = 8

Question 11.
For a Chi-square variate with 14 d.f, P (0 < χ² < 13.339) = 0.5 find median and mode
Answer:
df = 14, P(0 < χ²< 13.339)
Median = 13.339
Mode = 14 – 2 = 12.

Question 12.
If variance of a Chi-square variate is 14, what is the mean and mode?
Answer:
Var = 14
2n = 14 ⇒ n = \(\frac { 14 }{ 2 }\) = 7
Mean = n = 7
Mode = n – 2 = 7 – 2 = 5

2nd PUC Statistics Student’s T Distribution Exercise Problems

Question 1.
Define a Student t-distribution.
Answer:
A continuous probability distribution with probability density function.

Question 2.
Write down the p.d.f of a Students t-distribution
Answer:
p.d.f

Question 3.
Write the range and parameter of Students t-distribution.
Answer:
Range of‘t’ distribution = (-∞, ∞)
Parameter = n
T₂ = \(\frac{\lambda}{2}\).Tx(Using Recurrence relation)
= \(\frac{1.2}{2}\) × 180 = 108 T₂ = 108

Question 4.
What are the mean, variance and SD of Students t- distribution?
Answer:
Mean = 0 μ = 0
Variance = \(\frac{\mathrm{n}}{\mathrm{n}-2}\) ; SD = \(\sqrt{\frac{n}{n-2}}\)

Question 5.
Mention two features of a Student’s t-distribution.
Answer:

1. In t-distribution. n is the parameter
2. In t-distribution range is (-∞ < t < ∞)

Question 6.
If the parameter of a t-distribution is 8, find the mean and variance?
Answer:
n = 8
μ = 0
Variance = \(\frac{n}{n-2}=\frac{8}{8-2}\) = 1.33

Question 7.
If n = 10 for a Student’s t-distribution, find the median and SD
Answer:
n = 10
Median = 0, variance
= \(\frac{n}{n-2}=\frac{10}{10-2}\) = 1.25
SD = √variance = √1.25 = 1.118

Question 8.
If n = 4 for a Student’s t-distribution, find its mode and variance.
Answer:
n = 4
Mode = 0
Variance = \(\frac{n}{n-2}=\frac{4}{4-2}\) = 2

2nd PUC Statistics Theoretical Distributions : I (Binomial Distribution) Pratical Assignments

Question 1.
The Probability that a bomb hits the target is 1/4. Five bombs are aimed at the target. Find the probability that
i. 3 bombs hit the target,
ii. At the most two bombs hit the target,
Answer:
1. Given n = 5
p = \(\frac { 1 }{ 4 }\) = 0.25 and q = 1 – p = 1 – 0.25 = 0.75

2. Let X : number of bombs hitting a target.
X ~ B (n = 5, p = 0.25)

3. Then the p.m.f is:
p(x) = ⁿC_x p^xq^{n – x},x = 0,1,2….n
= ⁵C_x (0.25)^x (0.75)^{5 – x}; x = 0,1,2………5

(i) P (3 bombs hit the target)
= P (X = 3) = ⁵C₃(0.25)³ (0.75)^{5 – 3}
= 10 × 0.01563 × 05625 = 0.0879

(ii) P (atmost two bombs hit the target) P (X≤ 2)
= P (X = 0) + P(X = 1) + P (X = 2)
= ⁵C₀ (0.25)⁰(0.75)^{5 – 0} + ⁵C₁ (0.25)¹ (0.75)^{5 – 1} + ⁵C₂(0.25)² (0.75)^{5 – 2}
= (1 × 1 × 0.2373) + (5 × 0.25 × 0.3164) + (10 × 0.0625 × 0.4219)
= 0.2373 + 0.3955 + 0.2637 = 0.8964

Question 2.
In a college, 70% of the Student are boys. In a random sample of 6 students, find the probability of getting
(i) Two boys (ii) At least one boy
Answer:
1. Given n = 6, p = 70% = 0.7 , then q = 1-p = 0.3
2. Let X : number of boys
X~ B (n = 6, p = 0.7)
The p.m.f is:
p(x) = ⁿC_xp^xq^{n – x}, x = 0,1,2…….n
= ⁶C_x (0.7)^x (0.3)^{6 – x}, x = 0,1,2………..6

(i) P( Two boy) = P (X = 2)
= ⁶C₂(0.7)^x(0.3)^{6 – 2}
= 15 × 0.49 × 0.0081 = 0.0595

