Question

The shape of the distribution of the time required to get an oil change at a 20-minute facility is unknown. however, records indicate records indicate that the mean time is 21.4 minutes and the standard deviation 4.9 minutes.

B) what is the probability that a random sample of n=40 oil changes results in a sample mean time less than 20 minutes?

the probability is approximately _____

Answer 1

Solution :

Given that ,

mean = $\mu$ = 21.4

standard deviation = $\sigma$ = 4.9

n = 40

$\mu$_{$\bar x$} = 21.4

$\sigma$_{$\bar x$} =$\sigma$  / $\sqrt$ n = 4.9/ $\sqrt$ 40=0.7748

P($\bar x$ <20 ) = P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (20 - 21.4) /0.7748 ]

= P(z <-1.81 )

Using z table

= 0.0351