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## Karnataka 2nd PUC Statistics Previous Year Question Paper June 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’s -‘Longevity’?

Answer:

Life expectancy of new born baby is called ‘Longevity’.

Question 2.

Theoretically which is the best average for the construction of index numbers ?

Answer:

Geometric mean (GM).

Question 3.

Which weights are used in the construction of Paasche’s price index number ?

Answer:

Current year quantity (q_{1}).

Question 4.

What is ‘Time series’ ?

Answer:

Chronological arrangement of statistical data according to time is called ‘Time series’.

Question 5.

Write the relation between mean and variance of Binomial distribution.

Answer:

Mean > variance, ie : np > npq

Question 6.

If variance of Chi – square distribution is 10, find its mean.

Answer:

Variance = 2n = 10 : Mean = n = \(\frac{10}{2}\) = 5

Question 7.

Define ‘Parameter’ in testing of hypothesis.

Answer:

Statistical constants of the population such as mean – µ, S.D – σ are called ‘Parameters’.

Question 8.

Define sampling distribution of a Statistic.

Answer:

It is the distribution of the values of a statistic for different samples of same size.

Question 9.

What is an ‘Estimator’ ?

Answer:

The Statistic used to estimate the population parameter is an ‘Estimator’.

Question 10.

Which type of variations cannot be detected by Statistical Quality Control (SQC) ?

Answer:

‘Chance causes’ of variation.

Question 11.

What is a ‘Saddle Point’ in a game ?

Answer:

It is the position where maximin – minimax coincide.

Question 12.

What is ‘Carrying cost’ in an inventory ?

Answer:

If is the cost associated with carrying or holding the goods in stock until the goods are sold or used. Denoted by ‘C_{1}‘.

Section – B

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

Question 13.

In a village 2400 live births occurred in a year. The number of infant deaths was 60. Calculate infant mortality rate.

Answer:

Question 14.

Write any two considerations for the selection of base year in the construction of an index number.

Answer:

- It should be economically stable.
- It should not be too long or too short in distance from the current year.

Question 15.

If Σp_{0}q_{0} = 250, Σp_{1}q_{1} = 310, Σp_{1}q_{0} = 210 and ΣP_{0}q_{1} = 250, find P_{01}^{(DB)}

Answer:

Question 16.

Write any two demerits of least square method.

Answer:

- The procedure of calculation is difficult.
- If one of the value is added or removed the whole proce4ure has to be repeated.

Question 17.

Define ‘Inerpolatioh’ and ‘Extrapolation’

Answer:

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

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

Question 18.

If q = 0.4 for a Bernoulli distribution, find mean and variance of the distribution.

Answer:

If q = 0.4 .-. p = 1 – q = 1 – 0.4 = 0.6. Mean = p = 0.6 and Variance = pq = 0.6 × 0.4 = 6.24

Question 19.

Under what conditions does the Hypergeometric distribution tends to Binomial distribution?

Answer:

Hypergeometric distribution tends to Binomial distribution when

- a + b is very large i.e. (a + b) → ∞ and
- \(\frac{a}{a+b}\) = p

Question 20.

A random sample of size 36 is drawn from a population whose standard deviation is 4. Compute standard error of the sample mean.

Answer:

Given n = 36, 6 = 4 S.E (x̄) = \(\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{36}}=\frac{4}{6}\) = 0.67

Question 21.

Define ‘Type-I Error’ and ‘Type-II Error’.

Answer:

‘Type-I error’ is taking a wrong decision to reject the null – hypothesis when it is actually true.

‘Type-II error’ is taking a wrong decision to accept the null hypothesis when it is actually not true.

Question 22.

Mention any two advantages of ‘Acceptance Sampling Plan’.

Answer:

Advantages of Acceptance Sampling are :

- This method is more suitable when the items are of destructive in nature.
- It is less expensive, requires few staffs as compared to 100 inspection.

Question 23.

Given R = 1800/year, C_{3} = 300, C_{1} = Rs.2/Unit/year, find the optimum lot size of an inventory.

Answer:

Question 24.

Using Maximin-Minimax principle, find the value of the following game

Answer:

By maximin – minimax principle.

1. Maximin (x) – Minimax payoff in each row is circled

2. Minimax – Maximin payoff in each colümn is boxed.

3.Saddle point is at (1,3): ∴ V = 3 is the value of the game.

Section – C

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

Question 25.

For the following data compute:

(i) General Fertillity Rate.

(ii) ASFR. for the age groups 15 – 19 years and 25 – 39 years.

Answer:

= \(\frac{1955}{49100}\) × 100 = 39.82 children born per 1000 women of (15 – 49).years in the year.