(ii) (ii) P (atleastone boy) = P (X ≤ 1) = 1 – P(X < 1)
= 1 – P (X = 0)
= 1 – 6 [⁶C₀ (0.7)⁰ (0.3)^{6 – 0}]
= 1 – [1 × 1 × 0.0007]
= 1 – 0.0007 = 0.9993

Question 3.
Player A has probability 3/5 of winning a game of chess with player B. If they play 4 games, find the probability that player A wins a least 3 games.
Answer:
Given n = 4, p = \(\frac { 3 }{ 5 }\) = 0.6 and q = 1 – p = 1 – 0.6 = 0.4
Let X : number of chess games won,
X~ B (n = 4, p = 0.6)
Then the p.m.f is :
p(x) = ⁿC_xp^xq^n-x,x = 0,1,2………n
= ⁴C_x(0.6)^x(0.4)^4-x, x = 0,1,2,3,4,
P (Wins atleast 3 games)
= P(X≤3) = p(X = 3)+ p(X = 4)
= ⁴C₃ (0.6)³ (0.4)^4-3 + ⁴C₄ (0.6)⁴ (0.4)^4-4
= (4 × 0.216 × 0.4) + (1 × 0.1296 × 1)
= 0.3456 + 0.1296 = 0.4752

Question 4.
The probability of an arrow hitting a tree is !4. if a 3 arrows are aimed at the tree, find the probability that
(i) Two arrows miss the tree
(ii) At least one arrow hit the tree
Answer:
Let X be the number of arrows hitting a tree
Given n = 3 , p = \(\frac { 1 }{ 4 }\) = 0.25 and q = 1 – p – 1 – 0.25 = 0.75,
X~ B (n = 3, p = 0.25)
Then the p.m.f is :
p(x) = ⁿC_xp^xq^{n – x} ,x = 0,1,2……n
= ³C_x (0.25)^x (0.75)^3-x, x = 0,1,2,3

(i) P (2 arrows miss the tree)
= P(one arrow hit the tree)
= P (X = 1) = ³C₁ (0.25)¹(0.75)^3-1
= 3 × 0.25 × 0.5625 = 0.4219

(ii) P (atleast one arrow hit)
= P (X > 1) = 1 – p (X < 1)
= 1 – P (X = 0) = 1 – ³C₀ (0.25)⁰ (0.75)^3-0
= 1 – (1 × 1 × 0.4219) = 0.5781

Question 5.
In a certain locality it is known that (1/5)^th of the taxis have defective meters. Assuming 50 investigators take a sample of 7 taxis each to see if their meters are defective,how many investigators will report.
(i) exactly 4 taxis have defective meters
(ii) at the most 3 taxis have defective Meters.
Answer:
Let X be the number of taxis having defective meters
Given n = 7: p = 1/5 = 0.2 : q = 1 – p = 1 – 0.2 = 0.8 and N = 50.
X ~ B (n = 7, p = 0.2)
The p. m. f is:
p(x) = ⁿC_xp^xq^n-x,x = 0,l,2….n
= ⁷C_x(0.25)^x(0.8)^{7 – x}; x = 0,1,2 ……. 7

(i) P (4 taxies have defective meters) = P (X = 4)
= ⁷C₄ (0.2)⁴ (0.8)^7-4
= 21 × 0.0016 × 0.8 = 0.0269
∴ No. of investigators to report atmost 3 taxies have defective meters = 0.0269 × 50 = 1.344 = 1

(ii) P(atmost 3 taxis have a defective meters)
= P (X≤ 3)
= p (X = 0) + p (X = 1)+ p(X = 2) + p (X = 3)
= ⁷C₀ (0.2)⁰ (0.8)^7-0 + ⁷C₁ (0.2)¹ (0.8)^{7 – 1} + ⁷C₂(0.2)² (0.8)^7-2 + ⁷C₃ (0.2)³ (0.8)^7-3
= (1 × 1 × 0.2097) + (7 × 0.2 × 0.2621) + (21 × 0.04 × 0.3277) + (35 × 0.008 × 0.4096)
= 0.2097 + 0.367 + 0.2753 + 0.1147 = 0.9667
∴ No. of investigators to report atmost 2 taxies have defective meters
= 0.9667 × 50 = 48.335 = 48

