Question

A normal population has a mean of 65 and a standard deviation of 13. You select a random sample of 25. Compute the probability that the sample mean is: (Round your z values to 2 decimal places and final answers to 4 decimal places): Greater than 69.

Answer 1

0.0618

Explanation:

z = (x - μ)/σ, where

x is the raw score = 69

μ is the sample mean = population mean = 65

σ is the sample standard deviation

This is calculated as:

= Population standard deviation/√n

Where n = number of samples = 25

σ = 13/√25

σ = 13/5 = 2.6

Sample standard deviation = 2.6

z = (69 - 65) / 2.6

z = 4/2.6

z = 1.53846

Approximately to 2 decimal places = 1.54

Using the z score table to determine the probability,

P(x = 69) = P(z = 1.54)

= 0.93822.

The probability that the sample mean is greater than 69 is

P(x>Z) = 1 - 0.93822

P(x>Z) = 0.06178

Approximately to 4 decimal places = 0.0618