ASFR (15 – 19) = \(\frac{400}{12000}\) × 1000 = 33.33

ASFR (25 – 39) = \(\frac{(510+300)}{(9400+9000)}\) × 1000 = \(\frac{810}{18400}\) × 1000 = 44 02

Question 26.

Calculate ‘Simple geometric mean price index number’ for the following data :

Answer:

Question 27.

Explain the steps involved in the construction of ‘Consumer price index number’.

Answer:

1. Object and scope : The purpose/object is to decide in which the index number bring constructed, as to Farmers, Govt employees, merchants etc, and he scope ie, geographical area, income group, the time limit for the completion of construction of index number should be decided.

2. Conducting family budget survey: A detailed survey regarding the average expenditure of the families, on various items of consumption, quality grade of items under the heads of

- Food.
- Clothing.
- Housing.
- Fuel and inght.
- Medicine, Educatcon entertaiment and
- Miscellaneous are conducted.

3. Obtaining the price Quotation : The price list of various items/commodities included in the index number of current year and base year are obtained from different agencies such as from marketing inspetors, rent control officers, super bazars, Janatha bazars or public distribution system of retail prices are obtained or averaging the prices of various shops in the respective area.

4. Metheas of construction : Decision had to be made on the basis of purpose, accuracy needed the availability of data, among

- Aggregative expenditure method.
- Family budget method.

Question 28.

Compute five yearly moving averages for the following data.

Answer:

5YMT – 5 yearly moving total; 5YMA – 5 yearly moving average.

Question 29.

‘Interpolate’ and ‘Extrapolate’ the sales for the years 2013 ang 2016 for the following data. ‘

Answer:

Let X and Y be the year and sales (000’s)

Here the number of known values of Y : n = 4

The binomical expansion (y – 1)^{4} = 0

ie, (y – 1)^{4} = y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0 → 1.

and increase the suffixes of y by equation 1 in

We get y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0 → 2.

From equation by substing values of y’s in (1)

we get 54 – 4(40) + 6y_{2} – 4(18) +10 = 0.

54 – 160 + 6y_{2} – 72 + 10 =

∴ 6y_{2} + 68 = 0 ; y_{2} = \(\frac{168}{6}\) = 28(000’s)

Similarlly from (2) by substituting values of ‘y’

we get y_{5} – 4(54) + 6(40) – 4(28) + 18 = 0

y_{5} – 216 + 240 – 112 + 18 = 0

y_{5} – 70 = 0 ∴ y_{5} = 70 (000’s)

Question 30.

The probability of a thermometer manufactured by a firm found to be defective is 0.02. Find the probability that a box containing 50 thermometers contain no defective ones. Among 1000 such boxes, how many contain exactly 2 defective thermometers ?

Answer:

Let x be the number of defective thermometers is a Binomnal variate with the parameters n = 50, p = 0.02. Here ‘n’ is large and ‘p’ is very small under these conditions tends to poission distribution with mean = λ = np = 50 × 0.02 = 1, Then the p.m.f is

p(x) = \(\frac{e^{-\lambda} \lambda^{x}}{x !}\) : x = 0, 1,2 oo

p(x) = \(\frac{e^{-1} 1^{x}}{x !}\) : x = 0, 1;2 v

P(contain no defective ones) = p(x = 0) = \(\frac{\mathrm{e}^{-1} 1^{0}}{0 !}\) = e^{-1} = 0.3679

No. of boxes contain exactly 2 defective ther mometers = P(x = 2) × N(=1000)

= \(\frac{\mathrm{e}^{-1} 1^{2}}{2 !}\) × 1000 = \(\frac{0.3679}{2}\) × 1000 = 183.95 = 184

Question 31.

Mean and variance of normal distribution are 20 and 9 respectively. Find Q_{1} , Q_{2} and Q_{3}

Answer:

Mean = µ = 20, variance = σ^{2} = 9. ∴ σ = 3

We know that: for a normal distribution: Q_{1} = µ – 0.6745 σ =20 – (0.6745 × 3) =17.9765 = 18

Q_{3} = µ + 0.6745 σ = 20 + (0.6745 × 3) = 22.02 = 22

And Q_{2}/M = Median = mode = 20

Question 32.

From the following data test whether there is any significant difference in I.Q of boys and I.Q. of girls at α = 5%.

Answer:

Given : n_{1} = 40, n_{2} = 36, x̄_{1} = 125, x̄_{2} = 128s s_{1} = 5 and s_{2} = 3.

