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

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## Question 135

Appeared in Year: *2011*

### Describe in Detail

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…

… (18 more words) …

## Question 136

Appeared in Year: *2015*

### Describe in Detail

Essay▾Let X_{1} , X_{2} , … , X_{12} be a random sample from a normal N (µ_{1} , σ_{1} ^{2}) and Y_{1} , Y_{2} , … Y_{10} be another random sample from normal N (µ_{2} , σ_{2} ^{2}) , independently to each other. Carry out an appropriate test for testing

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

### Explanation

Let X_{1} , X_{2} , … , X_{12} be a random sample from a normal N (µ_{1} , σ_{1} ^{2}) and Y_{1} , Y_{2} , … Y_{10} be another random sample from normal N (µ_{2} , σ_{2} ^{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…

… (40 more words) …

## Question 137

Appeared in Year: *2012*

### Describe in Detail

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 …

… (18 more words) …

## Question 138

Appeared in Year: *2013*

### Describe in Detail

Essay▾Compute the factorial moments µ_{(r)} and the cumulants k_{r} , 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

K_{1} = coefficient of t in = m

K_{2} = coefficient of in = m

Similarly, for others K՚s values.

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