Question

Question 16 of 21 5 Points Assume that the length of time needed to change the oil in a car at Fiona's Service Station is normally distributed with mean u = 15 minutes, and a standard deviation o = 5 minutes. If Fiona decides to offer a discount to those customers whose cars take the longest 20% of servicing time, what is the cutoff for the time required to receive the discount? OA. 11.8 minutes B. 21.4 minutes C. 17.6 minutes D. 19.2 minutes OE. None of the options are correct Reset Selection

Answer 1

Solution:

Given, X follows Normal distribution with,

$\mu$ = 15

$\sigma$ =  5

For top 20% data , let x be the required cut-off.

P(X > x) = 20%

P(X > x) = 0.20

P(X < x) = 1 - 0.20

P(X < x) = 0.80

For the standard normal variable z , P(Z < z) = 0.80

Use z table, see where is 0.80 probability and then see the corresponding z value.

P(Z < 0.842) = 0.80

So z = 0.842

Now using z score formula ,

x = $\mu$ + (z * $\sigma$ ) = 15 + (0.842* 5) = 19.2

Answer : D. 19.2 minutes