Question 6.
If on an average one ship in every 10 is sunk, find the chance that out of 6 ships at least 5 will arrive safely.
Answer:
Given n = 6, p (ship sunk) = \(\frac { 1 }{ 10 }\) = 0.1,
q = 1 – p = 1 – 0.1 = 0.9.
Let X be the number of ships arrive safely out of 6 ships
X ~ B (n = 6, p = 0.1)
Then the p.m.f is :
p(x) = ⁿC_xp^xq^{n – x},x = 0,1,2….n
= ⁶C_x(0.9)^x(0.1)^{6 – x}, x = 0,1,2………6
P (atleast 5 arrive safely)
= P (X ≥ 5) = p (X = 5) + p(x = 6)
= ⁶C₅ (0.9)⁵ (0.1)^{6 – 5} + ⁶C₆ (0.9)⁶ (0.1)^6-6
= (6 × 0.5905 × 0.1) + (1 × 0.5314 × 1)
= 0.3543 + 0.5314 = 0.8857

Question 7.
In a certain school 40 % of the students have opted for first language kannada. Assuming 50 teachers take a sample of student each, how many teachers will report that 2 or 3 students opted for first language kannada.
Answer:
Given n = 5, P = 40% = 0.4, q = 1 – p = 0.6 and N = 50.
Let X be the No. of student opted first language among a sample of 5 students
X~ B (n = 5, p = 0.4)
Then the p.m.f is :
p(x) = ⁿC_xp^xq^n-x,x = 0,l,2….n
= ⁵C_x (0.4)^x (0.6)^5-x, x = 0,1,2,3,4,5
P (2 or 3 student opted for first language kannada)
= P (X = 2 or X = 3) = P(X = 2)+ P(X = 3)
= ⁵C₂ (0.4)² (0.6)^{5 – 2} + ⁵C₃ (0.4)³ (0.6)^5-3
= (10 × 0.16 × 0.216) + (10 × 0.064 × 0.36)
= 0.3456 + 0.2304 = 0.576
Number of teachers report that 2 or 3 students opted for first language kannada
= 0.576 × 50 = 28.8 = 29.

2nd PUC Statistics Theoretical Distributioins : II (Possion Distribution) Pratical Assignments

Question 1.
In a service station, in a day, 3 trucks can be serviced. If the number of trucks coming in for servicing in a day is a poisson variate with mean 1.3, find the probability that on a particular day some trucks are returned without being serviced.
Answer:
1. Given X = 1.3
2. Let X be the number of trucks arriving for servicing
X ~ P (X=1.3)
Then the p.m.f is :
p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\); x = 0,1,2, ….. ∞
= \(\frac{\mathrm{e}^{-13}(1.3)^{\mathrm{x}}}{\mathrm{x} !}\); x = 0,1,2,……
P (Some trucks returned without servicing)
= P (More than 3 trucks arrived for servicing)
= P (X > 3) = 1 – p(X ≤ 3)
= 1 – [p(X = 0) + p(X = 1) + p(X = 2) + (X = 3)]

= 1 – e^-1.3 [1 + 1.3 + 0.845 + 0.3662]
= 1 – (0.2725 × 3.5112)
= 1 – 0.9568 = 0.0432

Question 2.
On an average a telephone operator receives 4 calls between 9.00 A. m and 9.10 A. M Find the probability that in a specified interval of 10 minutes he receives (i) No call, (ii) at the most two calls.
Answer:
Let X be the no. of telephone calls receives by the operator
Given λ = 4
X ~ p(λ = 1.4)
Then the p.m.f is
p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) ; x = 0,1,2,a; P(x) = \(=\frac{e^{-4} 4^{x}}{x !}\) ; x = 0,1,2….
(i) P (no call received ) = p(x = 0) = \(\frac{e^{-4} 4^{0}}{0 !}=\) = 0.0183
(ii) P(receives atmost 2 calls)
= P(X≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= e^-4 (1 + 4 + 8) = 0.00183 × 13) = 0.2379

