Question

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

(c) suppose the manager agrees to pay each employee a $50 bonus If they meet a certain goal. On a typical Saturday the oil change facility will perform 40 oil changes between 10am and 12 pm. Treating this as a random sample, there would be a 10% chance of the mean oil change time being at or below what value? This will be the goal established by the manager.

There is a 10% chance of being at or below a mean oil change time of ____ minutes.

I TRIED 15.5 and that is not the correct answer.

Answer 1

Solution,

Given that,

mean = $\mu$ = 21.4

standard deviation = $\sigma$ = 4.9

n = 40

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

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 4.9 / $\sqrt$ 40 = 0.77

Using standard normal table,

P(Z < z) = 10%

= P(Z < z) = 0.10  

= P(Z < -1.282) = 0.10

z = -1.282

Using z-score formula  

$\bar x$ = z * $\sigma$ _{$\bar x$}+ $\mu$ _{$\bar x$}

$\bar x$ = -1.282 * 0.77 + 21.4

$\bar x$ = 20.4 minutes

There is a 10% chance of being at or below a mean oil change time of 20.4 minutes.