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1. In determining automobile mileage ratings, it was found that the mpg in the city (X) for certain model is normally distributed, with a mean of 22.5 mpg and a standard deviation of 1.5 mpg. Find the following:
a. P(X < 22.5)
b. P(0 < X < 24)
c. P(X > 25)
d. P(22 < X < 22.5)
e. P(X < 21)
f. P(21.5 < X < 23)
g. P(X > 20)
h. The mileage rating that the upper 5% of cars achieve
For example: Suppose that the size of individual customer orders (in dollars), X from a major discount book publisher Web site is normally distributed with a mean of $36 and standard deviation of $8. The probability that the next individual who places an order at the Web site will purchase more than $40 can be found by a simple normal probability calculation after standardizing the value of 40:
P(X > 40) = P(Z > (40 - 36)/8 = 1 - P(Z < 0.5) = 1 - 0.6915 = 0.3085
Now suppose that a sample of 16 customers is chosen. What is the probability that the mean purchase for these 16 customers will exceed $40? To find this, we must realize that we must use the sampling distribution of the mean to carry out the appropriate calculations. The sampling distribution of the mean will have a mean of $36, but a standard error of $8/√16 = $2. Then the probability that /X exceeds $40 for a sample size of n = 16 is
P(/X > 40) = P(Z > (40 - 36)/2) = 1 - P(Z < 2) = 1 - .9772 = 0.0228
Since the mean is m=22.5 and the standard deviation is 1.5, we can assume that Y = (X-m)/d = (X-22.5)/1.5,
then Y follows the standard normal distributiion. From Y = (X-22.5)/1.5, we have X = 1.5Y + 22.5
Next we can use the standard normal distributiion ...
Probabilities are calculated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.