H_{0}: There is no significant difference in I.Q. of boy and girls (ie., µ_{1} = µ_{2})

H_{1}: There is a Significant difference in I.Q. of boys and girls (ie., µ_{1} ≠ µ_{2}) {Two tailed test}

At α = 5% the two tail critical values are ± k = ±1.96

Here Z_{cal} lies in rejection region.

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

Conclusion: There is a Significant difference

I.Q. of boys and girls (ie., µ_{1} ≠ µ_{2})

Question 33.

To test the effectiveness of vaccination against Tuberculosis, the following table was obtained.

Test at 1% level of significance whether vaccination and attack of Tuberculosis are independent.

Answer:

χ^{2} – Test for attributes :

H_{0}: Vaccination and attack of T.B. are independent.

H_{1}: Vaccination and attack of T.B. are dependent {uppertail test}

The given data can be written in a 2 × 2 – contingency table as below:

under H_{0}, the χ^{2} – test statistic is :

At α = 1% for 1.d.f the upper tail critical value k_{2} = 6.63

Here χ^{2}cal lies in rejection region

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

Conclusion: Vaccination and attack of T.B are dependent.

Question 34.

Calculate control limits for R-chart for the following data given n = 5.

Answer:

The control kmits are : C.L = R̄ = 5; L.C.L = D_{3} R̄ = 0 × 5 = 0; U.C.L = D_{4} R̄ = 2. 115 × 5 = 10.575.

Here table constants for n = 5; D_{3} = 0, D_{4} = 2.115

Question 35.

A manufacture produces 2 products – A and B which needs two machiness M_{1} and M_{2}. Product A requires 10 hours on machine M_{1} and 4 hours on machine M_{2}. product B requires 7 hours on machine M_{1} and 5 hours on machine M_{2}. There are 250 hours of time available on machine M_{1} and 300 hours on machine M_{2}. Profit earned by the manufacture on selling one unit of A is Rs 50 and on selling one unit of B is Rs 45. Formulate LPP.

Answer:

The given data can be written in the following table as below:

From the above table; the L.RP is : Maximize Z = 50x + 45y

Subject to constraints, 10x + 7y ≤ 250

4x + 5y ≤ 300 and

Non – negativity restrictions x ≥ 0, y ≥ 0.

Question 36.

The Purchase cost of a vehicle is Rs 50,000 and its resale value is Rs 15,000. The ‘maintenance costs in different years are as follows :

Find the optimal replacement period of the vehicle.

Answer:

Given: p = 50,000 S_{n} =15,000, Depreiotion: (P – S_{n}) = (50,000 -15,000) = 35,000.

From the above, the annual average Maintenance cost is minimum for n = 4, is the optimum period for replacement of the vehicle.

Section – D

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

Question 37.

From the following data, compute the standardized death rates for locality A and locality B and comment.

Answer:

Since locality B is takes as standand so compute CDR for locality B and S.D.R. for locality A.

= \(\frac{370}{32,000}\) \frac{370}{32,000} 1000 = 11.5625 Deaths/1000 population/year.

For locality A : S.D.R = \(\frac{\sum \mathrm{P}_{\mathrm{s}} \mathrm{A}}{\sum \mathrm{P}_{\mathrm{s}}}\) ; where A – ASDR of locality A and P_{s} – be the standard population (of locality B)

ASDR (0 – 4) = \(\frac{120}{6000}\) × 1000 =20, similarity A.S.D.Rs can be calculated for other age groups.

S.D.R (A) = \(\frac{5,28,000}{32,000}\) = 16.5 deaths/1000 population/year.

Here SD.R (B.) < S.D.R(A) . So, locality B is more healtheir than locality A.

Question 38.

From the following data, compute

i) Laspeyr’s quantity index number

ii) Show that Fisher’s index number satisfies Factor Reversal Test (FRT).

Answer:

Question 39.

Fit an equation of the type Y = a + bx for the following data and obtain trend values. Estimate the profit for the year 2016.

Answer:

Let X and Y be the year and Profit.

Here x̄ = 2012

From the normal equation : na + bΣx = Σy , since Σx = 0, ∴ a = \(\frac{\sum y}{n}=\frac{376}{5}\) = 75.2

and from aΣx + b Σx^{2} = Σxy; b = \(\frac{\sum x y}{\sum x^{2}}=\frac{95}{10}\) = 9.5

The fitted stranght line frend eqnation is: y = a + bx; ŷ = 75.2 + 95 x .

Trend values: ŷ (2010) = 75.2 + 9.5(-2) = 56.2

ŷ(2011) = 75.2 + 9.5(-1) = 65.7; ŷ(2012) = 75.2 + 9.5(0) = 75.2

ŷ (2013) = 75.2 + 9.5(1) = 84.7, ŷ (2014) = 75.2 + 9.5(2) = 94.2

Estimation for 2016: ŷ (2016) = 75.2 + 9.5 (4) = 75.2 + 36 = Rs. 113.2 lakh profit.

