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## Karnataka 2nd PUC Statistics Previous Year Question Paper March 2016

Time: 3 Hours 15 Minutes

Max. Marks: 100

Section – A

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

Question 1.

What is a life table ?

Answer:

life table is a tabular presentation of numerical data, describing the mortality experience of a cohort.

Question 2.

Define an ‘Index number’.

Answer:

Index number is a statistical measure designed to show an average change in a variable or groap of related variables over a period (or Area or Income).

Question 3.

Write the formula for computing Marshall – Edgeworth’s quantity number.

Answer:

Question 4.

What is a historigram ?

Answer:

The graph of actual values of time series is called historigram.

Question 5.

What are the values that a Bernoulli variate can take ?

Answer:

0,1.

Question 6.

Name the distribution for which standard deviation and variance are equal.

Answer:

Standard normal distribution.

Question 7.

What is confidence coefficient ?

Answer:

Probability that the confidence interval contains parameter is called, confidence co-efficient (1 – α ).

Question 8.

Define critical region.

Answer:

The set of all those values of the test statistic which leads to the rejection of the null hypothesis is called critical region.

Question 9.

Write the degrees of freedom while testing independence of attributes in a 2 × 2 contingency table.

Answer:

1 (one) degrees of freedom.

Question 10.

What is chance cause of variation ?

Answer:

A small amount of variation for which no specific cause can be observed in the quality of the product, this variation is called as chance cause of variation.

Question 11.

Define ‘Solution’ in an LPP.

Answer:

A set of real values of X = (x_{1}, x_{2}, x_{3} …. x_{n}) which satisfies the constraints Ax (≤=≥)b, is called a Solution.

Question 12.

What is a two – person zero sum game ?

Answer:

In a game of two players, the gain of one player is the loss of the other is called two – person zero sum game.

Section – B

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

Question 13.

There was a record of 4000 live births in a city in a year. The number of neonatal deaths was 100* Calculate the neonatal mortality rate.

Answer:

Question 14.

Write any two limitations of Index number.

Answer:

- Many formulae are used and gives different answers for the index.
- As the customs and habits of the people changes from time to time, the uses of the commodities also vary.

Question 15.

Answer:

Question 16.

Mention all the components of a time series.

Answer:

- Secular trend/trend.
- Seasonal variation.
- cyclical variation.
- Irregular/Random variation.

Question 17.

Define ‘Interpolation’ and ‘Extrapolation’.

Answer:

‘Interpolation’ is the technique of estimating the value of the dependent varialde (y) for any intermediate value of the independent variable (x ).

‘Extrapolation’ is the technique of estimating the value of dependent viable (y) for any value of the independent variable (x) which is outside the rarge of the given series.

Question 18.

For a chi-square variate with 9 degrees of freedom, obtain mean and mode of the disribution.

Answer:

Degrees of freedom : n = 9, mean = n = 9, Mode = (n – 2) = (9 – 2) = 7.

Question 19.

Write the values of β_{1} and β_{2} in a t-distribution.

Answer:

For t-distribution : β_{1} = 0 (Non – skewed) and β_{2} > 3 (Leptokurtic).

Question 20.

Computer SE(P_{1} – P_{2}), given n_{1} = 80, n_{2} = 40, P_{1} = 0.8 and P_{2} = 0.4.

Answer:

Question 21.

Define Null and Alternative hypothesis.

Answer:

‘Null Hypothesis’ is a hypothesis which is being tested for possible rejection, assuming that it is true.

A Hypothesis which is being accepted when the null hypothesis (H_{0}) is rejected is called Alternative hypothesis (H_{1}).

Question 22.

Compute the lower and upper control limits for X̄ – chart when X̄^{1} = 40, σ^{1} = 6 and A = 1.342.

Answer:

L.C.L = X̄ – Aσ^{1} = 40 – (1.342 × 6) = 31.948, U.C.L = X̄ + Aσ^{1} = 40 + (1.342 × 6) = 48.052

Question 23.

