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Question 2: The average price for a BMW 3 Series Coupe 335i is $39,368. Suppose these prices are also normally distributed with a standard deviation of $2,367. What percentage of BMW dealers are pricing the BMW 3 Series Coupe 335i at more than the average price ($44,520) for a Mercedes CLK350 Coupe? Round your answer to 3 decimal places.

User Vu Anh
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Answer:

0.015 = 1.5% of BMW dealers are pricing the BMW 3 Series Coupe 335i at more than the average price ($44,520) for a Mercedes CLK350 Coupe

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:


\mu = 39368, \sigma = 2367

What percentage of BMW dealers are pricing the BMW 3 Series Coupe 335i at more than the average price ($44,520) for a Mercedes CLK350 Coupe?

This is 1 subtracted by the pvalue of Z when X = 44520. So


Z = (X - \mu)/(\sigma)


Z = (44520 - 39368)/(2367)


Z = 2.18


Z = 2.18 has a pvalue of 0.985

1 - 0.985 = 0.015

0.015 = 1.5% of BMW dealers are pricing the BMW 3 Series Coupe 335i at more than the average price ($44,520) for a Mercedes CLK350 Coupe

User Satvik
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