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## Karnataka 2nd PUC Statistics Model Question Paper 2 with Answers

Time: 3 Hours 15 Minutes

Max. Marks: 100

Section – A

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

Question 1.

Mention a source of vital statistics.

Answer:

Registration or census method

Question 2.

What is price relative ?

Answer:

Price relative is the price in the current year expressed as the percentage of the price in the base year.

ie„ p = \(\frac{p_{1}}{p_{0}}\) × 100

Question 3.

State the relation between Lapeyere’s, Paasche’s and Fishers index nunbers.

Ansewr:

p_{01} (F) = \(\sqrt{\mathrm{p}_{01}(\mathrm{L}) \times \mathrm{p}_{01}(\mathrm{p})}\) ie., Fisher’s index number is the geometric mean of Laspeyre’s and Passche’s price index numbers.

Question 4.

Define seasonal variation.

Answer:

Seasonal variations are the variations occurring annually due to the seasons of the year. But also includes the variations of any kind which are periodic in nature and whose period is shorter than one year (t <) year).

Question 5.

Write down the probability mass function of a Bernoulli distribution.

Answer:

The p.m.f is: p(x) = p^{x} q^{1 – x} ; x = 0,1 ,0 < p < 1 and x is a bernoulli variate.

Question 6.

If mean of a poisson distribution is 4, find its S.D.

Answer:

If Mean = λ = 4 variance; SD (x) = \(\sqrt{\text { variance }}=\sqrt{\lambda}=\sqrt{4}=2\)

Question 7.

What is parameter space ?

Answer:

The set of all admissible values of the population parameters is called parameters space.

Question 8.

Define Acceptance Region?

Answer:

The set of all admissible values of the test statistic, which leads to the acceptance of the null hypothesis is called acceptance region (A.R.)

Question 9.

If P = 0.5 and n = 100 then find S.E (P)

Answer:

Question 10.

Mention one advantage of statistical Quality control.

Answer:

SQC improves the quality of goods and decreases the proportion of defectives.

Question 11.

Define Optimal Solution of an L.P.P.

Answer:

If a feasible solution which optimises the objectives function Z = CX is called optimal solution.

Question 12.

What is meant by a fair game?

Answer:

If the value of the game V = 0 the game is called fair.

Section – B

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

Question 13.

Write two uses of life table.

Answer:

- Life tables are used in computation of acturials such as bonus, premium, etc., of insurance policies by insurance agencies.
- They are used to study the population growth and forecast the size and sex distribution of the population.

Question 14.

If ΣP_{1}q_{o} = 375 and ΣP_{0}q_{o} = 300, calculate suitable index number.

Answer:

Question 15.

Mention the steps involved in the constrction of cost of living index number.

Answer:

- Object and the scope.
- Conducting a family budget survey
- Obtaining the price Quatation
- Methods of construction.

Question 16.

What is irregular variation ? Give an example.

Answer:

Irregular variations are those changes of time series, which are irregular in nature and do not show any pattern. The causes of irregular variations are due to accidental happenings such as wars, earth quacks, floods, famine, fire accidents, strikes etc., are the examples.

Question 17.

Expand the Binomial expansion (y – 1)^{5}

Answer:

(y – 1)^{5} = y_{5} – 5y_{4} + 10y_{3} – 10y_{2} + 5y_{1} – y_{0} = 0.

Question 18.

Under what conditions binomial distribution tends to normal distribution?

Answer:

When (i) ‘n’ is large ie., n → ∞

(ii) Neither p nor q are very small ie., p = q with Mean = np = µ and S.D : σ = \(\sqrt{\mathrm{npq}}\)

Question 19.

The first two frequency terms of a poisson distribution are 150 and 180 respectively, find the next frequency term.

Answer:

Given : Frequency term : T_{x} = p(x). N, ie.,T_{0} =150 and T_{1} = 180.