The objective function and two solutions of an LPP are Max Z = 200x + 100y and A(0, 18); B(12, 0). Find the optimal solution.

Answer:

Z_{A} = 200 (0) + 100 (18) = 1800, Z_{B} = 200 (12) + 100 (0) – 2400. Z_{B} is maximum so, the optimal solution is : x = 12, y = 0.

Question 24.

Define pure and mixed strategy in a game.

Answer:

‘Pure strategy’ of a player is the pre – determinad course of action/strategy adopted irrespestive of the coursa of action of the opponent.

‘Mixed strategy’ of a player is his pre – decision to choose his course of action according to certain pre – assigned probabilites.

Section – C

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

Question 25.

Computer the net reproduction rate from the following data :

Answer:

S. Survival rate and W.S.F.R – women specific fertility rate

W.S.F.R (15 – 19) = \(\frac{1250}{50,000}\) × 1000 = 25 Similarity Computer W.S.F.Rs for other age groups.

∴ N.R.R = 5 x 120 = 600 Female Children bom per 1000 women of (15 – 49) years in the year.

Question 26.

Calculate ‘Weighted GM price Index number for the year 2014 from the following data :

Answer:

Question 27.

What is a consumer price Index number ? Write any four of its uses.

Answer:

Consumer price index number is an index number of the cost met by a specified class of consumers in buying a basket of goods and services.

Uses are : Consumer price index numbers are

- Commonly used in fixation of salary and dearness allowances to the employees.
- Used by the governments in formulating Economic, price and wage policies.
- In evaluation of purchasing power of money and deflating the money.
- In comparing the cost of living of different classes of people.

Question 28.

Compute 4-yearly moving averages from the following data.

Answer:

Let X and Y be the year and Profit.

Here 4YMT – 4 yearly moving total 259.37

4YCMT – 4 yearly centered moving total.

4 YCMA – 4 yearly centered moving average.

Question 29.

Interpolate the number of workers earning wages Rs. 775 per day from the following table :

Answer:

Let X and Y be the weges and no of workers the number of knows values of y : ’n’ = 5, so prepare the leading diffemces upto Δ^{4}

‘y’ – value total determine for x = 775, the Newton’s formula of interplotion is :

x = \(\frac{755-400}{150}\) = 2.5,(x – 1) = (2.5 – 1) = 1.5,(x – 2) = (2.5 – 2) = 0.5,(x – 3) = (2.5 – 3) = -0.5

= 50 + 125 + 46.875 + 4.6875 – 1.5625

∴ Y_{775} = 225 are the number of earning woges Rs. 775 Perday.

Question 30.

40% of buttons manufactured by a firm are found to be defective. In a random sample of six buttons, find the probability of getting

(i) exactly 4 defective buttons.

(ii) more than 1 defective button.

Answer:

Let x be the number of defective buttons manufactured is a Binomnal variate with the parameters n = 6, p = 40/100 = 0.4 and q= 1 – p = 1 – 0.4 = 0.6.

The p.m.f is : P(x) = nC_{x}p^{x}q^{n-x}: x = 0, 1, 2, n

P(x) = 6C_{x}(4)^{x} (0.6)^{6-x}: x = 0, 1, 2, …………. 6

(i) P (getting exectly 4 defective buttons) = P(x = 4)

= 6C_{4} (0.4)^{4} (0.6)^{6-4} = 15 × 0.0256 × 0.36 = 0.1382.

(ii) P (getting more than 1 defective button) = P(x > 1)

= 1 – P (x ≤ 1) = 1 – P (x = 0) + P (x = 1)

= 1 – 6C_{0} (0.4)^{0} (0.6)^{6-0} + 6C_{1} (0.4)^{1} (0.6)^{6-1}

= 1 – (1.1.0.0466) + (6 × 0.4 × 0.0777)

= 1 – (0.0466) + (0.1865) = 1 – 0.2331 = 0.7669.