Question 3.
On an average 20 in 100 blades manufactured by a firm are defective. If blades are supplied in packets of 25 each, what is the probability that a packet has
(i) no defective blade
(ii) defective blades
Answer:
Let X be the no. of defective blades 20
Given n = 25, p = \(\frac { 20 }{ 100 }\) = 0.2,
λ = np = 25 × 0.2 = 5
X ~ P(X = 5)
Then the p.m.f is :
p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) x = 0,1,2 …….. ∞
= \(\frac{\mathrm{e}^{-4} 5^{x}}{\mathrm{x} !}\) : x = 0,1,2,….
(i) P (no defective blades) = P(X = 0) = \(\frac{\mathrm{e}^{-5} 5^{0}}{0 !}\) = 0.0067
(ii) P(defective blades) = P (X ≥ 1)
= 1 – P(X< 1) = 1 – [P(X = 0)]
= 1 – \(\left[\frac{\mathrm{e}^{-5} \mathrm{S}^{0}}{0 !}\right]\) = 0.0067
= 1 – e^-5 = 1 – 0.0067 = 0.9933

Question 4.
2% of hooks manufactured by a firm are found to be defective. Find the probability that a box containing 100 hooks have
(i) exactly 4 defectives
(ii) More than one defective.
Answer:
Let X be the no. of defective hooks
Given n = 100, p = 2% = \(\frac{2}{100}\) = 0.02
Here n is large and P is small , under these condition binomial distribution tends to poission distribution with
λ = np = 100 × 0.02 = 2.
Then the p.m.f is
p(x) = \(\frac{\mathrm{e}^{-\lambda} \lambda^{\mathrm{x}}}{\mathrm{x} !}\) x = 0,1,2 ……. ∞
= \(\frac{e^{-4} 5^{x}}{x !}\) : x = 0,1,2, …..
X ~ P (λ = 2)

(i) P(exactly 4 defective)
= P(x = 4) = \(\frac{e^{-2} 2^{4}}{4 !}=\frac{0.1353 \times 16}{4.3 .2 .1}\) = 0.902

(ii) P(more than one defective)
= P(X > 1) = 1 – P (X ≤ 1)
= 1 – [P(X = 0) + P(X = 1)]
= 1 – \(\frac{e^{-2} 2^{0}}{0 !}+\frac{e^{-2} 2^{1}}{1 !}\) = 1 – e^-2(1 + 2)
= 1- (0.1353 × 3) = 1 – 0.4059 = 0.5941

Question 5.
In a city on an average 12 accidents take place in 30 days. Find the number days in a year in which
(i) 2 accidents take place
(ii) at least 3 accident take place.
Answer:
Let X be the number accident occurred 12
Given λ = \(\frac{12}{30}\) = 0.4, N = 365
X ~ P (λ = 0.4)
Then the p.m f is
p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) : x = 0,1,2, ……. ∞
(i) P(2 accidents)
= P(x = 2) = \(\frac{\mathrm{e}^{-0.4} 0.4^{2}}{2 !}=\frac{0.6703 \times 0.16}{2 \times !}\) = 0.0536
No of days with 2 accidents in a year
= 0.05\(\)36 × 365 = 19.56 = 20

(ii) P(atleast 3 accidents) = P(X ≥ 3) = 1 – P(X < 3)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

= 1 – e^-0.4 (l + 0.4 + 08)
No. of days with atleast 3 accidents in a year
= 0.00796 × 365 = 2.9 = 3

Question 6.
In a text book, on an average 5 mistake per 10 pages are found. If there are 425 pages in that book, in how many pages there will be
(i) exactly 3 mistakes
(ii) less than or equal to 2 mistakes?
Answer:
Let X be the number of mistakes in the book
Given λ = \(\frac{5}{10}\) = 0.5,N = 425
X ~ P (λ = 0.5)
Then the p.m.f is
p(x) = \(\frac{-\lambda}{\mathrm{x} !}\) ; x = 0,1,2, ……. ∞, = \(\frac{\mathrm{e}^{-0.5} 0.5^{\mathrm{x}}}{\mathrm{x} !}\) ; x = 0,1,2 ……..

