The shape of the distribution of the time required to get an oil change at a 10-

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 11.5 minutes, and the standard deviation is 4.4 minutes. a) What is the probability that a random sample of n = 35 oil changes results in a sample mean time less than 10 minutes? b) 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 35 oil changes between 10 A.M. and 12 P.M. 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. (Round to one decimal place as needed.)

Explanation / Answer

mean = 11.5

s = 4.4

sd(Xbar) = s/sqrt(n) = 4.4/sqrt(35)= 0.74373574416

since n = 35 > 30 ,we can apply central limit theorem

Z = (Xbar - 11.5)/0.74373574416

P(Xbar < 10)

=P(Z< (10- 11.5)/0.74373574416)

=P(Z< -2.01684538)

=0.021900

b) z0.1 = 1.282

hence required value

= 11.5 -1.282*0.74373574416

= 10.546530776

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