Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 33 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 33 weeks and that the population standard deviation is 2.3 weeks. Suppose you would like to select a random sample of 69 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is greater than 33.
P(X > 33) = (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a sample of size n=69 is randomly selected with a mean greater than 33.
P(M > 33) =

Answer 1

Solution :

Given that ,

mean = $\mu$ = 33

standard deviation = $\sigma$ =2.3

P(x >33 ) = 1 - p( x<33 )

=1- p [(x - $\mu$ ) / $\sigma$ < (33-33) /2.3 ]

=1- P(z <0 )

= 1 -0.5 = 0.5

probability =0.5000

n = 69

$\mu$_{$\bar x$} =$\mu$ = 33  

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 2.3/ $\sqrt$ 69 = 0.2769

P(M > 33) = 1 - P(M < )

= 1 - P[(M - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (33-33) /0.2769 ]

= 1 - P(z <0 )

= 1-0.5 =0.5

Probability = 0.5000