Students can Download 2nd PUC Statistics Previous Year Question Paper June 2017, Karnataka 2nd PUC Statistics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka 2nd PUC Statistics Previous Year Question Paper June 2017
Time: 3 Hours 15 Minutes
Max. Marks: 100
Section – A
Answer any Ten of the following (10 × 1 = 10)
Question 1.
Define fecundity.
Answer:
Fecundity refers to “the capacity of a woman to bear children”.
Question 2.
Mention one characteristic of index numbers.
Answer:
Index number s study the effect of factors that cannot be measured directly. (Any such related characteristics)
Question 3.
Theoretically which average is considered as the best average in the construction of index number?
Answer:
Geometric mean.
Question 4.
What is a Historigram?
Answer:
A graphical presentation of time series data is called historigram.
Question 5.
If p = 0. 2 for a Bernoulli distribution, find variance.
Answer:
Variance = pq = 0.2 × 0.8 = 0.16
Question 6.
If Z is a SNV and (Z > k) = 0.05, find the value of k.
Answer:
p (Z > k) = 0.05 ⇒ k = 1.64
Question 7.
What is an estimator?
Answer:
Any statistic which is used to estimate an unknown parameter is called an estimator
Question 8.
What is test statistics?
Answer:
A statistic based on whose distribution the testing of hypothesis is conducted is called test statistic.
Question 9.
When is the pooling done in testing of goodness of fit?
Answer:
When the expected frequencies are less than 5.
Question 10.
In S.Q.C what is defect?
Ans.
A defect is a quality characteristic which does not conform to specifications.
Question 11.
The graphical solution to the L.P.P lies in the first quadrant. Give reason.
Answer:
Because of non- negativity restrictions.
Question 12.
Mention a property of a competitive game.
Answer:
Number of competitors must be finite. ( Any such related property)
Section – B
II. Answer any Ten of the following questions (10 × 2 = 20)
Question 13.
In a life table , I0 = 1,00,000 and T0 = 65,00 000 years then, find longevity.
Answer:
e0 = \(\frac{T_{0}}{1_{0}}=\frac{6500000}{100000}\) = 65
Question 14.
State two norms (considerations) for the selection of base year.
Answer:
Base period should be economically stable.
The base period should not be too distant from the current period.
Question 15.
If Σ poq1 = 1125 and Σpoq0 = 1250 then, find suitable quantity index number.
Answer:
Question 16.
What is seasonal variation? Give an example.
Answer:
The regular, periodic and short term variation in a time series is called seasonal variation. Umbrellas are sold more in rainy season (A S R Example)
Question 17.
Write down the two assumptions of interpolation and extrapolation.
Answer:
- There are no sudden jumps in the values of dependent variable from one period to another.
- There will be no consecutive missing values in the series.
Question 18.
Write down the range and mean of a hypergeometric distribution whose parameters are a = 4, b = 6 and n = 5.
Answer:
Range: X = 0,1 min (a,n) 0,1,……, min (4, 5) = 0,1,2,3,4
Mean = \(\frac{n a}{a+b}=\frac{5 \times 4}{4+6}=\frac{20}{10}\) = 2
Question 19.
For a chi – square (χ2) variate, with 10 d.f p(0 < χ2 < 9.34) = 0.5. Find median and mode.
Answer:
Median = 9.4, Mode = n – 2 = 10 – 2 = 8
Question 20.
Define size of the test and power of a test.
Answer:
The probability of rejecting H0, when it is true is called size of the test.
The probability of rejecting H0, when it is not true is called power of a test.
Question 21.
Write two applications of χ2 test.
Answer:
- To test whether population has a given variance.
- To test goodness of fit of theoretical frequencies to – observed frequencies.
Question 22.
Mention two advantages of acceptance sampling.
Answer:
- It is used when the items are of destructive in nature.
- It is less expensive, as 100% inspection is more expensive.
Question 23.
