Answer: 0.1037
Step-by-step explanation:
Given : Mean : [tex]\mu=80\text{ miles per day}[/tex]
Standard deviation : [tex]\sigma = 30\text{ miles per day}[/tex]
The formula to calculate the z-score is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x = 97 miles per day ,
[tex]z=\dfrac{97-80}{30}\approx0.57[/tex]
For x = 107 miles per day ,
[tex]z=\dfrac{97-80}{30}=0.9[/tex]
The P-value =[tex]P(0.57<z<0.9)=P(z<0.9)-P(z<0.57)[/tex]
[tex]=0.8159398-0.7122603=0.1036795\approx0.1037[/tex]
Hence, the probability that a truck drives between 97 and 107 miles in a day = 0.1037
