Question

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 12.3 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 12.3 weeks and that the population standard deviation is 7 weeks. Suppose you would like to select a random sample of 221 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is between 12.1 and 12.9.
P(12.1 < X < 12.9) =

Find the probability that a sample of size n=221n=221 is randomly selected with a mean between 12.1 and 12.9.
P(12.1 < M < 12.9) =

Answer 1

Solution :

Given that ,

mean = $\mu$ = 12.3

standard deviation = $\sigma$ = 7

a) P( 12.1< x <12.3 ) = P((12.1-12.3)/7 ) < (x - $\mu$ ) / $\sigma$ < (12.9-12.3) /7 ) )

= P(-0.0286 < z < 0.0857)

= P(z <0.0857 ) - P(z <-0.0286 )

= 0.5341 - 0.4886

= 0.0455

P(12.1 < X < 12.9) = 0.0455

b)

n = 221

$\mu$_{$\bar x$} = 12.3

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$n = 7 / $\sqrt$221 = 0.4709

P( 12.1 < $\bar x$ < 12.9) = P((12.1-12.3) /0.4709)<($\bar x$ - $\mu$_{$\bar x$}) / $\sigma$_{$\bar x$} < (12.9-12.3) /0.4709 ))

= P(-0.4271< Z <1.2742)

= P(Z < 1.2742) - P(Z < -0.4271

= 0.8987 - 0.3347

= 0.5640

P(12.1 < M < 12.9) = 0.5640