Question 31.

There are 20 marbles in a box, out of which 12 are red coloured antf the rest white coloured. A child picks 6 marbles at random. Find the probability that the child gets 4 red coloured marbles.

Answer:

Let X be the number of red coloured marbles picked is a hypergeometric variate with the parameters a = 12 (Red marbles), b = 20 – 12 = 8 (White marbles) and n = 6 (Picked). Then the

Question 32.

Among 400 randomly selected persons of a city, 280 like ice-cream. Test at 1% level of significance that 50% of the people in that city like ice-creams.

Answer:

Given : n = 400, x = 280 (ice-cream liked persons)

The sample proportion : P = \(\frac{x}{n}=\frac{280}{400}\) = 0.7 and

P_{0} (population proportion) = 0.5 (50% liked) and ∴ Q_{0} = 1 – P_{0} = 1 – 0.5 = 0.5, α = 5%

H_{0} : 50% people like ice-cream (P_{0} = 0.5)

H_{1} : 50% people does’t like ice-cream (P_{0} ≠ 0.5) {Two tailed test: ±K}

At α = 5% the two tail critical values are ±k = ±2.58

Here Z_{cal} lies in rejection region.

∴ H_{0} is rejected and H_{1} is accepeted..

Conclusion: 50% people does’t like ice-cream (P_{0} ≠ 0.5)

Question 33.

From the following data, test whether mean of the first Population is greater than the mean of the second population. (Use α = 5%)

Answer:

Given: n_{1} = 10, n_{2} = 12, x̄_{1} = 62, x̄_{2} = 55, s_{1}^{2} = 16, S_{1}^{2} = 25

Here n_{1}, n_{2} are less than 30, use small sample t- test for equality means.

H_{0} : means of I and II populations are same (i.e, μ_{1} = μ_{2})

H_{1} : means of I population is greater than II population (i.e, μ_{1} > μ_{2}) {upper tail test ±K} Under H_{0}, the test statistic is :

At α = 5% for (n_{1} + n_{2} – 2) = (10 + 12 – 2) = 20d.f

the upper tail critical values are K = 1.73

Here t_{cal} lies in rejection region

H_{0} is rejected and H_{1} is accepted.

Conclusion: Mean of I population is goeater than II (μ_{1} > μ_{2})

Question 34.

Twenty rolls of different types of cloth contained the following number of defects in each roll.

2, 4, 2, 3, 5, 6, 7,-2, 3, 6, 4, 8, 2, 4, 5, 2, 3, 1, 2, 4.

Calculate the control -limits for C-chart.

Answer:

Here defects are given use C-chart.

The control limitsare : C.L = C̄ = 3.75

= 3.75 – 5.8094 = -2.05 = 0 [Token as ‘0’]

Question 35.

For the following transportation problem, find the initial basic feasible solution by Matrix Minima and obtain the cost associated with the solution.

Answer:

Here Σai = Σbj, the T.P. is balanced. By Matri X minima method, first allocation is made in least cost cell at(3,3)/(C,III) as : X_{33} = min (110, 330) = 110.

Here ‘C’ is satisfied so delete, and replace

330 as (330-110) = 220.

The redued cost matri x is.

Next least cost is ‘5’ at (B.I): Allocete as

X_{BI} = min (100,80) = 80

‘I’ is sotisfied so deleti and replace 100 by (100 – 80) = 20.

The reduced cost matri x is :

The least cost is 8 : allocate as : min (20, 150) = 20.

‘B’ is satisfied and replace 150 by (150 – 20) = 130.

The least is allocate as : min (350, 220) = 220. Ill is satisfied and replace 350 by (350 – 220)= 130.

Allocate as : ‘130’ as the remaing allocation.

The suggested I.B.F.S and total cost is.

Question 36.

