2nd PUC Statistics Model Question Paper 3 with Answers

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Karnataka 2nd PUC Statistics Model Question Paper 3 with Answers

Time: 3 Hours 15 Minutes
Max. Marks: 100

Section – A

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

Question 1.
What is ‘Radix’in the life table?
Answer:
Size of cohort is called radix and size is usually taken as 100,000 individuals.

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

Question 3.
Write the relation between laspeyre’s, paasche’s and Dorbish- Bowley’s price index number.
Answer:

Question 4.
What is secular trend
Answer:
It refers to the tendency of the variable / values of the time series to show increase, decrease or to remain steady over a given period.

Question 5.
In a Binomial distribution if n = 6, and p = \(\frac { 1 }{ 3 }\) find mean.
Answer:
Mean = np = 6 × \(\frac { 1 }{ 3 }\) = 2

Question 6.
Write the probability density function of a Normal variable with mean = 5 and variance = 9.
Answer:

Question 7.
Mention one use of standard error.
Answer:
It is used in interval estimation to write down the confidence interval

Question 8.
What is Null hypothesis.
Answer:
A Hypothesis which is being tested for possible rejection assuming that it is true.

Question 9.
Write down the expression for S.E (P)
Answer:
SE(P) = \(\sqrt{\frac{P_{0} Q_{0}}{n}}\)

Question 10.
What is control chart.
Answer:
Control chart is a graphical device use to verify whether the production process is under control or not.

Question 11.
When L.P.P is said to have an unbounded solution?
Answer:
For some L.P.P the optimum value of Z may exists at infinity, then the L.P.P is said to have unbounded solution

Question 12.
What is the value of a game?
Answer:
The pay – off at the saddle point is called value of game denoted by ‘V’.

Section – B

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

Question 13.
In a population of 11,200 there were 212 deaths in an year. Find C.D.R
Answer:

= \(\frac{212}{11.200}\) × 1000 = 18.9286 Deaths / 1000 population / in a year.

Question 14.
If P₀₁ (L) = 137.1 and P₀₁(P) = 139.3. Find Fisher’s index.
Answer:
P₀₁ (F) = \(\sqrt{P_{01}(L) \times P_{01}(P)}\) = \(\sqrt{137.1 \times 139.3}\) = 138. 1956

Question 15.
Mention two methods of constructing a consumer price Index number.
Answer:

1. Aggregative expenditure method
2. Family budget method.

Question 16.
Give an example for each of seasonal variation and irregular variation.
Answer:
There is more sales of ice creams and cool drinks in summer season and increase in price of vegetables due to heavy rain over a week.

Question 17.
Write down the Newton’s formula of interpolation.
Answer:
The Newton’s formula of interpolation is :

Question 18.
In a poisson distribution p(x = 0) = 2 p (x= 1) then find p(x = 2)
Answer:

Question 19.
In a Normal distribution if Q₁ = 22cms and Q₂ = 34 cms, then find median.
Answer:
Mean / Median / Mode = \(\frac{Q_{3}+Q_{1}}{2}=\frac{34+22}{2}\) = 28

Question 20.
Define ‘estimator’ and ‘estimate’.
Answer:
The statistic used to estimate the unknown population parameter is called ‘estimtor’. The value of the estimator is the ‘estimate’.

Question 21.
From the following data find S.E (x̄₁ – x̄₂).

Answer:

Question 22.
In S.Q.C define ‘variable’ and ‘attribute’
Answer:
A measurable quality characteristic which varies from unit to unit is a variable.
A qualitative characteristic which cannot be measured and can only be identified its presence and absence is an Attribute.

Question 23.
Mention any two situations when replacement is carried out.
Answer:

1. The item that has become inefficient with passage of time.
2. When its maintenance cost is more as its age increases.

Question 24.
Write the formula for minimum average cost in E.O.Q model with shortage with all Meaning of the notations.
Answer:
C(A⁰,S⁰) = \(\sqrt{2 \mathrm{C}_{3} \mathrm{RC}_{1}} \times \sqrt{\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}}\)
where C₃ – setupcost, R -Demand rate, C₁ – Maintenance cost, C₂ – Shortage cost.

Section – C

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

Question 25.
Compute GF.R, T.F.R and average number of children per women from the data.

Answer:

= \(\frac{7230}{91(1000)}\) × 1000 = 80.33
T.F.R = 5 Σ Quin quennial ASFRs; here ASFRs are readily given.
= 5(475.83) = 2779.15 births/ 1000 women
Average number of children bom per women
= \(\frac{\text { T.F.R }}{1000}=\frac{2379.15}{1000}\) = 2.379 children bom per women.

