# ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 140 - 144 of 165

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## Question number: 140

» Probability » Moment Generating Functions

Appeared in Year: 2014

### Describe in Detail

X _{1}, X _{2}, …, X _{N} are independently, identically distributed random variables. Define S _{N} = X _{1} + X _{2} + … + X _{N }, where N is a random variable independent of X _{i}, i = 1,2, … N.

Show that the moment generating function (mgt) of S _{N} is

where My (t) is the mgf of a random variable Y. Hence find the mgf of S _{N} when N follows a Poisson distribution with parameter λ. and X _{i} follows an exponential distribution with mean parameter θ, i = 1 to N.

### Explanation

The moment generating function of S _{N} is where S _{N} =X _{1} +X _{2} +…+X _{N}, X _{i} are i. i. d random variable and N is also a random variable.

X _{i} ’s are independent identically distributed, so

In the question if X _{i} follows an exponential di

… (143 more words) …

## Question number: 141

» Statistical Methods » Association and Contingency

Appeared in Year: 2014

### Describe in Detail

The runs scored by two batsmen A and B in five cricket matches were as follows:

Batsmen A: 50 60 100 70 20

Batsmen B: 120 100 30 20 40

Discuss the consistency and efficiency of the batsmen.

### Explanation

The mean of batsmen A is

The variance of the batsmen A is

The coefficient of variation is

The mean of batsmen B is

The variance of the batsmen B is

Th

… (230 more words) …

## Question number: 142

» Probability » Standard Probability Distributions » Poisson

Appeared in Year: 2011

### Describe in Detail

Show that the sum of two independent Poisson random variables with parameters λ and µ respectively is a Poisson random variable with parameter λ+µ.

### Explanation

Let X and Y are independent Poisson random variables with parameters λ and µ respectively. We proof this by moment generating function. The moment generating function of Poisson distribution is

So, the sum of X and Y moment generating function is

because X and Y are independent

… (122 more words) …

## Question number: 143

» Probability » Conditional Probability

Appeared in Year: 2009

### Describe in Detail

The joint density of (X, Y) is

Find the conditional densities and E [X|Y = 1·5].

### Explanation

First find the marginal distribution of X and Y

The conditional density of X|Y is

The conditional density of Y|X is

… (235 more words) …

## Question number: 144

» Probability » Standard Probability Distributions » Normal

Appeared in Year: 2011

### Describe in Detail

Let the joint p. d. f. of (X, Y) be f (x, y) = e ^{-y}, 0 < x < y < ∞.

Obtain the probability P (X + Y ≤ 1).

### Explanation

The Joint p. d. f. is

f (x, y) = e ^{-y}, 0 < x < y < ∞.

Let assume X + Y =U and Y = V, then X = U-V

Using Jacobian technique

The range is 0 < ∞, u ≤ v < ∞

The joint p. d. f. of U and V is

The marginal distribution of u is

Find that

… (114 more words) …