Which of the two feasible solutions (12,10) and (14,5) of an L.P.P minimises the objective function Z = 5X + 4Y.
Answer:
- The value of Z at (12,10) is 100 and the value of Z at (14,15) is 90
- The solution (14,5) minimizes the objective function Z = 5X + 4Y
Question 24.
From the following pay – off matrix, obtain the value of game.
Answer:
Row minima: 2,5 Column maxima: 5,8
maximin : 5 and minimax : 5 ∴ The value of the game 5
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:
WSFR formula
WSFR: 30, 50, 60, 50, 40, 24, 6
GRR = i ΣWSFR = 5 × 260 = 1300
Question 26.
Calculate kelly’s price index number from the following data and comment on the result.
Answer:
p0q: 70, 88, 100, 96: Σp0q = 354
p1q: 75, 96, 120, 120: Σp1q = 411
Pk01 = \(\frac{\Sigma p_{1} q}{\Sigma p_{0} q}\) × 100 = 1161
There is 16.1% increase in the prke of items in the current year.
Question 27.
Compute the cost of living index number for the following data:
Answer:
P = \(\frac{p_{1}}{p_{0}}\) × 100 or \(\frac{2500}{2000}\) × 100 : 125, 150, 120, 80, 125: Total
WP: 1250, 600, 1440, 640, 750: 4680
Formula Ans = 117
Question 28.
For the following data draw a trend line by the method of semi averages.
OR
For visually challenged students only)
Explain semi – averages method of measuring trend.
Answer:
Question 29.
For the following data interpolate the value of Y, when X = 30.
Answer:
Question 30.
In a college 35% student are girls. Find the probability that:
(i) two are girls,
(ii) at least one is a girl in a random sample of 5 students.
Answer:
p = 35% = 0.35, n = 5, p (x) = nCx (p)x (q)n-x; x = 0,1, …. n
(i) P (x = 2) = 5C2 (0.35)2 (0.65)3 = 0.3364
(ii) p(x ≥ 1) = 11 – [0.1160] = 0.8840
Question 31.
State five properties of a normal distribution.
Answer:
- The normal curve is symmetrical about the mean (β1 = 0).
- The distribution is mesokurtic (β2 = 3).
- The total area under curve is unity.
- The quartiles Q3 and Q3 are equidistant from the median i.e., Q2 = (Q3 + Q1)/2
[Any such related property)
Question 32.
In an election the leaders of a party contend that they would secure more than 36% of votes. A pre- poll survey of 400 votes revealed that the percentage is 42. Does the survey support the leader’s claim? (use 5 % L.O.S.)
Answer:
H0: Survey does not support the leader’s claim and H1: P > 0.36
Here, n = 400, P = 0.42, Q = 0.64 and Test statistic zcal = \(\frac{p-p}{\sqrt{\frac{P Q}{n}}}=2.5\)
K = 1.65 Here , Zcal lies in rejection region. ∴ reject H0
Question 33.
The following is the data regarding family condition and examination result of 100 students. Test whether results deepens on family conditions, (use ∝ 0.01)
Answer:
k2 = 6.65 χ2cal < k2 ∴ Accept H0, Results and family conditions are independent
Question 34.
Find control limits for R – Chart with sample size 4.
Answer:
R̄ = \(\frac{\Sigma R}{k}=\frac{30}{6}\) = 5 = CL
L.C.L = D3R̄ = 0(5) = 0
U.C.L = D4R̄ = 2.282(5) = 11.41
Question 35.
Find the initial basic feasible solution to the following transportation problem by NWCR method and find transportation cost to this solution.
Answer:
X11 = 400, X12 = 100, X22 = 200, X23 = 100, X33 = 200
TC = ΣCijXij = 8(400) + 4(100) + 5 (200) + 6(100)+ 3(200) = 5800
Question 36.
Purchase cost of an item is Rs. 5,000. Its running costs and resale values in different years are as follows:
Find the optimum period of replacement of item.
