Question

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.
0.3867

B.
0.5832

C.
0.5531

D.
0.0301

Answer 1

Option C is correct.

The probability of maximum distance between 125 and 135 miles is , 0.5531

Explanation:

Let X be the distance traveled by car between 125 and 135 miles.

Also, given: The mean distance(
`\mu`) = 134 miles and the standard deviation(
`\sigma`) = 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:

`P(125\leq X\leq 135)`

Let
`Z = (X -\mu)/(\sigma)`

then the corresponding z-values need to be computed are:

`Z_(1) = (X_(1)-134)/(4.8) = (125-134)/(4.8) = -1.875`

and

`Z_(2) = (X_(2)-134)/(4.8) = (135-134)/(4.8) = 0.2083`

Therefore, the following is obtained:

`P(125\leq X\leq 135)` =
`P((125-134)/(4.8) \leq Z \leq (135-134)/(4.8) )` =
`P(-1.875 \leq Z \leq 0.2083)`

=
`P(Z \leq 0.2083) - P(Z\leq -1.875)`

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.