Question

A) A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 26 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 26 weeks and that the population standard deviation is 3 weeks. Suppose you would like to select a random sample of 30 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is less than 25. P(X < 25) = ______

Find the probability that a sample of size n = 30 is randomly selected with a mean less than 25. P(M < 25) = ____

B) A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 39 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 39 weeks and that the population standard deviation is 3.8 weeks. Suppose you would like to select a random sample of 81 unemployed individuals for a follow-up study.

Find the probability that a single randomly selected value is greater than 38.8.
P(X > 38.8) =______(Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a sample of size n=81n=81 is randomly selected with a mean greater than 38.8.
P(M > 38.8) = ________ (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Solution :

Given that ,

A) mean = $\mu$ = 26

standard deviation = $\sigma$ = 3

P(x < 25) = P[(x - $\mu$ ) / $\sigma$ < (25 - 26) / 3]

= P(z < -0.33)

Using z table,

= 0.3707

n = 30

$\mu$_{$\bar x$} =  $\mu$ = 26

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 3 / $\sqrt$ 30 = 0.5477

P($\bar x$ < ) = P(($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (25 - 26) / 0.5477)

= P(z < -1.83)

Using z table

= 0.0336

B) mean = $\mu$ = 39

standard deviation = $\sigma$ = 3.8

P(x > 38.8) = 1 - p( x< 38.8)

=1- p P[(x - $\mu$ ) / $\sigma$ < (38.8 - 39) / 3.8]

=1- P(z < -0.05)

Using z table,

= 1 - 0.4801

= 0.5199

n = 81

$\mu$_{$\bar x$} =  $\mu$ = 39

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 3.8 / $\sqrt$ 81 = 0.4222

P($\bar x$ > 38.8) = 1 - P($\bar x$ < 38.8 )

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (38.8 - 39) / 0.4222]

= 1 - P(z < -0.47)

Using z table,    

= 1 - 0.3192

= 0.6808