# ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 95 - 100 of 165

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

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## Question number: 96

» Numerical Analysis » Interpolation Formulae » Gauss

Appeared in Year: 2011

### Describe in Detail

Use mathematical induction to prove

### Explanation

Let ∆ ^{i} define the i ^{th} finite difference which is defined as

∆ =E-1 where E is a shift operator that is

E ^{n} f _{x} = f _{x + nh}

Then

… (-1 more words) …

## Question number: 97

Appeared in Year: 2011

### Describe in Detail

Show that E (X - a) ^{2} is minimized for a = E (X), assuming that the· first 2 moments of X exist.

### Explanation

Assume Y = E (X - a) ^{2}

Here we want to minimized Y for a that is

The value of E (X - a) ^{2} is minimum when the value of a is E (X).

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## Question number: 98

» Probability » Distribution Function » Standard Probability Distributions

Appeared in Year: 2015

### Describe in Detail

Let X have pdf

Obtain the cdf of Y = X ^{2}.

### Explanation

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## Question number: 99

» Statistical Methods » Standard Errors and Large Sample Tests

Appeared in Year: 2009

### Describe in Detail

A sample of 100 records on lengths of stay of patients in a hospital gave a standard deviation of days of stay as 4·9. In order to estimate the mean number of days of stay within 0·25 day with 95 % confidence, what should be the sample size?

### Explanation

Let the patients in a hospital gave a standard deviation of days of stay as 4.9. The mean number of days of stay within 0·25 day that is margin error is the estimate fraction days between the difference of estimated and the true number of days. So, the test statistic for 95 % confidence is

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## Question number: 100

» Statistical Methods » Association and Contingency

Appeared in Year: 2013

### Describe in Detail

Out of 200 persons appeared for an examination, 80 % were males and the rest were females. Among 30 married males, 14 were successful; 110 unmarried males were successful. In respect of 10 married females, 4 were successful; 20 unmarried females were successful. Compute Yule’s coefficient of association between success in the examination and the marital status for males as well as females.

### Explanation

Out of 200 persons appeared for an examination, 160 males and 40 are females.

The frequencies for various attributes be show in this contingency table

Male | Successful (A) | unsuccessful (a) |

married (B) | (AB) =14 | (aB) =16 |

unmarried (b) | (Ab) =110 | (ab) =20 |

Female | successful (A) | unsuccessful (a) |

marri… |

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