Answer:
Option C is correct.
The probability of maximum distance between 125 and 135 miles is , 0.5531
Step-by-step explanation:
Let X be the distance traveled by car between 125 and 135 miles.
Also, given: The mean distance([tex]\mu[/tex]) = 134 miles and the standard deviation([tex]\sigma[/tex]) = 4.8 miles.
To calculate the Probability that in a random test rum the car will travel a maximum distance between 125 and 135 miles i.e:
[tex]P(125\leq X\leq 135)[/tex]
Let [tex]Z = \frac{X -\mu}{\sigma}[/tex]
then the corresponding z-values need to be computed are:
[tex]Z_{1} = \frac{X_{1}-134}{4.8} = \frac{125-134}{4.8} = -1.875[/tex]
and
[tex]Z_{2} = \frac{X_{2}-134}{4.8} = \frac{135-134}{4.8} = 0.2083[/tex]
Therefore, the following is obtained:
[tex]P(125\leq X\leq 135)[/tex] = [tex]P(\frac{125-134}{4.8} \leq Z \leq \frac{135-134}{4.8} )[/tex] = [tex]P(-1.875 \leq Z \leq 0.2083)[/tex]
= [tex]P(Z \leq 0.2083) - P(Z\leq -1.875)[/tex]
Now, using Standard Normal distribution table we have:
= 0.5835 - 0.0304=0.5531
Therefore, the probability that in a random test run the car will travel a maximum distance between 125 and 135 miles is, 0.5531.
