2nd PUC Statistics Previous Year Question Paper June 2015

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Karnataka 2nd PUC Statistics Previous Year Question Paper June 2015

Time: 3 Hours 15 Minutes
Max. Marks: 100

Section – A

I. Answer any Ten of the following questions : (10 × 1 = 10)

Question 1.
Define cohort.
Answer:
Cohort is a group of individual who are born at the same time and who experience the same mortality conditions.

Question 2.
What is the value of index number for the base year?
Answer:
100.

Question 3.
Which index number shows upward bias?
Answer:
Laspeyre’s.

Question 4.
Give an example for random variation.
Answer:
There are more number of deaths in utharkand due to flood in 2014.

Question 5.
Write down the relationship between mean and variance of a Bernoulli distribution.
Answer:
Mean > varience, ie., p > pq

Question 6.
For a chi – square variate with 10d.f., find the variance.
Answer:
Varience = 2n = 2 × 10 = 20.

Question 7.
What is parameter space?
Answer:
The set of all admissible values of the parameter is called parameter space.

Question 8.
Given σ² = 16 cm² and n = 49, calculate standard error of sample mean.
Answer:
σ² = 16 ∴ σ = √l6 = 4 ; n = 49
S.E(x̄) = \(\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{49}}=\frac{4}{7}\) = 0.57

Question 9.
What is type I error?
Answer:
Rejecting Null Hypothesis (H₀₎ when it is actually true is called type 1 Error.

Question 10.
Statistical quality control helps in detecting which type of variation?
Answer:
Assignable causes of variation.

Question 11.
What do you mean by non – degenerate solution in a transportation problem?
Answer:
If the no of positive allocation are equal to (m + n – 1) the solution is called Non – degenerate.

Question 12.
Write down one advantage of inventory.
Answer:
Bulk production decreases of production and transportation cost.

Section – B

II. Answer any Ten of the following questions : (10 × 2 = 20)

Question 13.
In a locality out of 20,000 births 100 mothers died due to child birth complications. Calculation maternal mortality rate.
Answer:

Question 14.
If Σp₀q = 12000 and Σp₁q = 1500, compute suitable index number.
Answer:
Kelly’s method: P₀₁ = \(\frac{\Sigma p_{1} q}{\Sigma p_{0} q}\) × 100 = \(\frac{1500}{1200}\) × 100 = 125

Question 15.
Mention the steps involved in the construction of consumer price index number,
Answer:

• Object and the scape
• Conducting Family budget survey
• Obtaining the price Quotation
• Method of construction.

Question 16.
Write down the normal equations for fitting linear trend.
Answer:
The normal equation for fitting linear trend are: na + b Σ x = Σy and a Σ x + b Σx² = Σ xy

Question 17.
Write down the conditions for the application of binomial expansion method of interpolation and extrapolation.
Answer:

• The values of the independent variable should have a common difference.
• The value of ‘x’ for which the value of y is to be interpolate mtist be one of the value of ‘x’

Question 18.
Write down the Bernoulli distribution with parameter p = 0.25
Answer:
The Bernoulli distribution has the following p.m.f and the probability distribution as: p(x) = p^xq^1-x; x = 0, OR

Here x is a Bernouli variate

Question 19.
Mention two features of a students t-distribution
Answer:

1. Range : (- ∞, ∞),
2. n – is the parameter called degrees of freedom.

Question 20.
What are point estimation and interval estimation?
Answer:
While estimating the unknown parameter, if a ‘Specific value is proposed as an estimate, is called‘point estimation’.

While estimating the unknown parameter instead of a specific value ‘an interval’ is proposed which is likely to contain the parameter is called ‘interval estimation’

Question 21.
Write down two utilities of standard error.
Answer:

1. It is used in interval estimation, to write down the confidence intervals
2. It is used in testing of hypothesis, to test whether the difference between the sample statistic and population parameter is significant or not.

Question 22.
What do you mean by process control and product control in statistical quality control?
Answer:
Controlling the quality of the product during the manufacturing process itself is called “process – control”.
The process of inspection of manufactured lot for acceptability is called ‘product control’, it is called as acceptance sampling.

