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## Karnataka 2nd PUC Statistics Previous Year Question Paper March 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 expectation of life.

Answer:

It is the average number of years that a person aged ‘x’ can expected to live subject to mortality conditions.

Question 2.

If the general price level goes up by 80% between 2000 and 2012. What is the index number for 2012 with base 2000?

Answer:

180.

Question 3.

Name the index number which satisfies circular test.

Answer:

Kelly’s and simple geometric mean methods.

Question 4.

Which index is used for the measurement of seasonal variation?

Answer:

Seasonal indices method

Question 5.

If p = \(\frac { 1 }{ 4 }\) for a Bernoulli distribution, write down the p.m.f.

Answer:

Question 6.

What is the probability that a normal variate takes a value greater than its mean ?

Answer:

p (x > µ) = 0.5

Question 7.

Define Sampling distribution of a satistic.

Answer:

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

Question 8.

What is the power of a test ?

Answer:

power of a test is the probability of rejecting H_{0} when it is not true, denoted by (1 – β) ; where β = p ( Type II error).

Question 9.

If p = 0.02 and n = 64 then find SE (p).

Answer:

S.E (p) = \(\sqrt{\frac{P Q}{n}}\) = \(\sqrt{\frac{0.02 \times 00.98}{64}}\) = 0.0175 : Q = 1 – p = 1 – 0.02 = 0.98

Question 10.

Define acceptance sampling.

Answer:

The process of inspecting a random sample of goods and deciding whether to accept the lot based on the result.

Question 11.

What is meant by lead time ?

Answer:

It is the time between the placing the order and the arrival of goods at the Inventory [T].

Question 12.

When a T.P. balanced?

Answer:

If Σa_{i} = Σb_{j} T.P is balanced

Section – B

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

Question 13.

In a locality 10,000 live births occurred. The number of infant deaths was 500. Calculate infant mortality rate.

Answer:

Question 14.

State two characteristics of index numbers.

Answer:

- Index numbers are specialised averages
- They are expressed in ratios, for better comparision expressed in percentages.

Question 15.

Quantity index number of current year is 250. If the number of units produced in the current year is 120, then find the number of units produced in the base year.

Answer:

Quantity relative : Q = \(\frac{\mathrm{q}_{1}}{\mathrm{q}_{0}}\) × 100; 250 = n × \(\frac{120}{\mathrm{q}_{0}}\); q_{0} = \(\frac{250}{120}\) × 100 = 48

Question 16.

Diagramatically represent ‘Business Cycle’ with stages.

Answer:

Question 17.

Define ‘Interpolation’ and ‘Extrapolation’.

Answer:

“Interpolation” is the method of estimating a value of the dependent variable (y) for any intermediate value of independent variable (x).

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

Question 18.

For a hypergeometric distribution the parameters are a = 7, b = 5 and n = 8. Find the variance.

Answer:

Question 19.

If z_{1} and z_{2} are two independent SNVs, then name the distribution of z^{2}_{1} + z^{2}_{2} and find its mean.

Answer:

It is a chi – square distribution with n = 2d.f. Mean = n =2

Question 20.

Write two applications of t-test.

Answer:

- To test whether the population has the given mean.
- To test whether the two populations means are equal.

Question 21.

Mention two conditions for applicability of Chi – square test of goodness of fit.

Answer:

- Total no of frequency N should be large.
- Expected frequencies : E
_{i}≥ 5 , if any one is below 5, it should be pooled with the adjacent frequencies.

Question 22.

In a floor mat manufacturing company, the average number of defects per Square meter of floor mat is known to be 4. Find the upper control limit for the number of defeats.

Answer:

Given: λ^{1} = 4,U.C.L = λ^{1} + 3\(\sqrt{\lambda^{1}}\) = 4 + 3\(\sqrt{4}\) = 10

Question 23.

Under what conditions is E.O.Q. model with shortage applicable.

Answer:

- Uniform demand,
- Production is instantaneaus
- Lead time is zero
- Shortages are allowed.

Question 24.

From the following T.P. test whether the solution is non – degenerate:

Answer:

No allocations = 4 < m + n – 1 = 3 + 3 – 1 = 5 The solution ip degenerate.

Section – C

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

Question 25.

Compute the Gross reproduction rate from the following data:

Answer:

W.S.F.R (15 – 19) = \(\frac{19}{1658}\) × 1000 = 11.46 Similarlly for other age groups can be calculated.

∴ GR.R = 5 × 200.3 = 1001.5 Female births per 1000 women of (15 – 49) years.

Question 26.

Briefly explain TRT and FRT.

