American automobiles produced in 2012 and classified
as “large” had a mean fuel economy of 19.6 miles per
gallon with a standard deviation of 3.36 miles per gallon.
A particular model on this list was rated at 23 miles per
gallon, giving it a z-score of about 1.01. Which statement
is true based on this information?
A) Because the standard deviation is small compared to
the mean, a Normal model is appropriate and we can
say that about 84.4% of “large” automobiles have a
fuel economy of 23 miles per gallon or less.
B) Because a z-score was calculated, it is appropriate
to use a Normal model to say that about 84.4% of
“large” automobiles have a fuel economy of 23 miles
per gallon or less.
C) Because 23 miles per gallon is greater than the mean
of 19.6 miles per gallon, the distribution is skewed
to the right. This means the z-score cannot be used to
calculate a proportion.D) Because no information was given about the shape of
the distribution, it is not appropriate to use the z-score
to calculate the proportion of automobiles with a fuel
economy of 23 miles per gallon or less.
E) Because no information was given about the shape
of the distribution, it is not appropriate to calculate a
z-score, so the z-score has no meaning in this situation.