Question 23.
Mention two methods of obtaining initial basic feasible solution for a transportation problem. .
Answer:

1. North West Comer Rule (N.W.C.R)
2. Matrix Minima Method (M.M.M)

Question 24.
Given R = 1000 units/month, C₃ = Rs. 350 and C₁ = Rs. 0.20 / unit/ month, find Q⁰.
Answer:

Section – C

III. Answer any Eight of the following questions: (8 × 5 = 40)

Question 25.
From the following data calculate GFR, ASFR (20 – 24) and ASFR (30 – 39).

Answer:

= \(\frac{2260}{25000}\) × 1000 = 90.4 Births / 1000 women of (15 – 49) year in a year.

Question 26.
Define index number. Mention three limitation of index numbers.
Answer:
Definition : Index number is a statistical measure designed to show an average change in a variable or a group of related variables over a period.
Limitations :

1. Many formulae are used in the construction of index number which gives different answers.
2. As the customs and habits of the customers change from time to time, the uses of corn modities also vary.
3. Like other statistical techniques, index numbers cañ be misused to get desired results.

Question 27.
Compute cost of living index number.

Answer:

Question 28.
Compute the trend values by finding five yearly moving averages for the following data.

Answer:
Let x and y be the year and values

From the moving averages (ŷ) the time series shows upward trend

Question 29.
Using Newton’s forward difference method find the value of ‘y’ when x = 25.

The number of known values of ‘y’ n = 5 , so, prepare the leading differences upto Δ⁴
Answer:

x = \(\frac{25-15}{4}\) = 2.5 , (x – 1) = (2.5 – 1) = 1.5 ; (x – 2) = (2.5 – 2) = 0.5, (x – 3) = (2.5 – 3) = -0.5
The Newton’s formula of interpolation is;

y₂₅ = 17 + 2.5 + 5.625 – 0.3125 + 0 = 24.5625

Question 30.
The probability that a bomb hits the target is \(\frac { 1 }{ 4 }\). 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:
Let x be the number of bombs hitting a target is a Binomial varible with the parameters
n = 5, p = \(\frac { 1 }{ 4 }\) = 0.25 and q = 1 – p = 1- 0.25 = 0.75.
Then the p.m.f is ; p (x) = ⁿC_x p^xq^1-x; n = 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 × 0.01562 × 0.5625 = 0.0878

(ii) p ( at the most 2 bombs hit the target) = p(x < 2)
= p(x = 0) + p(n = 1) + p(x = 2)
= 5 C₀ (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 = 0.8965

Question 31.
In a Hyper – geometric distribution if a = 4, b = 6 and n = 4, find mean and variance.
Answer:

Question 32.
From the following data test whether mean daily wages of workers of Factory – A and Factory – B are equal. (Use 1% l.o.s)

Answer:
Given : n₁ = 200, n₂ = 450
x̄₁ = 195, x̄₂ = 200, S₁ = 20, S₂ = 30
H₀ : Mean daily wages of workers of factory A and B are equal ie., µ₁ = µ₂
H₁ : Mean daily wages of workers of factory A and B are not equal ie., µ₁ ≠ µ₂ {Two tail test}


At α = 1% the two tail critical values are ±k = ±2.58
Here Z_cal lies in (AR) acceptance region, i.e., Z_cal > – k ∴ H₀ is accepeted.
Conclusion : Mean daily wages of workers of factory A and B are equal ie., µ₁ = µ₂

Question 33.
From the following data, test whether literacy and smoking are independent at 5% level of , significance.

Answer:
The given-data can be written under contingency table as below:

H₀ : Literacy and smoking are independent.
H₁ : Literacy and smoking are dependent {upper tail test) .


At α = 5% for 1. d.f the upper tail critical value k₂ = 3.84
Here χ²cal lies in (A.R) acceptance region ie., χ²cal < k₂
∴ H₀ is accepted.
Conclusion : Literacy and smoking are independent.

Question 34.
The following data relates to the number of knitting defects per unit length of cloth manufactured by a textile mill.

Draw control chart with λ¹ = 3.
Answer:
Knitting defects so use ‘c’ – chart. Standard λ¹ = 3 is known : The control limits are :
C.L = λ¹ = 3 ; L.C.L = λ’ – 3\(\sqrt{\lambda^{\prime}}\) = 3 – 3\(\sqrt{3}\) = -2.196 = 0; is C.L λ¹ + \(3 \sqrt{\lambda^{1}}\) = 3 + \(3 \sqrt{3}\) = 8.196

Conclusion : All sample point lies with the control limits, so the production process is under control.

