Question

The shape of the distribution of the time required to get an oil change at a 10-Minute oil-change facility is unknown.​ However, records indicate that the mean time is 11.2 minutes, and the standard deviation is 4.9 minutes.

What is the probability that a random sample of n=35 oil changes results in a sample mean time less than 10 minutes? Please answer rounding to four decimal places.

Answer 1

Solution :

Given that ,

mean = = 11.2

standard deviation = = 4.9

n = 35

=    = 11.2

= / n = 4.9 / 35 = 0.83

P( < 10) = P(( - ) / < ( 10 - 11.2 ) / 0.83)

= P(z < -1.45 )

Using z table

= 0.0735