Usmg recurrence relation: T_{x} = \(\frac{\lambda}{x}\)T_{x-1}; T_{1} = \(\frac{\lambda}{1}\)T_{0}; 180 = λ × 150

∴ λ = \(\frac{180}{150}\) = 1.2; The next frequency term : T_{2} = \(\frac{1.2}{2}\) × 180 = 180.

Question 20.

What are 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 true.

Question 21.

What are the mean and variance of a chi – square distribution with n degrees of freedom ?

Answer:

Mean = n; Variance = 2n.

Question 22.

Mention any two types of causes of variation in a manufacturing process.

Answer:

- Chance causes and
- Assignable causes.

Question 23.

In a T.P define feasible solution.

Answer:

A set of values x_{ij} where i = 1,2,… .m and j = 1,2 …n which satisfies the following conditions

Question 24.

The following is the pay – off matrix of player – A. write the payoff matrix of player – B.

Answer:

The pay – off matrix of player B is:

Section – C

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

Question 25.

For the following data find TFR.

Answer:

Total fertility rate: T.F.R = 5 Σ Quinquennial ASFRs

ASFR (15 – 49) = \(\frac{200}{5,000}\) × 1000 = 40; similarity compute for other age groups.

T.F.R = 5 × 466.5 = 2332.5 children bom per 1000 women of (15 – 49) years.

Question 26.

Mention three uses and two limitations of Index number.

Answer:

Uses :

- Index numbers are useful to governments in formulating Economic policies such as Taxation, Imports, Exports, fixation of bank rates.
- Fixation of salary and grant of dearess allowances to employees.
- Comparing variations in prices, production, shares etc.

Limitations:

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

Question 27.

Compute consumer price Index number from the following data.

Answer:

Weights-’w are given, use Family budget Method: C.P.I : = I_{01} = \(\frac{\Sigma P w}{\Sigma w}\) = \(\frac{p_{1}}{p_{0}}\) × 100

I01 = \(\frac{27633.1}{200}\) = 138.165

The cost of living has increased by 38.165% in the current year as compared to base year.

Question 28.

For the following data find three yearly moving averages and indicate the trend.

Answer:

Let X and Y be the year and profit in the table 3Y MT : 3 yearly moving total

From the ŷ – trend values, the profit shows upward trend.

Question 29.

Use Newton’s method to find the number of employees whose wages is Rs. 400 day.

Solution :

Let x and y be the wages and number of employees. The number of known values of ‘y’ n = 5, so prepare the leading differences (Δs) upto Δ^{4}.

(x – 1) = (1.5 – 1) = 0.5; (x – 2) = (1.5 – 2) = -0.5 ; (x – 3) = (1.5 – 3) = -1.5

The Newton’s formula of interpolation is

y_{400} = 72 – 15 – 1.5 – 0.875 – 0.6562 = 54.84 = 55

∴ Number of employees whose wages is Rs. 400/-day = 55

Question 30.

The probability of an arrow hitting a tree is \(\frac { 1 }{ 2 }\) . If 5 arrows atfe aimed at the tree find the probability that i) 3 arrows hit the tree ii) at least 4 arrows hit the tree

Answer:

Given: Let x be the number of arrows hitting a tree is k Binomial variate with the parameters n = 5; p = \(\frac { 1 }{ 2 }\) = 0. 5 and q = 1 – p = 0.5. Then the p.m.f is

p(x) = ^{n}c_{x} p^{x} q^{n – x}; x = 0,1,2 …… n

= ^{5}c_{x} (0.5)^{x} (0.5)^{5 – x}; x = 0,1,2….5

(i) p ( 3 arrows hit) = p (x = 3) = ^{3}c_{3}(0.5)^{3} (0.5)^{5-3} = 1 × 0.125 × 0.25 = 0.03125

(ii) p (atleast 4 arrows hit) = p (x > 4) = p(x = 4) + p (x = 5) = ^{5}c_{4} (0.5)^{4} (0.5)^{5-4} + ^{5}c_{5} (0.5)^{5} (0.5^{5-5} = 5 × 0.0625 × 0.5 + 1 × 0.03125 × 1 = 0.15625 + 0.03125 = 0.1875

Question 31.

