The shape of the distribution of the time required to get an oil change at a 15 minute facility is unknown. however, records indicate that the mean time is 16.1 minutes and the standard deviation is 4.6 minutes.

a) to compute the probabilities regarding the sample mean using the normal model, what size sample would be required?

b) what is the probability that a random sample of n = 40 oil changes results in a sample mean time of less than 15 minutes?

a) choose the required sample size:

1. the normal model cannot be used if the shape is unknown

2. sample size need to be greater than 30

3. sample size needs to be less than 30

4. any sample size could be used.

b) the probability is approximately? (round to 4 decimal places as needed)

a) According to the central limit theorem the sampling distribution follows normal distribution if the size of the sample is large enough. X> or equal to 30.

Therefore the answer to the question is option 2. (Sample size need to be greater than 30)

Step-by-step explanation:

b) sample mean is 15, population mean is 16.1 , standard deviation is 3.7 and sample size is n=40

P[ z<( xbar -meu)/(sigma/sqrt(n))]

P[ z<( 15-16.1)/(3.7/√40)]

P[z< -1.1/0.5859]

P[z<-1.8803]

Checking through the standard normal distribution table

The probability to 4 d.p is 0.0300.

Also see attached picture for clearer calculation.

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )