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 11.8 minutes, and the standard deviation is 4 8 minutes. Complete parts (a) through (c) below records indicate that the mean time is Click here to view the standard normal distribution table (page 1) Click here to view the standard normal distribution table (page 2) (a) To compute probabilities regarding the sample mean using the normal model. what size sample would be required? Choose the required sample size below O A. The normal model cannot be used if the shape of the distribution is unknown. O B. Any sample size could be used ° C. The sample size needs to be greater than 30 O D. The sample size needs to be less than 30 (b) What is the probability that a random sample of n 45 oil changes results in a sample mean time less than 10 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 45 oil changes between 10 A M and 12 P M Treating this as a random sample at what mean oil-change time would there be a 10% chance of being at or belo? This will be the goal established by the manager There would be a 10% chance of being at or below Round to one decimal place as needed.) ] minutes.

Explanation / Answer

a)
C. The sample size needs to be greater than 30

b)
mean = 11.8 , s = 4.8 , n =45

P(x <10)

z = (x -mean) / (s/sqrt(n))
= (10 -11.8) / ( 4.8/sqrt(45))
= -2.5156
P( x <10) = p(z < -2.5156) = 0.0059 by using stanadrd normal table

c)

z value at 10% below = -1.2816

z =(x -mean)/(s/sqrt(n))
-1.2816 = (x -11.8)/(4.8/sqrt(45))

x = 10.8829

There would be a 10% chance of being ar o rbelow 10.9 minutes

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