Question 26.
Calculate the price index number for the following data by using : (a) simple aggregative method (b) simple average of price relatives method using arithmetic mean.

Answer:

(a) Simple aggregative price index number
p₀₁ = \(\frac{\Sigma p_{1}}{\Sigma p_{0}}\) × 100 = \(\frac{91}{81}\) × 100 = 112.346
(b) Simple average of price relatives:
P₀₁ = \(\frac{\Sigma p}{n}\) = \(\frac{700.83}{6}\) = 116.805

Question 27.
Construct consider price I. No. using family budget method

Answer:
Here expenses = W

Question 28.
Calculate 5 year moving averages for the following data regarding the number of industries in loss in India.

Answer:
Let x and y be the year and sales.

Question 29.
Interpolate and extrapolate the sales fqr the years 2010 and 2014.

Answer:

The number of known values of y : ‘n’ = 4 and so the Binomial expansion (y – 1)⁴ = 0
(y – 1)⁴ = y₄ – 4y₃ + 6y₂ – 4y₁ + y₀ = 0 → (1)
52 – 4(30) + 6 (20) – 4y₁ + 10 = 0.
52 – 120 + 120 – 4y + 10 = 0; -4y₁ + 62 = 0 ∴ y₁ = \(\frac{+62}{+4}\) = 15.5
To get y₅, the suffixes of y has increases by ‘ 1’ in equation (1)
y₅ – 4y₄ + 6y₃ – 4y₂ + y₁ = 0
y₅ – 4(52) + 6(30) – 4 (20) + 15.5 = 0
y₅ – 208 + 180 – 80 + 15.5 = 0
∴ y₅ – 92.5 = 0; ∴ y₅ = 92.5

Question 30.
A fair coin is tossed 4 times. What is the probability of getting (a) more than 2 heads (b) atleast one head.
Answer:
Let x be the number of heads obtained is a Binomial variate with-the parameters n = 4,
p = \(\frac { 1 }{ 2 }\) = 0.5 (Fair) ∴ q = 1 – p = 0.5.
Then the p.m.f is : – p(x) = ⁿc_x p^x q^n-x; x = 0,1,2, …….. n.
= ⁴c_x (0.5)^x (0.5)^4-x; x = 0,1,2,3,4.

(i) p(more than 2 heads ) = p (x > 2) = p(x = 3) + p(x = 4)
= ⁴c₃ 0.5³ 0.5^4-3 + ⁴c₄ 0.5⁴c₃ 0.5^4-4 = (4 × 0.125 × 0.5) + (1 × 0.0625 × 1)
= 0.25 + 0.0625 = 0.3125 .

(ii) p(atleast one head) = p(x ≥ 1) = 1 – p (x <1);
= 1 – p(x = 0) = 1 – ⁴c_o 0.5⁰ 0.5^4-0 = 1- (1 × 10.0625) = 1 – 0.0625 = 0.9375

Question 31.
It is known that 3% of screws manufactured by a firm are defective. What is the probability that one or two defective screws will be found in a box of 100 screws ?
Answer:
Let x be the number of defective screws is a Binomial variate with the parameters n = 100, p = 3% = 0.03
Here ‘n’ is very large and ‘p’ is small. Under these conditions Binomial distribution tends to poisson distribution with mean = np = λ ; λ = 100 × 0.03 = 3
Then the p.m.f is: p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) ; x = 0,1,2 ….. ∞ = \(\frac{e^{-3} 3^{x}}{x !}\) ; x = 0,1,2 ……..
p ( one or two defectives) = p(x = 1) + p(x = 2)
= \(\frac{e^{-3} 3^{1}}{1 !}+\frac{e^{-3} 3^{2}}{2 !}\) = e^-3 (3 + 4.5) = 0.0498 × 7.5 = 0.3 7235.

Question 32.
It is known that the standard deviation of marks of Bangalore university (B.U) students is 10 and that Mysore University (M.U) students is 8. The mean of 300 randomly selected B.U students is found to be 68 marks and mean of 400 randomly selected M.U students is found to be 65 marks. Test of 5% level of significance that marks scored by B.U students is more than that of M.U studen
Answer:
Given: σ₁ = 10, σ₂ = 8, n₁ = 300, x̄₁ = 68, n₂ = 400, x̄₂ = 65, α = 5%
H₀ : Mean marks of B.U students and M.U students are same, (µ₁ = µ₂)
H₁ : Mean marks of B.U students is more than m.u students (ie., µ₁ > µ₂) {upper tail test


At α = 5% the upper tail critical region value ±k = 1.65
Here z_cal lies in rejection region / z_cal> k
∴ H₀ is rejected and H₁ is accepted.
Conclusion : Mean marks of B.U students is more than M.U. student ( (µ₁ > µ₂)

The Standard Deviation Calculator is a tool to calculate the standard deviation from the data, the standard error, the range, percentiles.

