Suppose that the cost of auto repairs have a skewed distribu

Suppose that the cost of auto repairs have a skewed distribution, with a mean of $200. and a standard deviation of $300. The sample mean of the SRS of 100 repairs has a sampling distribution with a standard deviation of: Suppose that the time between busses (inter arrival time) at a bus stop has a mean of 10 minutes and a standard deviation of 4 minutes. Assume the distribution is Normal. What is the chance that when taking a SRS of 16 inter arrival times that the sample mean is larger than 11 minutes?

Solution

2) SAMPLE MEAN = 10 MINUTES

STD DEV = 4 MINUTES

WE NEED TO FIND P(X>11) =

For x = 11, z = (11 - 10) / 4 = 0.25

Hence P(x > 11) = P(z > 0.25) = [area to the left of 0.25]

= 0.5987 = 0.60

OPTION D IS CORRECT.