Answer:
Minimum annual average cost = Rs. 1160, Optimal replacement period is 4th year.
Section – D
IV. Answer any Two of the following questions: (2 × 10 = 20)
Question 37.
By taking locality ‘B’ as standard calculate the STDR’s for both localities. Mention which locality is more healthier?
Answer:
STDR formula, STDR (A) = 13.4
For Locality B: CDR = STDR, CDR formula,
12.25 OR
(B : 20,5,6,30; PB: 100000, 60000,90000,24000,490000 Ans:12.25)
Town B is healthier.
Question 38.
Compute Laspeyre’s paasche’s and Fisher’s price index numbers for the following data:
Answer:
P0q0 : 300, 200, 240, 300, 320: ΣP0q0 = 1360
P0q1 : 336, 240, 240, 240, 288: ΣP0q1 = 1344
P1q0 : 500, 200, 360, 360,480: ΣP1q0 = 1900
P1q1 : 560, 240, 360, 288, 432: ΣP1q1 = 1880
Question 39.
For the following data fit a straight line trend by least square method, obtain trend values. Estimate the production for the year 2008.
n = 7, ΣY = 357, ΣX = 0, ΣX2 = 28 and ΣXY = 42
Answer:
a = \(\frac{\Sigma Y}{n}\) = 51, b = \(\frac{\Sigma x Y}{\Sigma x^{2}}\) = 1.5
∴ The trend line is., Ȳ = 51 + 5x
Trend values: 46.5, 48, 49.5, 51, 52.5, 54, 55.5
Ȳ2017 = 58.5(000’s tons)
Question 40.
For the following data fit a poisson distribution and test for goodness of fit at 5% level of significance.
Answer:
N = 100, λ = 1.2, P(0)= e-y = e-1.2 = 0.3012 or T0 = N × p(0) = 30.12
Theoretical frequencies: 30, 36, 22, 9, 2, 1
H0: P.D. is a good fit and H1: P.D is not a good fit.
Test Statistic, χ2 = \(\Sigma \frac{\left(0_{i}-E_{i}\right)^{2}}{E_{i}}\) = 0.5232
k2 = 5.99, χ2cal Lies in acceptance region.
∴ Accept H0 i.e., P.D is a good fit.
Section – E
V. Answer any Two of the following questions. (2 × 10 = 20)
Question 41.
Marks scored by the students of a class follow normal distribution with mean 80 and S.D. 5. Find the probability that a student selected at random from the class scored
(i) more than 90 marks
(ii) Between 70 and 85 marks.
Answer:
(i) P(x > 90) = P\(\) = p(Z > 2) = 0.0228
(ii) p(-2 < Z < 1) = 0.9772 = -0.1587 = 0.8185
Question 42.
A company manufactures car tyres. Their average life is 40,000 km and standard deviation 5,000 km. A change in the production process is believed to result in a better product. A test sample of 100 new tyres has mean life of 41,000 km. Can you conclude at 1% L.Q.S. that the new product gives better result?
H0 : The average Life of tyres is 40000 km (µ = 40000) and H1 : µ > 40000.
Answer:
k = 2.33, here, Zcal lies in acceptance region.
∴ accept H0
Question 43.
I.Q. of 5 students before and after training is given below:
Is training is effective? (use ∝ = 0.05)
Ho: Training is not effective and H1: µo < µ2.
Answer:
Test statistic tcal lies in acceptance region. ∴ accept H0
Question 44.
There is a demand for 3600 units/year. The cost of placing an order is ₹ 50. maintenance cost ₹ 9/unit/ year and shortage cost is ₹ 3/unit/year.
Find:
(i) EOQ (ii) Minimum average cost year.
R = 3600/year, C3 = ₹ 50, C1 = ₹ 9/year, C2 = ₹ 3/year
Answer:
Formula and Ans = 400 units
Formula and Ans = ₹. 900