Write down five features of normal distribution.

Answer:

Features of Normal distribution are

- The normal curve is bell – shaped, symmetrical ie., Non – skew (Pj = 0) and so,

Mean = Median = mode. - The parameters are : Mean = µ, S.D = σ and variance = σ
^{2} - The curve is mesokurtic (β
_{2}= 3) - Area under the Normal curve = 1
- For the distribution : Q.D = \(\frac { 2 }{ 3 }\) σ; M.D = \(\frac { 4 }{ 5 }\)σ

Question 32.

The average monthly income of 50 families is found to be Rs. 44,950 can we Conclude that the average monthly income of the population is 45,000 with S.D Rs. 50? Test at 5% level of significance.

Answer:

Given : n = 50; x̄ = 44,950, µ = 45,000, σ = 50, α = 5%

H_{0 }: Mean monthly income is Rs 45,000 (ie., µ = 45,000)

H_{1} : Mean monthly income differs from Rs. Rs. 45,000 (ie., µ ≠ 45,000) (Two tailed test}

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

∴ Here Z_{cal} is rejected and H_{1} is accepted.

Conclusion : Mean monthly income differs from Rs. 45,000 (µ ≠ 45,000)

Question 33.

From the following 2 × 2 contingency table, test whether result and areas of the students are independent ? [use 1% L.O.S].

Answer:

H_{0} : Attributes result and areas of the students are independent.

H_{1} : Attributes result and areas of the students are dependent, [upper tail test]

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

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

χ^{2}_{cal} = \(\frac{200(62 \times 61-28 \times 49)^{2}}{90 \times 110 \times 111 \times 89}=\frac{200(2410)^{2}}{97802100}\) = 11.88

At α = 1% for 1.d.f. the upper tail critical region K_{2} = 6.65

Here χ^{2}_{cal} > k_{2} / χ^{2}_{cal} lies in rejection region (R.R)

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

Conclusion : Result and areas of the students are dependent.

Question 34.

The following table gives the number of defectives found during the inspection of 8 samples of size 50 each. Calculate control limits for np chart.

Answer:

Given : No. of defectives ae given, so use ‘d’ or np – chart. Standard is not known. So,

we need p̄ = \(\frac{\Sigma \mathrm{d}}{\mathrm{nk}}\)

n – sample size = 50 ; k – Sample no. = 8.

∴ p̄ = \(\frac{12}{50 \times 8}\) = 0.03. ∴ q̄ = 1 – p̄ = 1 – 0.03 = 0.97

The control limits are: C.L = np = 50 × 0.03 = 1.5

Question 35.

Solve the foUowing L.P.P graphically:

Maximize z = 5x + 4y

Subject to 3x + 2y ≥ 30

2x + 3y ≤ 60

and x, y ≥ 0

Answer:

Consider the constraints as equalities and obtain the co – ordinates to get a straight line on x,y – plane as below

Consider, 3x + 2y ≥ 30 as

3x + 2y = 30; put x = 0 ; we get y = 15; (0,15) as co – ordinate

put y = 0, weget x = 10; (10,0)

and from 2x + 3y = 60 ;

put x = 0; we get y = 20 ; (0,20)

put y = 0; we get x = 30 ; (30,0)

Also : the non – negative restrictions as x = 0 ; y = 0.

Now plot the points on an x,y – plane.

From the graph the feasible region / common shaded region exists at the comer points A(0,15) B(0,20), C(30,0) and D( 10,0)

Determination of optimum value of ‘z’- objective function at the comer points:

From the above table the objectivity Maximises at the comer point C(30,0).

The suggested solution is : x = 30, y = 0.

The optimum value of Z_{c} = 150.

For visually challenged students

Following are the steps involved in graphical method of solving an L.P.P:-

- Consider the constraints as equalities
- Find the co – ordinates for each constraints as (x, y)
- Draw a graph and represent by straight line for each constraints
- Identify the feasible region – Which satisfies the constraints and Non – negative restrictions simultaneaasly (The common shaded region)
- Locate the corner points of the feasible region
- Find the value of the objective function for each of the comer points of the feasible region
- Suggest the solution to the L.P.P with its optimum value.

