MATH SOLVE

2 months ago

Q:
# 30 POINTS!! please help! After recording the maximum distance possible when driving a new electric car, the study showed the distances followed a normal distribution. The mean distance is 134 miles and the standard deviation is 4.8 miles. Find the probability that in a random test run the car will travel a maximum distance between 125 and 135 miles.A. 0.3867B. 0.5832C. 0.5531D. 0.0301

Accepted Solution

A:

Answer:Option C is correct.The probability of maximum distance between 125 and 135 miles is , 0.5531Step-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=0.5531Therefore, the probability that in a random test run the car will travel a maximum distance between 125 and 135 miles is, 0.5531.