Question

Let the shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is unknown. However, records indicate that the mean time is 21.6 minutes, and the standard deviation is 4.5 minutes. Round to four decimal points, as needed. Include Excel formulas. *

What is the probability that a random sample of 40 oil changes results in a sample mean time of less than 20 minutes? *

What is the mean time required to get an oil change when the probability that it takes longer than that time is 5%? *

Answer 1

Solution :

Given that ,

mean = = 21.6

standard deviation = = 4.5

n = 40

= 21.6

=  / n = 4.5/ 40=0.71

P( <20 ) = P[( - ) / < (20-21.6) / 0.71]

= P(z <-2.25 )

Using z table  

= 0.0122

2.

Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05

= P(Z < z) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.65) = 0.95

z =1.65

Using z-score formula,

x = z * +

x = 1.65* 0.71+21.6

x = 22.77