ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 135 - 138 of 164

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Question 135

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Appeared in Year: 2011

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Essay▾

Let X have the continuous c. d. f. F (x) . Define U = F (x) . Show that both - log U arid -log (1 - U) are exponential random variables:

Explanation

Let U = F (x) , then the distribution function of G of U is given by

Since F is non-increasing and its continuous.

G (u) = F (F -1 (u) ) implies G (u) = u

Then the p. d. f is

Since F is a distribution function takes value in range [0,1] . Hence

This is a uniform distribution on [0,1] .

Let Y =-log U. Then

The probability density function

Y =-logU has expon…

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Question 136

Appeared in Year: 2015

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Essay▾

Let X1 , X2 , … , X12 be a random sample from a normal N (µ1 , σ1 2) and Y1 , Y2 , … Y10 be another random sample from normal N (µ2 , σ2 2) , independently to each other. Carry out an appropriate test for testing

at 5 % level of significance. It is given that .

Explanation

Let X1 , X2 , … , X12 be a random sample from a normal N (µ1 , σ1 2) and Y1 , Y2 , … Y10 be another random sample from normal N (µ2 , σ2 2) , independently to each other. The test of hypothesis is

The sample size of X is n = 12 and Y is m = 10

The test is depend upon the population variance is known and unknown. Here the variance is known, so the tes…

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Question 137

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Appeared in Year: 2012

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Essay▾

You arc given the following information:

(i) In random testing, you test positive for a disease.

(ii) In 5 % of cases, the test shows positive even when the subject does not have the disease.

(iii) In the population at large, one person in 1000 has the disease. What is the conditional probability that you have the disease given that you have been tested positive, assuming that if someone has the disease, he will test positive with probability 1?

Explanation

Let X denotes the test is positive and Y denotes the person has disease. Given that 5 % of cases, the test shows positive even when the subject does not have the disease that is

where denotes the person has no disease. Also given the test positive with probability 1

Then the conditional probability that you have the disease given that you have been …

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Question 138

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Appeared in Year: 2013

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Essay▾

Compute the factorial moments µ(r) and the cumulants kr , r = 1,2, … , of Poisson distribution with parameter m.

Explanation

Let X follows Poisson distribution with parameter m. The density function is

The r th factorial moment of Poisson distribution is

The cumulants of Poisson distribution is

we known that the moment generating function of X is

So, the cumulants is

K1 = coefficient of t in = m

K2 = coefficient of in = m

Similarly, for others K՚s values.

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