Question

Per the Barron's report the average weeks unemployed is 21.5 (population mean) with a population standard deviation of 5 weeks and a sample of 55. In the next five questions we will be calculating the probability that the sample of 55 will provide a mean within 1 week of the population mean. What is the Z-value to 2 decimals for (x<20.5)?

Answer 1

Solution :

Given that ,

mean = = 21.5

standard deviation = = 5

= / n = 5 / 55 = 0.6742

P( < 20.5) = P(( - ) / < (20.5 - 21.5) / 0.6742)   

P(z < -1.48)

= 0.0694

Probability = 0.0694