Answers ADepartment of Mathematics, IIT Madras

Quiz-2 MA204 Statistics

Question-cum-Answer Sheet

Date: March 17, 2008 Time: 8:00-8:50 a.m. Max. Marks: 20

Answer all the questions.

Blanks marked with ‘NA’ must be filled with numerical answers.

All numerical answers must be in decimals, correct to two decimal places.

Roll No. Name: Teacher’s Name:

1. The life length, in hours, of a certain type of electronic device is a random variable X having

the probability density

f(x) =

0.001 e−0.001x for x > 0

0 otherwise.

If 100 such devices are chosen at random for testing and X denotes the sample mean, then

E(X) = 1000 (NA), V ar(X) = 10000, (NA), and

P (950 < X < 1100) = .0.532 (NA). [1+1+2]

2. Let X1, X2, . . . , X9 constitute a random sample from a normal distribution with mean µ and

variance 1. If S2 denotes the sample variance, then

P

(−2.306

3≤ X − µ

S≤ 2.306

3

)= .0.95 (NA). [2]

3. The probability that the sample variance of a random sample of size 5 from a normal population

with σ = 12 exceeds 180 is .0.001925 (NA). [2]

4. Let X1, X2, . . . , Xn constitute a random sample taken from a discrete distribution having

probability distribution function as f(x; p) = p (1− p)x−1, x = 1, 2, 3, · · · , 0 < p < 1.

The maximum likelihood estimator of the parameter p is 1/X . [2]

[P.T.O.]

A

5. Let X1, X2, X3 constitute a random sample from a distribution with unknown mean µ and

variance 1. Define Y =X1 + 2X2 + 3X3

6and W =

2X1 + 3X2 + 4X3

9. Then,

V ar(Y ) = 7/18 = 0.389 (NA)

Among Y and W as estimators of µ, W is more efficient than Y [1+1]

6. A discrete probability distribution is given by P (X = x) =1

2

[e−α · αx + e−β · βx

x!

]. Then, in

terms of α and β, µ′1 = (α+ β)/2., and µ′2 = (α2 + α+ β2 + β)/2.

The estimate of α obtained by the method of moments, expressed in terms of m′1 and m′2, is

m′1 ±√m′2 −m′1 − (m′1)

2. [1+1+2]

7. If a random sample from a normal population has the values 2.3, 1.9, 2.1, 2.8, 2.3, 3.6, 1.4, 1.8,

2.1 and 3.1, then the sample standard deviation s = 0.655. (NA), and a 95% confidence interval

for the population mean is 1.8713 < µ < 2.8087. (NA) [2+2]

Relevant Tabular Values (ν denotes the degree of freedom.)

Standard Normal (Cumulative Distribution Function):

F (−0.05) = 0.4801, F (−0.5) = 0.3085, F (0.1) = 0.5398, F (1) = 0.84134

t-distribution:

tα, ν : t0.005, 8 = 3.355, t0.025, 8 = 2.306, t0.025, 9 = 2.262, t0.005, 9 = 3.250

χ2-distribution:

χ2α, ν : χ2

0.0019, 4 = 5, χ20.0019, 5 = 3.99, χ2

0.207, 4 = 0.995