SOLUTION: Suppose that when a Honda dealership sells a Honda Accord, the profit the dealership makes is normally distributed with a mean of $1200 and standard deviation of $350. Suppose that

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Question 1205489: Suppose that when a Honda dealership sells a Honda Accord, the profit the dealership makes is normally distributed with a mean of $1200 and standard deviation of $350. Suppose that over a 2 month period of time, the dealership sells 50 Honda Accords. What is the probability that the dealership averages at least $1300 profit per car over that period of time?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
assumed population profit has a mean of 1200 and a standard deviation of 350.
sample size is 50.
standard error = standard deviation / sqrt(sample size) = 350 / sqrt(50) = 49.49747 rounded to 5 decimal digits.
z-score formula is z = (x-m)/s
z is the z-score
x is the sample mean.
m is the population mean.
s is the standard error.
formula becomes:
z = (1300 - 1200) / 49.49747 = 2.0203 rounded to 4 decimal digits.
area under the normal distribution to the right of that z-score = .0217 rounded to 4 decimal digits.
that's the probability that the mean of the sample of 50 sold honda accord profits will be greater than 1300, assuming the profits are normally distributed.

here are the z-score results on a normal distribution graph.

here are the raw score results on the same normal distribution graph.

z-score results use a mean of 0 and a standard deviation of 1.
raw score results use the population mean and standard error.