Question

Barron's reported that the average number of weeks an individual is unemployed is 21.5 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 21.5 weeks and that the population standard deviation is 5.5 weeks. Suppose you would like to select a sample of 65 unemployed individuals for a follow-up study. Use z-table.

What is the probability that a simple random sample of 65 unemployed individuals will provide a sample mean within 1 week of the population mean? (to 4 decimals)

What is the probability that a simple random sample of 65 unemployed individuals will provide a sample mean within 1/2 week of the population mean? (to 4 decimals)

Answer 1

1)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	21.5
std deviation   =σ=	5.5000
sample size       =n=	65
std error=σx̅=σ/√n=	0.6822

probability that a simple random sample of 65 unemployed individuals will provide a sample mean within 1 week of the population mean:

probability =

P(20.5<X<22.5)

=

P(-1.47<Z<1.47)=

0.9292-0.0708=

0.8584

2)

probability that a simple random sample of 65 unemployed individuals will provide a sample mean within 1/2 week of the population mean:

probability =

P(21<X<22)

=

P(-0.73<Z<0.73)=

0.7673-0.2327=

0.5346