(i) P(exactly 3 mistakes)
= p(x = 3) = \(\frac{e^{-0.5} 0.5^{3}}{3 !}\)
= \(\frac{0.6065 \times 0.125}{6}\) = 0.0126
No. of pages with 3 mistakes
= 0.0126 × 425 = 5.37 = 5

(ii) P(less than or equal to 2 mistakes) = P(X < 2)
= P(X = 0) + P(X = 1) + P(X = 2)

= e^0.5 (1 + 0.5 + 0.125) = 0.6.65 × 1.625 = 0.9856
No. of pages with less than or equal to 2 mistakes
= 0.9856 × 425 = 418.99 = 419

2nd PUC Statistics Hyper Geometric Distribution Pratical Assignments

Question 1.
In a hyper – geometric distribution with a = 14, b = 4 and n = 3, find mean and variance.
Answer:

Question 2.
In a hyper geometric distribution if a = 4, b = 6 and n = 4 find P[X = 3].
Answer:
In a p.m.f. of hypergemetric variate, by substituting all values:

Question 3.
There are 30 boys and 40 girls in a class. A committee is to be formed by selecting 5 students at random from the class. Find the probability that committee formed has 2 boys.
Answer:
Let X be the no. of boys from a sample of 5 students
Given a = 30, b = 40 and n = 5,
X~ H (a = 30, b = 40, n = 5)
Then the p.m.f. is :

Question 4.
In a hostel, there are 20 vegetarian and 30 non vegetarian inmates. For a statistical survey, a random sample of 6 inmates are selected. Find the probability that there are 4 vegetarians.
Answer:
Let X be the no. of vegetarians from a sample of 6 inmates
Given a = 20, b = 30 and n = 6,
X ~ H (a = 20, b = 30, n = 6)
Then the p.m.f. is:

2nd PUC Statistics Theorttical Distributions : III (Fitting Binomiai Distuibution) Pratical Assignments

Question 1.
Five unbiased coins are tossed 512 times, Calculate the expected frequencies for the number of heads obtained.
Answer:
Let X be the no. of heads obtained when fire unbiased coins are tossed
Given n = 5, p = 0.5, and q = 1 – p = 0.5, N = 512
X ~ B (n = 5, p = 0.5)
Then the p.m.f is
p(x) = ⁿC_xp^xq^{n – x},x = 0,1,2 …… n
= ⁵C_x (0.5)^x (0.5)^{5 – x} ;x = 0,1,2 …… 5
Theoretical frequency = expected frequency : T_x = N × p(x)
When x = 0,
T_o = p(x = 0) × N = ⁵C0(0.5)⁰(0.5)^{5 – 0} × 512
= 1.1.0.03125 × 512 = 16
Using Recurrence relation

The distribution of theoretical frequencies of no. of heads is

Question 2.
Six unbiased coins are tossed 64 times. Find the theoretical frequencies for the number of tails obtained.
Answer:
Let X be the no. of tails obtained when 6 unbiased coins are tossed
Given n = 6; p = 0.5 and q = 1 – p = 0.5, N = 64.
X ~ B (n = 6, p = 0.5)
Then the p.m.f is
p(x) = ⁿ C_xp^xq^{n – x},x = 0,1,2……n
T₀ = p(x = 0) × N = ⁶C₀ × 0.56 0 × 64 = 1
Using recurrence relation:

The distribution of theoretical frequencies of tails is :

Question 3.
Fit a binomial distribution for the following data and obtain the theoretical frequencies.

Answer:
Let X be no of defective screws given n = 6 and p is to be estimated.

6p = 3.354; p = \(\frac{3.354}{6}\) = 0.56 andq = 1 – p = 0.44
X ~ B (n = 6, p = 0.56)
Then the p.m.f is
p(x) = ⁿ C_xp^xq^{n – x}, x = 0,1,2…….n
= ⁶C_x(0.56)^x(0.44)^{6 – x}; x = 0,1,2,…….6
Theoretical frequency: T_x = N × p(x)
When x = 0,
T_o = p(x = 0) × 127 = ⁶C_o(0.56)⁰(0.44)^{6 – 0} × 127 = 0.0073 × 127 = 0.927 = 1
The remaining theoretical frequencies are obtained using the recurrence relation Recurrence relation:

The distribution of frequencies theoretical is

Question 4.
Fit a binomial distribution for the following data and obtain the theoretical frequencies.

