Answer: 0.1824
Step-by-step explanation:
Given : The mileage per day is distributed normally with
Mean : [tex]\mu=110\text{ miles per day}[/tex]
Standard deviation : [tex]\sigma=38\text{ miles per day}[/tex]
Let X be the random variable that represents the distance traveled by truck in one day .
Now, calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 132 miles per day.
[tex]z=\dfrac{132-110}{38}\approx0.58[/tex]
For x= 159 miles per day.
[tex]z=\dfrac{159-110}{38}\approx1.29[/tex]
Now by using standard normal distribution table, the probability that a truck drives between 132 and 159 miles in a day will be :-
[tex]P(132<x<159)=P(0.58<z<1.29)\\\\=P(z<1.29)-P(0.58)\\\\= 0.9014747-0.7190426=0.1824321\approx0.1824[/tex]
Hence, the probability that a truck drives between 132 and 159 miles in a day =0.1824