Answer:

Time Reversal test (T RT): It requires that the Index number computed time backward (P_{01}) is the reciprocal of the Index number computed time forward (P_{01}) except the factor 100.

ie., p_{01} = \(\frac{1}{\mathrm{p}_{10}}\) ; ie.,P_{01} × p_{10} = 1 ; Except 100

Factor Reversal Test (F.R.T) : It requires that the product of price I. No. and the Quantity I. No. both computed time backward is equal to the total change in the value.

i e., P_{01} × Q_{01} = \(\frac{\Sigma p_{1} q_{1}}{\Sigma p_{0} q_{0}}\) ⇒ v_{01}; Except 100

Question 27.

Compute the cost of living index number by using the following data:

Answer:

C.P.I = \(\frac{\Sigma p w}{\Sigma w}=\frac{3560}{27}\) = 131.85

Question 28.

Compute trend values by using 4 – yearly moving averages method:

Answer:

Question 29.

Interpolate and extrapolate the production for the year 2006 and 2010 for the following data:

Answer:

The number of known values of y : n = 4. Then the Binormal expension (y – 1)^{4} = 0. we get interpolation value.

y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{1} = 0 – (1)

ie„ 26 – 4(15) + 6(10) – 4y_{1} + 5 = 0

26 – 60 + 60 – 4y_{1 }+ 5 = 0 ∴ y_{1} = 7.75 Tons.

To get extrapolated value for the year 2010, the suffines of ‘y’ in equation (1) has increased by ‘1’

we get; y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0

i e., y_{5} – 4(26) + 6 (15) – 4(10) + 7.75 = 0

y_{5} – 104 + 90 – 40 + 7.75 = 0 ∴ y_{5} = 46. 25 Tons.

Question 30.

In a certain school 40% of the students have opted for first language kannada. Assuming 20 teachers take a sample of 4 students each, how many teachers will report that 2 or 3 students opted for first language kannada?

Answer:

x – No . of students opted first language as kannada is Binomial variate, n = 4, y = 20

p = 40 . 1 = \(\frac{40}{100}\) = 0.4

∴ q = 1 – p = 1.04 = 0.6

Then the Binomial distribution is ; p(x) = ^{n}C_{x} p^{x}q^{n-x}; x = 0,1,2 …….. n

= 4C_{x} 0.4^{x} 0.6^{4-x}; x = 0, 1,2, 3,4.

p(2 or three students opted kannada) = p(x = 2) + p(x = 3)

= 4C_{2} (0.4)^{2} (0.6)^{4-2} + 4C_{3} (0.4)^{3} (0.6)^{4-3} = 0.3456 + 0. 1536 = 0.4992

No of teachers who report 2 or 3 students opted first language kannada = 0.4992 × 20 = 10

Question 31.

In. a text book, on an average 0.3 mistakes per page is found. If there are 500 pages in that text book, in how many pages will there be at the most two mistakes?

Answer:

x ~ No of print mistakes is a poisson variate, λ = 0.3, N = 500. The poisson distribution is :

p ( at most 2 mistakes) = p(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

= e^{-0.3} (1 + 0.3 + 0.045) = 0.7408 × 1.345 = 0.9964

No of pages having at most 2 mistakes = 0.9964 × 500 = 498 pages.

Question 32.

A random sample of 400 tins of Vanaspathi has mean weight 4.96 kg and S.D of 0.4 kg. Test at 1% L.O.S. that the average weight of tins of Vanapathi is less than 5 kg.

Answer:

Given: n = 400; x̄ = 4.96, S = 0.4, α = 1.1, µ = 5. (n > 30 use z – test for mean)

H_{0} : Mean weight of tins is 5 kg (µ = 5kg) .

H_{1} : Mean weight of tins is less than 5kg (µ < 5kg) (Lower tail test)

under H_{0} the test statistic is : Z = \(\frac{\bar{x}-\mu}{^{s} / \sqrt{n}}\) ~ N(0,1)

Z_{cal} = \(\frac{4.96-5}{0.4 / \sqrt{400}}\) = -2

At α = 1% the lower tail critical value – k = -2. 33

Here Z_{cal} lies in A.R. ie., Z_{cal} > -k.

∴ H_{0} is accepted.

Conclusion: Mean weight of tins is 5kg

Question 33.

Following is the data regarding I.Q. of five students before and after performing yoga.

Is trainig effective? (Use 5% L.O.S).

Answer:

Before and after paired t – test.