Question 36.

A firm is considering the replacement of machine whose cost price is Rs. 12,800 and scrap value is Rs. 300. The‘maintenance costs are as followes.

Determine the optimal period for replacement of the machine.

Answer:

Given : p = 12, 800, Sn = 300, ∴ the depreciation cost (P – S_{n}) = (12,800 – 300) = 12, 500 is fixed for all the years.

The given data is represented in the following table:

From the above table Annual average maintenance cost A(n) is minimum for n = 6^{th} year. So

the optimum period for the replacement of the machine is 6^{th} year. The optimum cost:

Section – D

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

Question 37.

Calculate standardardized Death Rates from the following data and write your conclusion

Answer:

Let Ps – the standards population, A, B – be A.S.D.R.S of city A and city B.

For cityA: ASDR (Below 10) = \(\frac{32}{4,000}\) × 1000 = 8

For city B: ASDR (Below 10) = \(\frac{120}{12,000}\) × 1000 = 10,

Similarlly ASDRs can be calculate for other age groups.

S.D.R (A) = \(\frac{\Sigma p_{s} A}{\Sigma p_{s}}=\frac{226,000}{26,000}\) = 8.6,923

S.D.R (B) = \(\frac{\Sigma p_{s} B}{\Sigma p_{s}}=\frac{247,000}{26,000}\) = 9.5 deaths/1000 population in a year.

Conclusion : Here SDR(A) < SDR (B)

City A is more healthier than city B.

Question 38.

Construct Marshall- Edgeworth’s, Fisher, and Dorbish Bowley’s price index numbers from the following data.

Answer:

Question 39.

Fit a quadratic trend for the following time series by the method least squares and estimate the sales in 2008.

Answer:

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

From the normal equations obtain the values of a, b and c as below : Here n = 7 years,

na + bΣx + cΣx^{2} = Σy; 7a + b.0 + c.28 = 64

ie., 7a + 28c = 64 → (1)

a Σx + b Σx^{2} + c Σx^{3} = Σxy; a.0 + b. 28 + c.0 = 46

28b = 46; b = \(\frac{46}{28}\) = 1.6429 28

a Σx^{2} + b Σx^{3} + c Σx^{4} = Σx^{2}y; a.28 + b.0 + c.196 = 262,

we get, 28a + 196 c = 262 → (2)

from (1) & (2)

∴ c = \(\frac{6}{84}\) = 0.0714;

put c = 0.0714 in (1) 7a + 28(0.0714) = 64

7a + 1.9992 = 64; a = \(\frac{64-1.9992}{2}\) = 31

The fitted quadratic trend equation is : y = a + bx + cx^{2}; ŷ = 31 + 1.6429 x + 0.0714 x^{2}

Estimation for the year 2008; x = 5; x^{2} = 25

ŷ(2008) = 31 + 1.6429 (5) + 0.0714(25) = 40.9995 = 41(000’s)

Question 40.

Fit a binomial distribution to the following data and test at 5% level of significance that whether the distribution is good fit.

Answer:

The parameters are n = 5, and p is obtained as: Mean = np = x̄ = \(\frac{\Sigma \mathrm{fx}}{\mathrm{N}}\) from the given distribution ∴ 5p = \(\frac{0 \times 10+1 \times 29+2 \times 55+3 \times 71+4 \times 28+5 \times 7}{10+29+55+71+28+7}\)

5p = \(\frac{499}{200}\) ∴ p = \(\frac{2.495}{5}\) = 0.499 = 0.5 and q = 1 – p = 0.5 and N = 200

The p.m.f is ; p(x) = ^{n}c_{x} p^{x} q^{n – x}, x = 0,1,2 …… n; 5(x 0.5x 0.5 5-x; x = 0,1,2 …….. 5)

∴ T_{0} = 0(x = 0) × 200 = ^{5}c_{0} 0.5^{0} 0.5^{5 – 0} × 200 = 0.03125 × 200 = 6.25

The successive theoretical frequencies are obtained using the recurrence :

The observed and theoretical frequency distribution is: – (approximated)

CHI – SQUARE TEST;

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

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

∴ χ^{2}_{cal} = 5.629

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

Here χ^{2}_{cal} < k_{2} / χ^{2}_{cal} lies in Acceptance region (A.R.)

