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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%2022%2C%20%5Csigma%20%3D%205)
What is the probability that a car will get less than 21 miles-per-gallon?
This probability is the pvalue of Z when
. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{21 - 22}{5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B21%20-%2022%7D%7B5%7D)
![Z = -0.2](https://tex.z-dn.net/?f=Z%20%3D%20-0.2)
has a pvalue of 0.4207.
So the probability that a car will get less than 21 miles-per-gallon is 0.4207.