Question

A population has a mean of 400 and a standard deviation of 800 . Suppose a sample of size 100 is selected and is used to estimate . Use z-table. a. What is the probability that the sample mean will be within +-8 of the population mean (to 4 decimals)? .3830 b. What is the probability that the sample mean will be within +-16 of the population mean (to 4 decimals)?

Answer 1

= 400

= 800

n = 100

For sampling distribution of mean, P( < A) = P(Z < (A - )/)

= = 400

=

=

= 80

a) P(the sample mean will be within 8 of the population mean) = P(Z < 8/) - P(Z < -8/)

= P(Z < 8/80) - P(Z < -8/80)

= P(Z < 0.1) - P(Z < -0.1)

= 0.5398 - 0.4602

= 0.0796

b) P(the sample mean will be within 8 of the population mean) = P(Z < 8/) - P(Z < -8/)

= P(Z < 16/80) - P(Z < -16/80)

= P(Z < 0.2) - P(Z < -0.2)

= 0.5793 - 0.4207

= 0.1586