# ISS (Statistical Services) Statistics Paper I (Old Subjective Pattern): Questions 132 - 138 of 165

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## Question number: 132

» Statistical Methods » Bivariate Distributions » Bivariate Normal Distribution

Appeared in Year: 2010

### Describe in Detail

Show that for discrete distribution β _{2} > 1

### Explanation

We have to prove that

By definition of Kurtosis

Let x _{1}, x _{2}, …, x _{n} are n observations in a set have frequency f _{1}, f _{2, } …, f _{n} and the mean of observation is , then

… (77 more words) …

## Question number: 133

» Statistical Methods » Tests of Significance » Z-Test

Appeared in Year: 2010

### Describe in Detail

How large a sample must be taken in order that the probability will be at least 0·95 that X _{n} will be within 0·5 of µ (µ is unknown and σ = l).

### Explanation

Here, you would know the standard deviation of a population but not know the mean of that population. So, you want to estimate the mean μ to within a given margin of error in a given confidence interval. The required sample size is

Margin of error E = 0.5,

1-α=0.95 →α=0.05

σ=1 and z _{α/2} =1.96

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## Question number: 134

» Statistical Methods » Measures of Location

Appeared in Year: 2013

### Describe in Detail

Find the weighted arithmetic mean of the first ‘n’ natural numbers, the weights being the corresponding numbers.

### Explanation

If x _{1}, x _{2}, …, x _{n} are the values having weights w _{1}, w _{2}, …, w _{n} respectively, the weight mean is

Here X is first ‘n’ natural numbers that is 1,2, 3, …, n

… (77 more words) …

## Question number: 135

» Numerical Analysis » Interpolation Formulae » Lagrange

Appeared in Year: 2015

### Describe in Detail

Fit the exponential curve y = a + bx to the following data

x: 0 2 4

y: 5.01 10 31.62

### Explanation

his is not exponential equation, it is a straight line y = a + bx

S. no | x | y | Xy | x |

1 | 0 | 5.01 | 0 | 0 |

2 | 2 | 10 | 20 | 4 |

3 | 4 | 31.62 | 126.48 | 16 |

Sum | 6 | 46.63 | 146.48 | 20 |

The normal equations is

… (50 more words) …

## Question number: 136

» Probability » Standard Probability Distributions » Uniform

Appeared in Year: 2011

### Describe in Detail

Let X have the continuous c. d. f. F (x). Define U = F (x). Show that both - log U arid -log (1 - U) are exponential random variables:

### Explanation

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

Since F is non-increasing and its continuous.

G (u) =F (F ^{-1} (u) ) implies G (u) =u

Then the p. d. f is

Sin

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## Question number: 137

» Statistical Methods » Tests of Significance » T-Test

Appeared in Year: 2015

### Describe in Detail

Let X _{1}, X _{2}, …, X _{12} be a random sample from a normal N (µ _{1}, σ _{1} ^{2}) and Y _{1}, Y _{2}, …Y _{10} be another random sample from normal N (µ _{2}, σ _{2} ^{2}), independently to each other. Carry out an appropriate test for testing

at 5 % level of significance. It is given that .

### Explanation

Let X _{1}, X _{2}, …, X _{12} be a random sample from a normal N (µ _{1}, σ _{1} ^{2}) and Y _{1}, Y _{2}, …Y _{10} be another random sample from normal N (µ _{2}, σ _{2} ^{2}), independently to each other. The test of hypothesis is

The sample size of X is n = 12 and Y is m = 10

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## Question number: 138

» Probability » Conditional Probability

Appeared in Year: 2012

### Describe in Detail

You arc given the following information:

(i) In random testing, you test positive for a disease.

(ii) In 5 % of cases, the test shows positive even when the subject does not have the disease.

(iii) In the population at large, one person in 1000 has the disease. What is the conditional probability that you have the disease given that you have been tested positive, assuming that if someone has the disease, he will test positive with probability 1?

### Explanation

Let X denotes the test is positive and Y denotes the person has disease. Given that 5 % of cases, the test shows positive even when the subject does not have the disease that is

where denotes the person has no disease. Also given the test positive with probability 1

… (53 more words) …