Question 33.
From the following data test whether the difference between the proportions in the two samples is significant.

Answer:
Given: n₁ = 1000; P₁ = 0.02; n₂ = 1200, p₂ = 0.01
H₀ : There is no significant difference in the proportions (ie., p₀₁ = p₀₂)
H₁ : There is a significant difference in the proportions (ie., p₀₁ ≠ p₀₂)) { Two tail test}


At α = 5% the two tail critical values are ± k = ± 1.96
Here Z_cal lies in acceptance region
∴ H₀ is accepted.
Conclusion : There is no significant difference in the proportion (ie., p₀₁ = p₀₂).

Question 34.
The following data gives number of defects observed In assembling cars inspected everyday.

Construct a suitable control chart and draw your conclusion.(For visually challenged only)
What is single sampling plan? Mention the steps involved in S.S’.P
Answer:
Defects are given so use c- chart, standard is not known
We need; c̄ = \(\frac{\Sigma c}{k}\) defects = \(\frac{92}{10}\) = 9.2
The control limits are : C.L = c̄ = 9.2
L.C.L = \(\bar{C}-3 \sqrt{\bar{C}}\) = 9.2 – \(3 \sqrt{9.2}\) = 0.1006
U.C.L = \(\bar{C}+3 \sqrt{\bar{C}}\) = 9.2 + \(3 \sqrt{9.2}\) = 18.29

Conclusion : All sample points lies with in control limits and so, the number of defects in the production process are under control.

Question 35.
Solve the following game using maximin – minimax principle.

Answer:

By maximum-minimum principle:

• Minimum pay – off in each row is circled
• Maximum pay – off in each column is boxed.
• ‘-4’ at (2, 3) is circled as well as boxed is the saddle point
• The suggested optimal Strategies are : A₂, B₃
• The value of the game V = -4

Question 36.
A firm is considering the replacement of a Machine whose cost is Rs. 12,800 and the scrap value is Rs. 300. The maintenance costs at different year as given below:

When should the machine be replaced?
Answer:
Given: p = 12800, s_n = 300 and so (p – s_n) = (12800 – 300) = 12500 for all the years

From the above table, the annual average maintenance cost is minimum for the n = 6^th year. So, the Machine is suggested to replace for 6^th year. The optimum cost A(6) = Rs. 3016.67

Section – D

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

Question 37.
Compute crude death rates, standardised death rates for the two towns A and B, taking town A as the standard population. Hence find which town 4s more healthy.
Answer:

Answer:
Given : Let B be the ASDRs of town B. Since town A is taken as standard and so, calculating C.D.R is same as SDR.

Question 38.
Compute suitable price index number and show that it satisfies T.RT and F.R.T.

Answer:

Question 39.
Fit a Quadratic trend for the following data and hence find the production for the year 2002.

Answer:
Given :Let x and y be the year and production and x̄ in the table x̄ = 1997.5
x = 2 \(\left(\frac{x-\bar{x}}{c}\right)\) ; c = 1

Fromthenormalequations: na + b Σx + c Σx² = y
6a + b.0 + c.70 = 145; 6a + 70c = 145 → (1)
From: a Σx + b Σx² + c Σx³ = Σxy ; a.0 + b.70 + c 0 = 85 ie., 70b = 85 ; ∴ b = \(\frac{85}{70}\) = 1.2143
And from : a Σx² + bΣx³ + c Σx⁴ = Σx²y
a. 70 + b.0 + c. 1414 = 1793 ; 70a + 1414 c = 1793 → (2)

∴ c = \(\frac{608}{3584}\) = 0 = 1696 ; put c = 0.1696 in (1) we get;
6a + 70 (0.1696) = 145 ; 6a = 145 – 11.872 ; a = \(\frac{133.128}{6}\) = 22.188
The fitted quadratic equation is :- y = a + bx + cx²
ŷ = 22. 188 + 1.2143 x + 0. 1696 x².
Estimation of production for the year 2002 ; x = 9, x² = 81
∴ ŷ(2002) = 22. 188 + 1.2143 (9) + 0.1696(81) = 46.8543 tons.

Question 40.
The table below shows the number of road accidents qccured iiu50 days period. Fit a poisson distribution and test whether it is good fit.

