# ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 60 - 64 of 165

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## Question number: 60

» Probability » Standard Probability Distributions » Uniform

Appeared in Year: 2010

### Describe in Detail

Let X be a random variable with a continuous distribution function F. Show that F (X) has the uniform distribution on (0,1).

### Explanation

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

The inverse exists. Since F is non-increasing and its continuous

G (u) =F (F ^{-1} (u) ) {F is a distribution of X}

G (u) =u

Then the p. d. f of U = F (x) is given b

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## Question number: 61

» Probability » Standard Probability Distributions » Cauchy

Appeared in Year: 2012

### Describe in Detail

For the Cauchy distribution given by

where k is a constant to be suitably chosen, derive the expression for the distribution function. Hence obtain a measure of central tendency and a measure of dispersion. What are the points of inflexion of the distribution?

### Explanation

We choose k as constant that gives the integral over the range x for the density function is equal to one.

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## Question number: 62

Appeared in Year: 2011

### Describe in Detail

A fair die is thrown until a 6 appears. Specify the sample space. What is the probability that it must be thrown at least 3 times?

### Explanation

A fair die content the value {1,2, 3,4, 5,6}

So, the probability of getting 6 is p = 1/6, then probability of getting other than 6 is q = 5/6

If we throw a die, the six is appear. Then the probability is p and the experiment end. However, if the six is not appear, the dice is throwing again and this time the six appear. Then the probability i

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## Question number: 63

» Probability » Conditional Probability

Appeared in Year: 2011

### Describe in Detail

Consider the following bivariate p. m. f. of (X, Y):

p (0,10) = p (0,20) = 2/18;

p (l, 10) = p (l, 30) = 3/18;

p (1,20) = p (2,30) = 4/18;

Obtain the conditional mass functions p (y lx = 2), and p (y lx = 1).

### Explanation

The joint probability is given

- | x | Total | |||

- | 0 | 1 | 2 | ||

Y | 10 | 2/18 | 3/18 | 0 | 5/18 |

20 | 2/18 | 4/18 | 0 | 6/18 | |

30 | 0 | 3/18 | 4/18 | 7/18 | |

Total | 4/18 | 10/18 | 4/18 | 1 |

First find the marginal probability of x at x = 0,1, 2

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## Question number: 64

» Statistical Methods » Bivariate Distributions » Bivariate Normal Distribution

Appeared in Year: 2014

### Describe in Detail

The marginal distributions of X and Y are given in the following table:

X | 1 | 2 | Total | |

Y | 3 4 | ? ? | ? ? | 1/4 3/4 |

Total | 1/2 | 1/2 | 1 |

If the co-variance between X and Y is zero, find the cell probabilities and see whether X and Y are independent.

### Explanation

Given that covariance between X and Y is zero

This gives that f (xy) =f (x) f (y)

So, the joint probability density function of x = 1, y = 3 is

We know that if the covariance

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