Question

A population has a mean of 400 and a standard deviation of 50. 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 +/- 4 of the population mean (to 4 decimals)?

b. What is the probability that the sample mean will be within +/- 11 of the population mean (to 4 decimals)?

Answer 1

Solution :

Given that,

mean = = 400

standard deviation = = 50

= / n = 50 / 125 = 4.4721

a)

= P[(-4) / 4.4721< ( - ) / < (4) / 4.4721)]

= P(-0.89 < Z < 0.89)

= P(Z < 0.89) - P(Z <-0.89 )

= 0.8133 - 0.1867

= 0.6266

Probability = 0.6266   

b)

= P[(-11) / 4.4721< ( - ) / < (11) / 4.4721)]

= P(-2.46 < Z < 2.46)

= P(Z < 2.46) - P(Z < -2.46)

= 0.9931 - 0.0069

=0.9862

Probability = 0.9862