H_{0} : Trainig is not effective (ie., µ_{1} = µ_{2})

H_{1} : Training is effective (ie, µ_{1} < µ_{2}) {Lower tail test}

Under H_{0}, the paired t – test statistic is

∴ d̄ = \(\frac{-24}{5}\) = -4.8

At α = 5% for (n – 1) = 5 -1 = 4.d.f the lower tail critical value – k = -2.13

Here t_{cal} < k / t_{cal} lies in rejection region

∴H_{0} is rejected H_{1} is accepted

Conclusion : Training is effective ( µ_{1} < µ_{2})

Question 34.

For the following data find out the control limits for x̄ – chart. (Given A_{2} = 0.577)

Answer:

Given : Here standards are not known, For x̄ – chart,

We need ; x̿ = \(\frac{\Sigma \bar{x}}{k}\) and R̄ = \(\frac{\Sigma R}{K}\) ;

Given: k = 6, Σx̄ = 297, ΣR = 61 ; x̿ = \(\frac{297}{6}\) = 49.5, R̄ = \(\frac{61}{6}\) = 10.16

The control limits for x̄ – chart are;

C.L = x̿ = 49.5

L. C. L = x̿ – A_{2}R̄ = 49.5 – 0.577 × 10.16 = 43.63

U .C.L = x̿ + A_{2}R̄ =49.5 + 0.577 × 10.16 = 55.36

Question 35.

Graphically solve the given L.P.P.

Maximise z = 12x + 40y

S.t. 3x + 3y ≤ 12

6.x + 3y ≤ 18

and x, y ≥ 0.

Answer:

Coordinates for each constraints are obtained as below:

3x + 3y = 12 : put x = 0 weget y = 4; (0,4) as (x,y)

put y = 0, x = 4, (4,0)

6x + 3y = 18, put x = 0, y = 6, (0,6)

put y = 0, x = 3, (3,0)

and Non- negativity; x = 0, y = 0

From the graph the feasible region exists at the comer points 0(0,0), A (0,4), B(2,2), C(3,0).

The optimum value for the corresponding comer points are obtained as below:-

From the above table L.P.P, maximise at the comer point A(0,4); The suggested solution for the L.P.P. is : x = 0 y = 4. The optimum value of Z_{A} = 160

(For Blind Students Only)

Finding a solution for the L.P.P. by graphical method :

- Consider the given constraints as equation (=)
- Find the co – ordinates for each constraints as (x,y) as x = 0, and y = 0, to get respective co-ordinates
- Plot the points on a graph and obtain straight line for each constraints
- Identify the feasible region, which satisties the constrains simultaneously ie., common region
- Locate the comer points of the feasible region
- Find the value of the objective function for each comer points of the feasible region.
- Suggest the solution as x, y and the optimum value.

Question 36.

The following are the maintenance and depreciation costs per year of a vehicle whose purchase, price is Rs. 50,000

When should the’ vehicle be replaced?

Answer:

Given: Depreciation cost: (P-Sn) and Maintenance cost: C_{i}P = 50, 000

From the above table the annual average maintaince cost is minimum for the 6^{th} year, the optimum cost: A (6) = 17,083.33

Section – D

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

Question 37.

For the following two towns compute standardized death rates and comment.

Answer:

Let A and B be ASDRs of Town A and B

For Town A: ASDR (0 – 20) = \(\frac{128}{8000}\) × 1000 = 16

For Town B: ASDR (0 – 20) = \(\frac{72}{4,000}\) × 1000 =18; similarlly all ASDRs can be calculated for

S.D.R (Town A) = \(\frac{\Sigma P_{s} A}{\Sigma P_{s}}\) = \(\frac{19,05,000}{100,000}\) = 19.05

S.D.R (Town B) = \(\frac{\Sigma P_{s} B}{\Sigma P_{s}}\) = \(\frac{16,75,000}{100,000}\) = 16‘75

Here S.D.R (B) < S.D.R (A)

Therefore Town B is healthier than town A.

Question 38.

Compute Marshall – Edgeworth’s and Dorbish – Bowley’s price index numbers from the following data:

Answer:

Question 39.

Population figures for a city are given below:

Fit a curve of the type y = ab^{x} and estimate the population for the year 2014.

Answer:

Let x and y be the years and the population.

x̄ = 2010. From the Normal equations:

n log a + log b. Σx = Σ logy. since Σx = 0

log a = \(\frac{\Sigma \log \mathbf{y}}{\mathbf{n}}\) = \(\frac{10.9802}{5}\) = 2.1960

∴ a = A.L(log a) = AL(2.1960) = 157.036

And from: log a.Σx + log b. Σx^{2} = Σx.logy

ie., logb = \(\frac{\Sigma x \cdot \log y}{\Sigma x^{2}}\) = \(\frac{0.3991}{10}\) = 0.03991

∴ b = A.L(log b) = AL (0.0399 1) = 1.0962

The fitted exponential trend equation is :

y = ab^{x}; ŷ = 157. 036(1.0962)^{x}

Estimation for the year 2014 : x = 4

ŷ(2014) = 157.036(1.0962)^{4} = 157.036 × 1.4440 = 226. 75(000’s)

Question 40.