A machine costs Rs. 8,000. Its maintenance cost and resale value per year are given in the following table :

Determine the optimal replacement age of the machine.

Answer:

Given: p = 8000 cost of machine.

The A(n) is minimum for 5^{th} year, is the optimal age of replacement of the machine. A(5) = Rs. 2976/-

Section – D

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

Question 37.

From the following data, compute the standardized death rates. Hence find the locality with healthier population.

Answer:

Let A and B be the Age specific death rates of locality A and B and P_{s} be the standard populatioa

Question 38.

For the following data, compute Fisher’s Price Index number, Show that Fisher’s Index number satisfies Time reversal and Factor reversal tests.

Answer:

Here Expenditures means : under base year – p_{0}q_{0} and under current year – p_{1} q_{1} . The given data is written as below :

Question 39.

Fit a parabolic trend of the from Y = a + bX + cX^{2} for the following rime series :

Answer:

Let x and y be the year and students strength

Here x̄ = 2012 middle most year.

From the normal equation: na + b Σx + c Σx^{2} = Σxy

5a + b.0 + c.10 = 53; 5a + 10c = 53 ………. (1)

From: aΣx + bΣx^{2} + cΣx^{3} = Σxy ; a.0 + b. 10 + c.0 = 0; 10b = 0; ∴ b = 0.

And From: aΣx^{2} + bΣx^{3} + cΣx^{4} = Σx^{2}y ; a. 10 + b.0 + c.34 = 120

i.e, 10a + 34c = 120 …………… (2)

Tx = \(\frac{1.9}{2}\) C = \(\frac{14}{14}\); C = 1, Put C = 1 in equation …….. (1)

We get 5a + 10(1) = 53; 5a = 53 – 10; a = \(\frac{43}{5}\) = 8.6

Hence the titted parabolic trend equation is : Y = a + bx + cx^{2}; ŷ = 8.6 + 0.x + 1x^{2} = 8.6 + x^{2}

Question 40.

Fit a Poisson distribution to the following data and test the goodness of fit at 5% level of significance.

Answer:

Let x be the number of mistakes is a poisson variate the Parameter λ is obtained as below :

Let‘t’ be the number of pages, then from the treghenly distribution : mean = x̄ = \(\frac{\sum_{\mathrm{fx}}}{\mathrm{N}}\) = λ

∴ λ = \(\frac{285}{150}\) = 1.9

Then the p.m.f is : P(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) : x = 0,1,2, …… ∞

p(x) = \(\frac{\mathrm{e}^{-1.9}(1.9)^{x}}{\mathrm{x} !}\) ; x = 0,1,2, …….

Theoretical/Expected frequency = T_{x} / E_{x} = P(x)N.

T_{x}/E_{x} = p(x = 0) .150 = \(\frac{\mathrm{e}^{-1.9}(1.9)^{0}}{0 !}\) × 150 = 0.1496 × 150 = 22.44

The fitted observed theoretical frequency disribution is:

χ^{2 }TEST :

H_{0} : Poisson distribution is good fit ie., (O_{i} = E_{i})

H_{1} : Poisson distribtion is not good fit (O_{i} > E_{i}) {upper tail only}.

Under H_{0}, The χ^{2} – test statistic is:-

Here d.f is (n – 2) because ‘λ’ computed from the data and so d.f. will be (n – 1 – 1) = (n – 2)

∴ χ^{2}_{cal} = 1.8309

At α = 5% for (n – 2) = (6 – 2) = 4. d.f the upper tail critical value k_{2} = 9.49.

Here χ^{2}_{cal} lies in acceptance region

∴ H_{0} is accepted.

Conclusion: Poisson distribution is not a good fit i.e., (O_{i} = E_{i})

Section – E

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

Question 41.

The distribution of monthly income of 5,000 workers of a factory follows Normal distribution with mean Rs. 10,000 and standard deviation Rs. 1,000. Find probability of the number of workers.

i) having income more than Rs. 9,000.

ii) having income between Rs. 8,500 and Rs. 12,000.