Answer:
Given n = 5 and p is to be estimated
Here, mean = x̄ = np = \(\frac{\Sigma \mathrm{fx}}{\mathrm{N}}\)

5p = \(\frac{197}{104}\) = 1.894; p = \(\frac{1.894}{5}\) = 0.38
And q = 1 – p = 1 – 0.38 = 0.62; N = 104
X ~ B (n = 5, p = 0.38)
Then the p.m.f is
p(x) = ⁿ C_xp^xq^{n – x},x = 0,1,2…….n
P(x) = ⁵C_x(0.38)^x(0.62)^{5 – x}; x = 0,1,2,..5
Theoretical frequency: T_x = N × p(x)
When x = 0,
T_o = p(x = 0) × 104
= ⁵C₀(0.38)⁰(0.62)^{5 – 0} × 104 = 9.53
The remaining theoretical frequencies are obtained
Using recurrence relation:

The fitted theoretical and observed frequencies is : (approximated)

2nd PUC Statistics Fitting Poisson Distribution Pratical Assignments

Question 1.
Fit a poisson distribution and obtain theoretical frequencies:
Answer:
From the frequency distribution: Mean = x̄ = λ = \(\frac{\Sigma f x}{N}\)

λ = \(\frac{120}{100}\) = 1.2.
Then the p.m.f is

Theoretical frequency
T_x = Nxp(x)
∴ T₀ = p(x = 0)(N = 10) = \(\frac{\mathrm{e}^{-1.2} 1.2^{0}}{0 !} \times 100\)
= 0.3012 × 100 = 30.12
Using recurrence relation:

The fitted observed and theoretical frequency distribution is: (Approximated)

Question 2.
Below are given the number of vacancies of lecturers occurring in a college over a period of 100 years.

Fit a Poisson distribution and obtain expected frequencies.
Answer:
Let x be the number of vacancies is a poisson variate and f be the number of years then
λ is obtained as :
Mean = x̄ = λ = \(\frac{\Sigma f x}{N}\)

∴ λ = \(\frac{98}{100}\) = 0.98
The p.m.f is :

Theoretical frequency: T_x = p(x)N
T₀ = p(x = 0) × 100 = \(\frac{\mathrm{e}^{-0.98} 0.98^{0}}{0 !} \times 100\)
= 0.3753 × 100 = 37.53

The fitted poisson distribution is

2nd PUC Statistics Theoretical Distuibutions : IV (Normal Distribution) Pratical Assignments

Question 1.
The mean and variance of a normal distribution are 173 and 36 respectively. Find the probability that the value of the variable selected at random will be
(i) below 158
(ii) between 173 and 185
(iii) above 175
Answer:
Given = mean : μ = 173
Variance : σ² = 36
S.D: σ = √var = √36 = 6
Then, S.N.V Z = \(\frac{X-\mu}{\sigma}\) – N(0,1); Z = \(\frac{x-173}{6}\)

(i) P(X is below 158) = P(X< 158)
= P(z < \(\)) = p(z < -2.5)
= Area from (-2,.5) to (-∞)
= 1 – Area from (-2.5) to ∞
= 1 – 0.9938 = 0.0062

(ii) P (173 < X < 185)
= P(\(\frac{173-173}{6}<\mathrm{Z}<\frac{185-173}{6}\)) = P(0 < Z < 2)
= Area from 0 to 2
= Area from 0 to ∞- Area from 2 to ∞
= 0.5 – 0.0228 = 0.4772

(iii) p(X > 175) = p[z > \(\frac{175-173}{6}\) = P(z > 0.33)
= Area from 0.33 to ∞ = 0.3707

Question 2.
is a normal variate with parameters μ = 50 and σ² = 16. Find
(i) P[X > 48]
(ii) P[X < 56]
(iii) P[52 < X < 55]
Answer:
Given μ = 50: σ² = 16,
G = √l6 = 14 then S.N.V
Z = \(\frac{X-\mu}{\sigma}\) = \(\frac{x-50}{4}\) ~ N(0,1)

(i) P(X > 48) = P(z > \(\frac{48-50}{4}\)) = p(z > -0.5)
= Area from -0.5 to ∞ = 0.6915

(ii) P(X < 56) = P(Z < 1.5) = Area from 1.5 to (-∞)
=1 – Area from 1.5 to ∞
= 1 – 0.0668 = 0.9332.

