Question

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.

Answer 1

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{44520 - 39368}{2367}$

$Z = 2.18$

$Z = 2.18$ 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