Answer:
Given : Let x be the number of road accidents is a poisson variate, the parameters obtained as below: Let f be the no. of days, then from the distribution

Then the p.m.f is : p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) ; x = 0,1,2, ……… ∞ p(x) = \(\frac{\mathrm{e}^{-0.9} 0.9^{\mathrm{x}}}{\mathrm{x} !}\) ; x = 0,1,2, ……… ; N = 50
Theoretical / Expected frequency : T_x = p ( x) N
∴ T₀ = p(x = 0) × 50 = \(\frac{\mathrm{e}^{-0.9} 0.9^{0}}{0 !}\) × 0.4066 × 50 = 20.33
Using recurrence relation, the remaining theoretical frequencies can be obtained as : T_x = \(\frac{\lambda}{x} T_{x-1}\)

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

CHI – SQUARE TEST
H₀ : Poisson distribution is good fit (O_i = E_i)
H₁ : Poisson distribution i not good fit (O_i ≠ E_i) {upper tail test}
under H₀, the χ² – test statistic is

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


∴ χ²_cal = 0.13
At α = 5%. for (n – 2) = (3 – 2) = 1 d.f the upper tail critical value k₂ = 3.84
Here χ²_cal < k2 χ²_cal llies in A.R.
∴ H₀ is accepted
Conclusion: – Poisson distribution is good fit (Oi = Ei)

Section-E

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

Question 41.
If the heights of 500 solutions are found to be Normally distributed with mean 70” and S.D 2”. The find the probability of soldiers whose height is (a) above 73” (b) between 66”and 73”.
Answer:
Let x be the height is a Normal variate with Mean: µ = 70” and S.D : σ = 2”, then S.N.V is
z = \(\frac{x-\mu}{\sigma}\) ~ N(0,1); z = \(\frac{x-70}{2}\)

(i) p (height is above 73) = p(x > 73) = p\(\left(z>\frac{73-70}{2}\right)\)

= p (z > 1.5) = Area from 1.5 to ∞ = 0.0668 from the table

(ii) p ( between 66” & 73”) = p\(\left(\frac{66-70}{2}<z<\frac{73-70}{2}\right)\)

= -p(-2 < z < 1.5)
= Area from (-2) to ∞ – Area from 1.5 to ∞
= 0.9772 – 0.0668 = 0.9104

Question 42.
A Milk filling machine fills sachets with Milk. The contention is that the standard deviation of quantity of Milk filled is more then 3ml. To test this, 24 sachets are randomly selected and their contents noted. If the standard deviation of these observations is 3.8 ml. What is year conclusion?
Answer:
Given : σ = 3 ml. n = 24, s = 3.8 ml
H₀ : SD of milk filling is 3ml (ie., σ = 3ml)
H₁ SD of milk filling is more than 3 ml (ie., σ > 3ml) (upper tail test}
under H₀ the χ² – test statistics is : χ² = \(\frac{\mathrm{ns}^{2}}{\sigma^{2}}\) ~ χ²( n – 1) d.f
χ²_cal = \(\frac{24 \times(3.8)^{2}}{3^{2}}\) = 38.5067

At α = 5% for (n – 1) = (24 – 1) 28 d.f the upper tail critical value k₂ = 35.17
Here χ²_cal > k₂ / χ²_cal lies in rejection region.
H₀ is rejected and H₁ is accepted.
Coclusion: SD. of milk filling is more than 3m1 ( ie.,σ > 3m1)

Question 43.
The mean percentage pass of all the school of a town was found to be 83. To test this a random sample of 17 schools revealed that 86% passed with a.S.D of 3% can we conclude that mean percentage of pass Is more than 83.
Answer:
Given: Mean percentage: µ = 83 n = 17, x̄ = 86 s = 3, α = 1%
(n < 30 use small sample t- test)
H₀ : Mean pass percentage is 83 (le., µ = 83)
H₁ : Mean pass percentage is more than 83 (ie., µ > 83) { upper tail teset) under H₀, the test statistic is:


At α = 1% for (n – 1) = 17 – 1= 16 d.f the upper tail critical, value k = 2.58
Here t_cal > k / t_cal lies in rejection region (RR)
∴ H₀ is rejected and H₁ is accepted.
Conclusion : Mean pass percentage is more than 83 (µ > 83)

Question 44.
A manufacturer has to maintain a stock of 10800 unit of a product per year to his customer. The shortage cost is assumed to be infinite. The demand is known and fixed. The inventory holding cost is Rs. 0.25per unit per month? The time to be a month Find (i) the total optimum lot size (ii) the minimum t0tal monthly cost
Answer:
Given R = 100 / year c₁ = 0.25/month c₃ = 300. Here R = \(\frac{10800}{12}\) = 900/month
c₂ – shortage not allowed use model I.