Question

A manufacturer knows that their items have a lengths that are skewed right, with a mean of 8.9 inches, and standard deviation of 0.6 inches. If 49 items are chosen at random, what is the probability that their mean length is greater than 9.1 inches? (Round answer to four decimal places)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 8.9

standard deviation = $\sigma$ = 0.6

n = 49

$\mu$_{$\bar x$} =  $\mu$ = 8.9

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 0.6 / $\sqrt$ 49 = 0.086

P($\bar x$ > 9.1) = 1 - P($\bar x$ < 9.1)

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < ( 9.1 - 8.9 ) / 0.086 ]

= 1 - P(z < 2.33 )

Using z table

= 1 - 0.9901

= 0.0099