# ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 89 - 95 of 165

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## Question number: 89

» Probability » Standard Probability Distributions » Binomial

Appeared in Year: 2010

### Describe in Detail

Let X _{1}, X _{2}, …, X _{m} be i. i. d. random variables with common p. m. f.

obtain the p. m. f. of S _{m} = X _{1} + X _{2} + …. + X _{m}.

### Explanation

Let X _{1}, X _{2}, …, X _{m} 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 S _{m} = X _{1} + X _{2} + …. + X _{m}, sum of random variables are found by moment generating function. The moment generating function of binomial distribution is

… (95 more words) …

## Question number: 90

» Statistical Methods » Tests of Significance » Chi-Square

Appeared in Year: 2011

### Describe in Detail

Explain how to carry out the chi-squared test for on the basis of a random sample

X _{1}, X _{2}, …, X _{n} from N (µ, σ ^{2}) population.

### Explanation

Given that a random sample X _{1}, X _{2}, …, X _{n} from N (µ, σ ^{2}) population. The hypothesis is

The null hypothesis is test by chi-square test only assume the sample size is less than 30.

For this we use the likelihood ratio test

Under the model, the parameter space is

Under the null hypothesis, the p

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## Question number: 91

» Probability » Standard Probability Distributions » Gamma

Appeared in Year: 2011

### Describe in Detail

Let

Show that f (x) is a probability density function. Obtain V (X).

### Explanation

if X is a continuous random variable and f (x) is a continuous function of X, then f (x) is a probability density function if

Assume but limit is same

This integral is a gamma function

So,

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## Question number: 92

» Numerical Analysis » Summation Formula » Euler-Maclaurin's

Appeared in Year: 2014

### Describe in Detail

Using Euler’s method, compute the values of y correct upto 4 places of decimal for the differential equation with initial condition x _{0} = 0, y _{0} = 1, taking h = 0.05.

### Explanation

The given differential equation is

with initial condition x _{0} = 0, y _{0} = 1, taking h = 0.05.

Using Euler’s method,

where

So, putting the initial condition, when n = 0

and

Now first modification of y _{1}

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## Question number: 93

» Probability » Definitions and Axiomatic Approach

Appeared in Year: 2010

### Describe in Detail

Let X be a random variable defined on (Ω, A, P). Define a point function F (x) =P {ω: X (ω) ≤x}, for all xϵR. Shoe that the function F is indeed a distribution function.

### Explanation

Let x _{1} < x _{2}. Then (-∞, x _{1}] ( (-∞, x _{2}] and we have

Since F is non decreasing, it is sufficient show that for any sequence of numbers x _{n} ↓x, x _{1} > x _{2} > … > x _{n} > …x, F (x _{n}) →F (x). Let A _{k} = {ω, X (ω) ϵ (x, x _{k}] }. Then A _{k} ϵS and also

Since none of the interval (x, x _{k}] contains x. It fo

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## Question number: 94

» Statistical Methods » Correlation Coefficient » Partial Correlation

Appeared in Year: 2009

### Describe in Detail

The two regression lines between X and Y are 8X - 10Y + 66 = 0,40X - 18Y = 214. The variance of X is 9. Find X, Y, σ _{Y} and ρ.

### Explanation

Given that

We know that the mean value of the given series satisfies the regression line that is

The means of X and Y is calculate by multiply equation (1) by 5 to subtract equation (2), we get

The correlation coefficient is defined by

The first equation is writ

… (96 more words) …

## Question number: 95

» Statistical Methods » Bivariate Distributions » Bivariate Normal Distribution

Appeared in Year: 2013

### Describe in Detail

Let (X, Y) have a joint probability mass function

= 0, elsewhere

Find the marginal mass functions of X and Y.

### Explanation

Let (X, Y) have a joint probability mass function, the individual distribution of either X or Y is called the marginal distribution. So, the marginal mass functions of X is

and other values of f (x, y) is 0

The marginal mass functions of Y is

and ot

… (94 more words) …