Question 40.

Fit a Binomial distribution for the following data and test at 5% level of significance – that Binomial is a good fit.

Answer:

Let x be the number of defective items is a Binomial variate with the n = 5, and ‘P’ is obtained as below: let be the no. of samples.

Mean x̄ = np = \(\frac{\sum_{\mathrm{fx}}}{\mathrm{N}}\) ; 5p = \(\frac{300}{100}\) = 3; p = 0.6 and q = 1 – p = 1 – 0.6 = 0.4

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

p(x) = 5C_{x} (0.6) (0.4) 5 – x, x = 0,1,2 …… 5.

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

To = P(x = 0) 100 = 5C_{0} (0.6)^{0} (0.4)^{5-0} × 100 = 1.024 = 1

χ^{2} TEST:

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

H_{1} : Binomial distribtion is not good fit ie., O_{i} ≠ E_{i} {upper tail test}.

Here ‘P’ is estimated from the data and so the degrees of freedom will be (n – 1 – 1) = (n – 2)

∴ χ^{2}_{cal} = 9.743

At α = 5% for (n – 2) = 5 – 2 = 3. d.f the upper tail critical value K_{2} = 7.81

Here χ^{2} and H_{1} is accepted.

Conclusion: Bin0mial distribution is not a good fit (O_{i} ≠ E_{i})

Section – E

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

Question 41.

Heights of PU students is normally distributed with mean 155 cm. and standard deviation 5 cm. find the

i) Probability that a randomly selected PU student has height more than 155 cm.

ii) Percentage of PU students having height between 150 am and 160 am.

Answer:

Let x be the heights is a Normal variable with mean ‘ μ ’ = 155 = 500 and S.D ‘ σ ’ = 5, then the S.N.V z = \(\frac{x-\mu}{\sigma}=\frac{X-155}{5}\) ~ N(0,1)

(i) p(x > 55) = p\(\left(\mathrm{Z}>\frac{155-155}{5}\right)\) = p(z > 0)

= Area from 0 to ∞ = 0.5

(ii) p(150<X< 160)

= p\(\left(\frac{150+155}{5}<Z<\frac{160-155}{5}\right)\)

= P (-1 < Z < 1)

= Area from (-1) to‘1’ =0.8413 – 0.1587 = 0.6826.

The required percentage = 0.6826 × 100 = 68.26%.

Question 42.

A machine produced 6 defective articles among 50 articles. Test whether the proportion of defective articles is less than 10% (Use α = 5%).

Answer:

Given : (for Test for proportion) x = 6, n = 50

∴ The sample proportion: P = \(\frac{x}{n}=\frac{6}{50}\) = 0.12

Population proportion : P_{0} = 10% = 0.1 and α = 5%

H_{0} : proportion of defective articles is 10% (ie., p_{0} = 0.1)

H_{1} : proportion of defective articles is less than 10% (ie., p_{0} < 0.1) {Lower tailed test}

At α = 5% the Lower tail critical

values are -k = -1.65

Here Z_{cal} lies in acceptence region ∴ H_{0} is Accepted.

Conclusion : Proportion of defective articles is 10% (ie, P_{0} = 0.1)

Question 43.

Following data given the readings of sugar level of 5 diabetic patients before and after taking insulin.

Using 1% level of significance, test whether insulin has reduced sugur level in patients.

Answer:

Given: (for paired test) : n = 5, let x and y be the sugar level before and after taking insulin.

H_{0}: Mean sugar level is same before and after taking insulin (ie., µ_{1} = µ_{2})

H_{1}: Mean sugar level is less/redued after taking insulin (µ_{1} > µ_{2}) {upper tail test}

Under H0, the test statistic is; t = \(\frac{\overline{\mathrm{d}}}{\mathrm{S} / \sqrt{\mathrm{n}-1}}\) ~ t (n-1) d.f.

At α = 1% for (n – 1) = (5 – 1) = 4. d.f the upper tail

Here t_{cal} lies in rejection region

∴ H_{0} is accepted.

Conclusion : Sugar level is same before and after taking insulin (µ_{1} = µ_{2})

Question 44.

For the following transportation problem. obtain basic feasible solution using North – West Corner rule. Also find transportation cost.

Answer:

Here Σai = Σbj = 650 T.P is balanced.

By N.W.C.R first allocation is made at (1,1) cell as : X_{11} = m,n (200, 100) = 100.

Similarly all allocations can be made. The suggested Initial basic feasible solution (I.B.F.S) and total transportation cost is: