Question

A random sample of size n = 25 is obtained from a normally distributed population with population mean μ =200 and variance σ^2 = 100.
a) What are the mean and standard deviation of the sampling distribution for the sample means?
b) What is the probability that the sample mean is greater than 203?
c) What is the value of the sample variance such that 5% of the sample variances would be less than this value? d) What is the value of the sample variance such that 5% of the sample variances would be greater than this value?

Answer 1

Mean = $\mu$ = 200

Variance = σ^2 = 100.

Standard deviation = $\sigma$ = 10

a)

Mean of the sampling distribution for the sample means =  $\mu$ = 200

The standard deviation of the sampling distribution for the sample means is:

10 o (n 10 。 =5=

b)

We have to find the probability that the sample mean is greater than 203.

We have to find P( $\bar{x}$ > 203)

For finding this probability we have to find a z score.

urlo

203 - 2003 | 10/05 = 1.5

That is we have to find P(Z > 1.5)

P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668

( From z table)

c)

5% of the sample variances would be less than this value:

P(?>k) = pp (n = 1) * 5?

24,0.95 = 13.85

( From chi-square table)

246 LooC 13.85

57 , 702 > دی

d)

5% of the sample variances would be greater than this value:

P(?>k) = pp (n = 1) * 5?

124.0,05 = 36.42

( From chi-square table)

2434 100> 36.42

في 151 . 879