Five unbiased coins are tossed loo times and the frequencies of number heads are noted below. Fit a Binomial distribution. Test at 5% L.O.S that it is a good fit.

Answer:

Answer:

Given: Let x be the no. of heads is a Binomial variate with n = 5, p = \(\frac { 1 }{ 2 }\) = 0.5 and q

1 – p = 1 – 0.5 = 0.5 . N = 1oo [Here the corn is unbiased so p = 0.5]

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

= 5C_{x}(0.5)^{x} (0.5)^{5-x} ;x = 0,1,2,3,4,5

Expected frequency: Ex = p(x) N

∴ To/Eo = p(x = 0)100 = 5C_{0} (0.5)^{0} (0.5)^{5-0} × 100 3.125

using recurrence relation the successive expected frequencies can be obtained as:

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

CHI-SQUARE TEST:

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

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

{ONLY UPPER TAIL TEST k_{2}}

χ^{2}_{cal} = 15.6163.

At α = 5% for (n-1) = 4 – 1 = 3.d.f the uper tail critical value k_{2} = 7.81

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

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

Conclusion: Binomial distribution is not good fit. (ie., O_{i} ≠ E_{i})

Section – E

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

Question 41.

The weights of 1000 students are normally distributed with mean 55 kgs and S.D. 3kgs. Find the number of students with weight between 57 kgs and 60 kgs.

Answer:

Let x be the weight of students is a normal variate with the parameters Mean : μ = 55kgs, and S.D : σ = 3kgs, N = 1000.

Then the standard Normal variate is SNV is z = \(\frac{x-\mu}{\sigma}\) ~ N (0,1); Z = \(\frac{x-55}{3}\)

(i) p(57 < × < 60) = p\(\left(\frac{57-55}{3}<z<\frac{60-55}{3}\right)\)

= p(0.67 < z < 1.67)

= Area from 0.67 to 1.67

= Area from 0.67 to ∞ – Area from 1.67 to ∞

= 0.2514 – 0. 0475 = 0.2039

No. of students with weight between 57 kgs and 60kgs = 0. 2039 × 1000 = 2039 =204 students.

Question 42.

A random sample of 100 people from a city in the year 2011 revealed that 65 were cricket match viewers. Another random sample of 100 people from same city in the year 2013 revealed that 75 were cricket match viewers. Examine whether there is a significant increase in proportion of cricket match viewership at 1% level of significance.

Answer:

Given : n_{1} = 100, x_{1} = 65 . The sample proportion p_{1} = \(\frac{x_{1}}{n_{1}}=\frac{65}{100}\) = 0.65,

n_{2} = 100, x_{2} = 75; p_{2} = \(\frac{x_{2}}{n_{2}}=\frac{75}{100}\) = 0.75

H_{0} : Proportion of cricket viewers in two years is same (ie., P_{01} = P_{02})

H_{1} : Proportion of cricket viewers has increased in 2013 as compared to 2011 (ie., P_{01} < P_{02}) {Lower tail test}

At α = 1% the lower tail critIcal value -k = -2.33 .

Here Z_{cal} lies acceptance region (A.R)

∴ H_{0} is accepted

Conclusion : Proportion of cricket viewers in two years is same (ie., P_{01} = P_{02})

Question 43.

From the.followlng data, test whether ‘education’ and ‘employment’ are indepentent at 5% level of significance:

Answer:

Given: χ^{2} test for independence of attributes

H_{0} : Education and on1polyment are independent

H_{1} : Education and employment are dependent

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

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

At α = 5% for 1 .d.f the upper tail critical value k_{2} = 3.84

Here χ^{2}_{cal} < K_{2} / χ^{2}_{cal} – lies in acceptance region (A.R)

∴ H_{0} is accepted.

Conclusion: Education and employment are independent

Question 44.

Obtain the value of the game and mention the best strategies of the players.

Answer:

The given pay – off matrix is

Using maximin – minimax principle:

- Maximin (α): Minimum payoff in each row is / are circled
- Minimax (β): Maximum payoff in each column is / are boxed.
- The pay off ‘3’ is circled as well as boxed is the saddle point at (2,3)
- The suggested optimal or best strategies for the players are : for player A : A
_{2}and player B : B_{3} - Value of the gave v = 3