∴ H_{o} is accepted

Conclusion : Binomial distributon is good gift

Section – E

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

Question 41.

The heights of a group of soldiers are Normally distributed with mean 170 cms and S.D. of 4 cms. Then Rnd the probability that a randomly selected soldier has height (i) More than 180 cms (ii) between 162cms and 172cms.

Answer:

Let x be the height is a Normal variate with the parameters µ = 110 and σ = 4, then the S.N.V is: z = \(\frac{x-\mu}{\sigma}\) ~ N(0,1) : z = \(\frac{x-170}{4}\)

(i) p (more than 180) = p(x >180) = p\(\left(z>\frac{180-170}{4}\right)\) = p(Z > 2.5)

= Area from 2. 5 to ∞

p(z > 2.5) = 0.0062 (From the table)

(ii) p(between 162 and 172) p (162 < × < 172)

= p\(\left(\frac{162-170}{4}<z<\frac{172-170}{4}\right)\) = P(-2 < Z < 0.5)

= Area from (-2) to 0.5

= Area from (-2) to ∞ – Area from 0.5 to ∞

= 0.9772 – 0.3085 = 0.6687

Question 42.

For the following data test whether there is any significant difference between the proportions of two populations? (use – k = -196, k = 1.96)

Answer:

Given : p_{1} = 0. 26, p_{2} = 0.2; n_{1} = 1000, n_{2} = 500.

H_{0} : There is no significant different between the proportions the two populations (ie., P_{01} = P_{02})

H_{1}: There is a significant difference between the proportions of the two populations (ie., P_{01} ≠ P_{02}) { Two tail test}

Q_{0} = 1 – p_{0} = 1 – 0.24 = 0.76

Given : k = -1.96, k= 1.96

Here Z_{cal} lies in rejection region (R.R) ie., Z_{cal} > 1.96 (k). ∴ H_{0} is rejected and H_{1} is accepted

Conclusion: There is a significant difference between the proportions of the two populations. (ie., P_{01} ≠ P_{02}).

Question 43.

A sample of 15 values shows that S.D to be 6.4 Does this agree with the hypothesis that the population S.D is more than 5 (α = 0.01)

Answer:

Given: n = 15; s = 6.4, σ = 5, α = 0.01 (1%)

H_{0} : population S.D is 5 (ie., σ = 5)

H_{1} : population S.D is more than 5 (ie., σ > 5) { upper tail tfest}

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} = \(\frac{15 \times 6.4^{2}}{5^{2}}\) = 24.576

At α = 1% for (n – 1) = (15 – 1) = 14 df

The upper tail critical value k_{2} = 29. 14

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

∴ H_{0} is accepted.

Conclusion : population S.D is (ie., σ = 5).

Question 44.

There is a demand for 10,000 items per year. The setup cost is Rs.1600 and Maintanance cost is Rs. 4 per item per year. Shortages are not allowed. Find

(i) E.O.Q

(ii) Number of orders per year

(iii) Minimum average cost.

Answer:

Given: R = 10,000/year, C_{3} = 1600, C_{1} = 4/year, C_{2} = shortages not allowed, use EOQ model I

(ii) No. of orders : n° = \(\frac{1}{t^{0}}=\frac{R}{Q^{0}}=\frac{10,000}{2828.4271}\) = 3.535 orders / year

(iii) Min. Average cost: C (Q°) = \(\sqrt{2 \mathrm{C}_{3} \mathrm{RC}_{1}}=\sqrt{2 \times 1600 \times 10,000 \times 4}\) = Rs. 11313.71