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

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($\mu$) = 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 $Z = \frac{X -\mu}{\sigma}$

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

$Z_{1} = \frac{X_{1}-134}{4.8} = \frac{125-134}{4.8} = -1.875$

and

$Z_{2} = \frac{X_{2}-134}{4.8} = \frac{135-134}{4.8} = 0.2083$

Therefore, the following is obtained:

$P(125\leq X\leq 135)$ = $P(\frac{125-134}{4.8} \leq Z \leq \frac{135-134}{4.8} )$ = $P(-1.875 \leq Z \leq 0.2083)$

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

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.