Question

A leading magazine (like Barron's) reported at one time that the is 13.2 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 13.2 weeks and that the population standard deviation is 8.8 weeks. Suppose you would like to select a random sample of 29 unemployed individuals for a follow-up study average number of weeks an individual is unemployed Find the probability that a single randomly selected value is between 8 and 14.2 Find the probability that a sample of size n = 29 is randomly selected with a mean between 8 and 14.2. Enter your answers as cumbers accurate to 4 decimal places. Points possible: 1 Unlimited attempts Submit Next . 11:35 PM 1/23/2019 BANG & OLUFSEN 6 8
Right answer please.

Answer 1

$\mu$ = 13.2, $\sigma$ = 8.8.

P(X < x) = p( z < x - $\mu$ / $\sigma$)

p( 8 < X < 14.2) = p( X < 14.2) - p( X < 8)

= p( z < 14.2 - 13.2 / 8.8) - p( z < 8 - 13.2 / 8.8)

= p( z < 0.1136) - p( z < -0.5909)

= p( z < 0.1136) - (1 - p( z < 0.5909) )

= 0.5442 - ( 1 - 0.7226)

= 0.2668

p( 8 < X < 14.2) = 0.2668

For sample size of n = 29,

p( $\bar{x}$ < x ) = p( z < x -  $\mu$ / $\sigma$/$\sqrt n$))

p( 8 < $\bar{x}$ < 14.2) = p( $\bar{x}$ < 14.2) - p( $\bar{x}$ < 8)

=p( z < 14.2 - 13.2 / (8.8 / sqrt(29))) - p( z < 8 - 13.2 / (8.8 / sqrt(29)))

= p( z < 0.612) - p( z < -3.182)

= p( z < 0.612) - ( 1 - p( z < 3.182) )

= 0.7293 - ( 1 - 0.9905)

= 0.7198

p( 39.2 < $\bar{x}$ < 40.4) = 0.7198