Answer:

Let x be the monthly income is a Normal variable with the parameters ‘ µ ’ = 10,000 and ‘σ’= 1,000 and N = 5,000.

Then the S.N.V is Z = \(\frac{x-\mu}{\sigma}\) = \(\frac{x-10,00}{1000}\) ~ N(0,1)

(i) P (income is more then Rs.9000) =

p(x > 900) = p\(\left(\mathrm{Z}>\frac{9000-10,000}{1000}\right)\) = p(z > -1)

= Area from 1 to ∞ = 0.8413

No. of workers = 0.84 13 × 5000 = 4206.5

(ii) P (income between Rs. 8500 and Rs. 12,000) = p (8500 < X < 12000)

= p \(\left(\frac{8500+10,000}{1000}<\mathrm{Z} \frac{12000-10,000}{1000}\right)\)

= P(-1.5 < Z < 2) = Area from (-1.5) to ‘2’

= Area from (-1.5) to ∞ – Area from 2 to ∞

= 0.9332 – 0.0228 = 0.9104

No. of workers = 0.9104 × 5000 = 455.2

Question 42.

Samples of electric lamps manufactured by two firms gave the following results :

Test at 5% level of significance whether the average life of bulbs manufactured by – firm A is less than the average life of bulbs manufactured by firm B.

Answer:

Given : n_{1} = 120, n_{2} = 160, X̄_{1} = 2,540, X̄_{2} = 2600

S_{1} = 96, S_{2} = 108. and α = 5% (Here n_{1}, n_{2} > 30, use large sampletest)

H_{0}: Average life of bulbs of firm A and B is same firm (µ_{1} = µ_{1})

H_{1} : Average life of bulbs of firm A and is less than firm B (i,e. µ_{1} < µ_{2}) {Lower tailed test}

At α = 5% the Lower tail

critical values are -k = -1.65

Here Z_{cal} lies in rejection region ∴ H_{0} is Accepted H_{1} and is accepted.

Conclusion : Average life of bulbs of firm A and is less than firm B (i,e.µ_{1} < µ_{2}).

Question 43.

A normal variate has variance 100. Twenty random observations of the variate has variance 144. Test at 1% level of significance whether the variance of the population is 100.

Answer:

Given: variance (population): σ^{2} = 100, n = 20,

variance (sample): s^{2} = 144 and σ = 1%.

χ^{2} test for variance.

H_{0} ; Population variance is 100 (i,e. σ^{2} = 100)

H_{1} ; Population variance differ from 100 (i,e. σ^{2} ≠ 100) (Two tail test}

Under H_{0}, the χ^{2} – test statistic is; χ^{2} = \(\frac{\mathrm{ns}^{2}}{\sigma^{2}}\) ~ χ^{2}(n – 1) d.f.

∴ χ^{2}_{cal} = \(\frac{20 \times 144}{100}\) = 28.8

At α = 1% for (n – 1) = (20 – 1) = 19.d.f the two tail.

Critical values are K_{1} = 6.84 and K_{2} = 38.6

Here χ^{2}_{cal} lies in acceptance region ∴ H_{0} is accepted.

Conclusion: Population variance =100 (i,e. σ^{2} = 100)

Question 44.

The demand for an item is 7,000 units per year. The cost of placing an order is Rs. 300 per year. The holding cost is Rs. 10 per unit per year. The storage cost is Rs. 3 per unit per year. Find

i) Economic Order Quantity, ii) Minimum annual average inventory cost.

Answer:

Given : R = 7000/year, C_{3} = 3000, C_{1} = 10/year and C_{2} = 3/year – Shortages given use

= 648.074 × 2.0816 = 349.03 units/year.

ii) Minimum average inventory cost :

= 648.074 × 0.4804 = 3113.35 units/year.