Question

The miles-per-gallon obtained by the 1995 model q cars is normally distributed with a mean of 22 miles-per-gallon and a standard deviation of 5 miles-per-gallon. what is the probability that a car will get less than 21 miles-per-gallon? (4 decimal format = 0.0000)

Answer 1

Answer:

The probability that a car will get less than 21 miles-per-gallon is 0.4207.

Step-by-step explanation:

Problems of normally distributed samples can be solved using 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 problem, we have that:

$\mu = 22, \sigma = 5$

What is the probability that a car will get less than 21 miles-per-gallon?

This probability is the pvalue of Z when $X = 21$. So

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

$Z = \frac{21 - 22}{5}$

$Z = -0.2$

$Z = -0.2$ has a pvalue of 0.4207.

So the probability that a car will get less than 21 miles-per-gallon is 0.4207.

Answer 2

That probability is about 0.4207.