Question

A population has a mean of 400 and a standard deviation of 40. Suppose a sample of size 125 is selected and x is used to estimate μ.

a. What is the probability that the sample mean will be within +/- 9 of the population mean (to 4 decimals)?

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

Answer 1

Solution :

Given that,

mean = = 400

standard deviation = = 40

n = 125

= = 400

= / n = 40 / 125 = 3.78

a) P(391< < 409 )  

= P[(391 - 400) / 3.78 < ( - ) / < (409 - 400) / 3.78 )]

= P(- 2.38 < Z < 2.38)

= P(Z < 2.38) - P(Z <- 2.38 )

Using z table,  

= 0.9913 - 0.0087

= 0.9826

n = 125

= = 400

= / n = 40 / 125 = 3.78

b) P(390< < 410 )  

= P[(390 - 400) / 3.78 < ( - ) / < (410 - 400) / 3.78 )]

= P(- 2.65 < Z < 2.65)

= P(Z < 2.65) - P(Z <- 2.65 )

Using z table,  

= 0.9960 - 0.0040  

= 0.9920