OR
(For Visually challenged Students only)

Briefly explain single sample plan.
Answer:
Single sampling plan (S.S.P): In S.S.P a random sample of size ‘n’ is drawn from the lot size ‘N’ Depending on the number of defectives the conclusion is made as below.
Let ‘d’ be the no of defectives in the samples; c – be the acceptance no. then,

• If d ≤ c, lot is accepted and ‘d’ no. of defectives are replaced by good ones.
• If d > c, lot is rejected.The entire lot is inspected and all defectives are replaced by good ones.

Question 35.
A tailor gets a profit of Rs. 100 from a shirt and Rs. 170 from a pant. In a week of 56 hours, he uses 20 hours for stitching and 36 hours for cutting. He requires one hour to stitch a shirt and two hours to stitch a pant. For cutting he requires two hours for a shirt and three hours for a pant. Formulate L.P.P.
Answer:
Given : The given statement is written as below:

The L.P.P. is: Maximize ‘z’ = 100x + 170y
Subject to constraints 2x + 3y ≤ 36
1x + 2y ≤ 20 and
Non – negativity x ≥ 0, y ≥ 0

Question 36.
The cost of a scooter is Rs. 36,000. Its resale value and maintenance cost years is given below:

Determine the optimal year for replacement of the scooter.
Answer:
Given : p = 36,000 S_n – Resale value, C_i – Maintenance cost and n – be the years.

From the above, the annual average Maintenence cost is minimum for the 5th year. So n = 5 is the optimum period for replacement of the scooter and A(5) = Rs. 5920.

Section – D

IV. Answer any Two of the following questions: (2 × 10 = 20)

Question 37.
For the following data compute standardized death rates and comment.

Answer:
Let A and B be the ASDRs of Localitis A and B, Ps the standard Papulation.


for Locality B:
ASDR (0 – 20) = \(\frac{128}{8000}\) × 1000 = 16
Similary compute ASDRs can be calculated for the age groups.
S.D.R(A) = \(\frac{\Sigma P_{s} A}{\Sigma P_{s}}\) = \(\frac{464000}{30,000}\) = 15.47
S.D.R(B) = \(\frac{\Sigma P_{s} B}{\Sigma P_{s}}\) = \(\frac{428000}{30,000}\) = 14.27
Here SD.R (B) < S.D.R(A). So locality B is healtheir than locality B.

Question 38.
Compute Fischer’s price index number. Show that it satisfies time reversal test and factor reversal tests.

Answer:

Question 39.
Fit an exponetial trend line of the form y = ab^x Also estimate the value for 2008.

Answer:
Let x and y be the year and value.
Exponential trend equation: y = ab^x Taking logarithms : logy = log a + x logb
Normal equations are : n log a + log b. Σ x = Σ logy – (1)
log a. Σx + log b. Σ x² = Σ x log y – (2)

log a = \(\frac{\Sigma \log \mathbf{y}}{\mathbf{n}}\) = \(\frac{7.8530}{7}\) = 1.1219
∴ a = Antilog (loga) = A.L(1.1219) = 13.24
From (2) logb = \(\frac{\Sigma x \log y}{\Sigma x^{2}}\) = \(\frac{4.207}{28}\) = 0.15025
∴ b = Antilog (log b) = A.L (0.15025) = 1.4135
The fitted exponential trend equation is; y = ab^x; ŷ = 13.24 (1.4135)^x
Estimation for the year 2008 ; x = 5 : ŷ (2008) = 13.24 (1.4135)5 = 74.7079

Question 40.
Consider the following data:

Fit a Poisson distribution to the data and test the goodness of fit (Use 5% L.o.s)
Answer:
Given : From the given frequency distribution ‘ λ ’ is obtained as: Mean = x̄ λ = \(\frac{\Sigma \mathrm{fx}}{\mathrm{N}}\)

∴ λ = \(\frac{99}{100}\) = 0.99
The p.m.f is ; p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x 1}\) ; x = 0,1,2 …….. ∞
= \(\frac{\mathrm{e}^{-0.99}(0.99)^{x}}{x !}\) ; x = 0,1,2, …….
Theoretical frequency; T_x = p(x) N
To = P(x) = 0 × 100 = \(\frac{\mathrm{e}^{-0.99}(0.99)^{0}}{0 !}\) × 100 = 0.3716 × 100 = 37.16
Using recurrence relation : T = \(\frac{\lambda}{x}\) T_x-1

