Question

The shape of the distribution of the time required to get an oil change at a 15-minute oil change facility is unknown. However, records indicate that the mean time is 16.3 minutes, and the standard deviation is 48 minutes. Complete parts (a) through (c) (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? O A. The sample size needs to be greater than or equal to 30. O B. Any sample size could be used OC. The normal model cannot be used if the shape of the distribution is unknown OD. The sample size needs to be less than or equal to 30 (b) What is the probability that a random sample of n 40 oil changes results in a sample mean time less than 15 minutes? The probability is approximately (Round to four decimal places as needed.) (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 10 AM, 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 minutes There is a 10% chance of being at or below a mean oil change time of (Round to one decimal place as needed.)

Answer 1

Answer

(A) To apply central limit theorem, we must have a minimum sample size of 30 or more.

So, requirement is to have a sample size that is greater than 30

option A is correct

(B) using normalcdf

enter lower = -1E99

upper = 15

mean = 16.3

standard deviation = 4.8/sqrt(40)

we get

$P(X \le 15) = normalcdf(-1E99,15,16.3,4.8/\sqrt{40}) \\ P(X \le 15)= 0.0434$

(C) We have to find sample mean for which there is 10% chances that mean time is at or below

Using InvNorm(area, mean, standard deviation)

where area = 0.10

mean = 16.3

standard deviation = 4.8/sqrt(40)

we get

Required sample mean = invNorm(0.10,16.3,4.8/sqrt(40))

= 15.3