# ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 74 - 81 of 165

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## Question number: 74

» Probability » Tchebycheffs Inequality

Appeared in Year: 2011

### Describe in Detail

Let X be a positive valued random variable. Prove that

Hence deduce the Chebychev’s inequality.

### Explanation

The expectation of X is define as

For any x≥r

This implies that

Hence, any function of X, assume g (x) the Chebychev’s inequality is

… (88 more words) …

## Question number: 75

» Probability » Distribution Function » Standard Probability Distributions

Appeared in Year: 2015

### Describe in Detail

Let Y _{1} denote the first order statistic in a random sample of size n from a distribution that has the pdf

Obtain the distribution of Z _{n} =n (Y _{1} -θ).

### Explanation

To find the distribution of order statistic, the pdf is

Here denote the first order statistic that is

The cdf of x is

The pdf of Y _{1} is

Then, the distribution of Z

… (195 more words) …

## Question number: 76

» Numerical Analysis » Interpolation Formulae » Newton (Dividend Difference)

Appeared in Year: 2010

### Describe in Detail

If f (x) =1/x ^{2}, find the divided differences f (a, b) and f (a, b, c).

### Explanation

we can write the Newton’s divided difference formulas as

First find f (b, c)

Then, from (1) and (2)

… (237 more words) …

## Question number: 77

» Statistical Methods » Tests of Significance » Z-Test

Appeared in Year: 2015

### Describe in Detail

If X follows binomial b (n _{1}, p _{1}) distribution and Y follows binomial b (n _{2}, p _{2}), provide an appropriate test at level α for H _{0}: p _{1} =p _{2} against H _{1}: p _{1} > p _{2}.

### Explanation

The hypothesis is equivalent to testing the null hypothesis that p _{1} −p _{2} = 0 against the alternative is p _{1} -p _{2} > 0. The statistic on which we base our decision is the random variable . Suppose the samples of sizes n _{1} and n _{2} are selected from binomial populations are independent.

For large sample size, the estimator was approximately norm

… (56 more words) …

## Question number: 78

» Statistical Methods » Correlation Coefficient » Partial Correlation

Appeared in Year: 2014

### Describe in Detail

Given the following correlation matrix of order 3 x 3

Calculate (i) r _{12.3} (ii) r _{1.23}

### Explanation

In general, the correlation matrix is in these notations

(i)

(ii)

… (211 more words) …

## Question number: 79

» Probability » Distribution Function » Standard Probability Distributions

Appeared in Year: 2013

### Describe in Detail

Find the distribution of the ratio of two iid random variables with density function:

### Explanation

Let consider x and y are two iid random variable with density function is same. Then find the distribution of their ratio that is x/y. The joint pdf of x and y is

To find this we use Jacobian transformation technique, assume

X/Y = u, Y = v

X = uv and Y = v

So, the joint density function of u and v

… (82 more words) …

## Question number: 80

» Probability » Standard Probability Distributions » Normal

Appeared in Year: 2012

### Describe in Detail

12·3 % of the candidates in a public examination score at least 70%, while another 6·3 % score at most 30%. Assuming the underlying distribution to be normal, estimate the percentage of candidates scoring 80 % or more.

### Explanation

Let total marks obtain is 100. Assuming the underlying distribution to be normal, the mean µ and variance σ ^{2}. It is given that

The value of z corresponding to an area

0.500 - 0.123 = 0.377

We can write

Similarly, . It is given that

The value of z corresponding to an area

0.50

… (100 more words) …

## Question number: 81

» Probability » Probability of M Events Out of N

Appeared in Year: 2013

### Describe in Detail

A fair die is rolled twice. Let A be the event that the first throw shows a number ≤ 2, and B be the event that the second throw shows at least 5. Show that P (AUB) =5/9.

### Explanation

A fair die is rolled twice, the sample space consists of thirty six outcomes. The sample space is

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (5,5), (5,6)

Total outcome

… (65 more words) …