ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 85 - 89 of 164

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Question 85

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Appeared in Year: 2014

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Essay▾

Suppose that all the four outcomes 01 , 02 , 03 and 04 of an experiment are equally likely. Define A = (01 , 04) , B = (02 , 04) and C = (03 , 04) . What can you say about the pairwise independence and mutually independence of the events A, B and C?

Explanation

Given that there are four outcomes. Define A = (O1 , O4) , B = (O2 , O4) , C = (O3 , O4)

Their intersection is and Union is AUBUC = (O1 , O2 , O3 , O4)

P (A) = P (B) = P (C) =

P (AUBUC) = 1

The events A, B and C are not pairwise independent because

But the events A and B are pairwise independent because

Similarly the events B and C or A and C are pairw…

… (59 more words) …

Question 86

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Appeared in Year: 2013

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Essay▾

Obtain 100 (1 -α) % confidence interval for the ratio of population variances by using two independent random samples from N (µ1 , σ1 2) and N (µ2 , σ2 2) under the assumption that the population means are (i) known and (ii) unknown.

Explanation

Let X1 , X2 , … , Xn and Y1 , Y2 , … Ym are the samples taken from independent N (µ1 , σ1 2) and N (µ2 , σ2 2) .

In this question the hypothesis for testing is

Let s1 2 and s2 2 be the estimates variances of σ1 2 and σ2 2 based on sample sizes n and m.

and

Then

This two chi-square random variable are independent. So, H0 can be test by F-test

where s1 2…

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Question 87

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Appeared in Year: 2015

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Essay▾

Let X be a random variable with E [X] = 3 and E [X 2] = 13. Use Chebyshev՚s inequality to obtain P [-2 < X < 8] .

Explanation

Let X be a random variable with mean µ and variance σ 2 . Then any k > 0, the Chebyshev՚s inequality is

or ‎

σ 2 = E [X 2] - (E [X] ) 2 = 4

Then, a lower bound for the probability

Using Chebyshev՚s inequality

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Question 88

Appeared in Year: 2012

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Essay▾

Of three independent events A, Band C, A only happens with probability , B only happens with probability and C only happens with probability . Find the probability that at least one of these three events happens.

Explanation

Given that P (A) = , P (B) = , P (C) =

then probability that at least one event of these three events happens is

The events are independent because only one event happens

Question 89

Appeared in Year: 2010

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Essay▾

Let X1 , X2 , … , Xm be i. i. d. random variables with common p. m. f.

obtain the p. m. f. of Sm = X1 + X2 + … . + Xm .

Explanation

Let X1 , X2 , … , Xm i. i. d. random variables with common p. m. f. is P (X = k) which is a binomail random variables with common parameters n and p respectively. Then, the p. m. f. of Sm = X1 + X2 + … . + Xm , sum of random variables are found by moment generating function. The moment generating function of binomial distribution is

So, the moment g&#8230;

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