Similary T₇ = N – [T₀ + T₁ + T₂ + ……. + T₆] = 100 – 100 = 0
The fitted poisson distribution is:

CHI – SQUARE TEST :
H₀ : Poisson distribution is good fit ie., O_i = E_i
H₁ Poisson distribtion is not good fit ie., O_i ≠ E_i (UPPER TAIL TEST &NLY)

λ is estimated from the data and so (n – 1 – 1) = (n – 2) d.f

Here χ² = 11.098
At α = 5% for (n – 2) = 4 – 2 = 2 d.f the upper tail critical value k₂ = 5.99
Here χ²_cal lies in (RR) rejecting region ie χ²_cal > K₂
∴ H₀ is rejected and H₁ is accepted.
Conclusion: Poisson distribution is not a good fit.

Section – E

V. Answer any Two of the following questions: (2 × 5 = 10)

Question 41.
The weekly wages of workers are normally distributed with mean Rs. 3,000 and S.D. Rs. 500. Find the probability of workers whose’ weekely wages will be i) More than Rs. 3,400 ii) Between Rs. 2,506 and Rs. 3,500
Answer:
Given : Let x be the weekly wages is a Normal variable with the parameters µ = 3000,
5 = 500, then S.N.V : z = \(\frac{x-\mu}{\sigma}\) ~ N(0,1) z = \(\frac{x-3000}{500}\)

(i) p(x > 3400) = P\(\left(z>\frac{3400-3000}{500}\right)\) = p(z > 0.8)

= Area from 0.8 to ∞ = 0.2119 from the table

(ii) p(2500 < x < 3500) = p\(\left(\frac{2500-3000}{500}<z<\frac{3500-3000}{500}\right)\) = P(-1 < Z < 1)

= Area from (-1) to 1 .
= Area from (-1) to ∞ – Area from 1 to ∞
= 0.8413 – 0. 1587 = 0.6826

Question 42.
In a random sample of 1000 persons from a large population, 470 are females. Can it be concluded that males and females are in eqaul proportion in the population? (use α = 0.05).
Answer:
Given : n = 1000, x = 470, P₀ = 0.5 (Male & female are equal in proporation)
H₀ : Male and female are equal in proportion ie., p₀ = 0.5
H₁: Male and female are not in equal proportion ie., p₀ ≠ 0.5 { Two tailed test}
Under H₀ the test statistics; z = \(\frac{p-P_{0}}{\sqrt{P_{0} Q_{0}}}\) ~ N(0,1)
And sample proportion ; p = \(\frac{x}{n}=\frac{470}{1000}\) = 0.47 and
Q₀ = 1 – p₀ = 1 – 0.5 = 0.5


At α = 0.05 or 5% the two tail critical values are ±k = 1.96. Here Z_cal lies in acceptence region (A.R) ie, – k < Z_cal ≤ k. Therefore H₀ is Accepted.
Conclusion : Male and female are equal in proportion ie, p₀ = 0.5

Question 43.
It is required to test whether those who practice yoga have average blood sugar less than 120. A sample consisting of 17 persons who practice yoga is observed. If their mean blood sugar is 108 and S.D. is 8, what would you conclude ? (Use 5% level of significance).
Answer:
Given: µ = 120, n = 17, x̄ = 108, s = 8, α = 5% (n < 30, small sample t – test for mean)
H₀ ; Average blood use sugar is 120 ie., µ = 120
H₁ ; Average blood sugar is less than 120, ie, µ < 120 (Lower tail test).

At α = 5% the lower tail critical value -k = -1.75 for (n – 1) = 17 -1 = 16. d.f
Here t_cal lies in rejection region (R.R)
∴ H₀ is rejected ans H₁ is accepted.
Conclution : Average B – P. is Less than 120 (ie, µ < 120).

Question 44.
Find the solution of the game by the principle of dominance for the following pay – off matrix of A.

Answer:

By principle of dominance, all the pay-off A₄ are more or equal to all the payoff of A₁ and A₂. A₄ dominates A₁ and A₃, so delete A₁, and A₃.

All the payoff of B₂ are less than all the payoff of B₁ and B₃ B₂ dominates B₁ and B₃ So delete B₁ and B₃

A₄ dominates A₂, So delete A₂

B₂ dominates B₄ delete B₄

Saddle point occurs at (4, 2) or A₄, B₂
The suggeste optimal strategies for the player are : A₁, B₂. The value of the game v = 2.