(iii) p(52 < X < 55) = p\(\left(\frac{52-50}{4}<Z<\frac{55-50}{4}\right)\)
= P(0.5 < Z < 1.25) = Area from 0.5 to 1.25 = Area from 0.5 to ∞- Area from 1.25 to ∞ = 0.3085 – 0.1056 = 0.2029

Question 3.
Heights of a group of boys who attend the army selection camp is normally distributed with mean 170 cm and S.D. 3.9cm. Minimum eligible height for army is 175cm. Show that only 10% of the group of boys eligible.
Answer:
Let x be the height of boys is a normal variate with the parameters µ = 170, = 3.9
Then S.N.V. Z = \(=\frac{X-\mu}{\sigma}\) = \(\frac{X-170}{3.9}\) ~ N(0,1)
P(minimum height is 175 cms) = p(x ≥ 175)
= P(Z ≥ \(\frac{175-170}{3.9}\)) = P(Z ≥ 1.28)
= Area from 1.28 to ∞ = 0.1003
The required percentage is 0.1003 × 100 = 10.03%

Question 4.
1200 students took an examination. Their mean marks is 55 and the S.D. is 10. Assume that marks scored by students follow normal distribution.
(i) How many of the students are expected to score above 50?
(ii) If 80% of the students are to be promoted, what should be the marks for promotion?
Answer:
Let X be the marks of students is a Normal variate with parameters ,u= 55 and a = 10
And N = 1200.Then S.N.V: Z = \(\frac{X-\mu}{\sigma}\) = \(\frac{x-55}{10}\) ~ N(0,1)
(i) p(X > 50) = P( Z > \(\frac{50-55}{10}\)) = p(Z > -0.5)
= Area from (-0.5) to ∞ = 0.6915
No. of students score above 50
= 0.6915 × 1200 = 829.8

(ii) Let k be the marks for promotion of 80% student
i.e., P(X ≥ k) = 80% = 0.8
p(Z ≥ \(\frac{\mathrm{k}-55}{10}\)) = 0.8
Let \(\frac{\mathrm{k}-55}{10}\) = Z₁ since Area from (-Z) to ∞ is 0.8
∴ P(Z ≥ -Z₁) = 0.8

Area from -Z₁ to = 0.8
∴ From the table Z₁ = 0.84 (Table value 0.7995)
(i.e., in the table 0.8 under 4)
i.e., \(\frac{k-55}{10}\) = o.84; k – 55 = 0.84 × 10
K = -8-4 + 55 = 46.6 is the minimum marks.

Question 5.
The marks scored by the candidates appearing for a competitive examination follow normal distribution with mean 60 and S.D. 5. If only 10% of the top most scores are to be provided with jobs, what is the leastmarks a candidate has to be score for securing the job.
Answer:
Let x be the marks of students is a normal variate with the parameters µ = 60 and σ = 5
Then S.N.V Z = \(\frac{x-\mu}{\sigma}\) = \(\frac{X-60}{5}\) ~ N(0,1)
Let k be the least marks, then,
P(X ≥ k) = 10% = 0.1; p(Z ≥ \(\frac{k-60}{5}\)) = 0.1
Let Z₁ = \(\frac{k-60}{5}\) ;P(Z ≥ Z₁) = 0.1
From the tabe Z₁ = 1.28
i.e., \(\frac{k-60}{5}\) = 1.28; k – 60 = 1.28 × 5

Question 6.
Daily wages of 60 workers are normally distributed with mean ₹ 599 and S.D. ₹ 40. Find the number of number of workers getting wages
(i) more than ₹ 530
(ii) more than ₹ 490
(iii) between ₹ 380 and ₹ 460
Answer:
Let x be the daily wages of workers is a normal variate with µ = 500; σ = 40, N = 60
Then S.N.V : Z = \(\frac{X-\mu}{\sigma}\) = \(\frac{x-500}{40}\) ~ N(0,1)

(i) P(X > 530) = P( Z > \(\frac{530-500}{40}\)) = P(Z > 0.75)
= Area from 0.75 to ∞ 0.2266
No. of workers with wage more than ₹ 530
= 0.2266 × 60 = 13.596 = 14

(ii) p(X > 490) = P(Z > \(\frac{470-500}{40}\)) = P(Z > 0.25)
Area from [-0.25) to ∞ = 0.5987
No. of workers with wage more than ₹ 470
= 0.5987 × 560 = 35.92 = 36

(iii) P(380 < X < 460) = \(\left(\frac{380-500}{40}<z<\frac{460-500}{40}\right)\)
= P[-3 < Z < – 1) = Area from [-3) to [-1)
= Area from [-3) to ∞- Area from (-1) to ∞
= 0.9987 – 0.8413 = 0.1574
No. of workers with wages between ₹ 380 and ₹ 460
= 0.1574 × 60 = 9.44 = 9