Question

The time required for an automotive center to complete and change service on an automobile primately follows anomalion, with a man of 15 minutes and a standard deviation (a) The automotive center quanto customers that the service will take no longer than 20 minutes ir does take longer the customer will receive the service for all price. What percent of customers the si half price (1) of the automotive Center does not want to give the discount to more than 2 of its customers, how long hou make the water Click here to view the standardnom bution togel Click here to view the standard memabution at age 2 (a) The percent of customers that receive the service for all price is 0% (Round to two decimal places as needed) The guaranteed time limit is minutes (Round up to the nearest minute)

Answer 1

Solution :

Let X be the time required for an automotive centre to change service on an automobile (in minutes).Here X is approximately normally distributed with mean 15 minutes and standard deviation 3.5 minutes.

(a). The automobile Centre guarantees customer that the service will take no longer than 20 minutes. If it does take longer the customer will receive the service at half price. The the probability of customers that will the service for half-price is given by; P(X > 20).

To find the probability P(X > 20) we need to find the Z score for the value 20. These are given by,

Z1 = (20-15)/3.5 = 1.43. Therefore, problem becomes finding probability, P(Z > Z1).

This is given by; P(Z > Z1) = 1 - P(Z < Z1) = 1 - 0.9236 ...{from Z table}.

We get, P(Z > Z1) = 0.0764.

The the percent of customers that will the service for half-price are, 0.0764*100% = 7.64%.

(b). The automotive centre do not want to give the discount to more that 2% of customers. That is, we need to find the value of X1 such that, P(X > X1) = 0.02, this gives, P(X < X1) = 1 - 0.02 = 0.98.

Form the Z table we get, P(Z < 2.05) = 0.9798  $\approx$  0.98. This gives, 2.05 is the Z score corresponding to the value X1. This gives, 2.05 = (X1-15) / 3.5. Solving the equation for "X1" we get, X1 = 15 + 2.05*3.5 = 22.175  $\approx$ 22.

That is, if the guaranteed time limit is 22 minutes then no more than 2% will receive discount.

Now let Y be the incubation time of fertilized eggs in days.Here Y is approximately normally distributed with mean 19 days and standard deviation 1 day.

(a). The 13th percentile for incubation time be Y+, then we get P(Y < Y0) = 0.13. And from the Z table wet, P(Z < -1.13) = 0.1292 $\approx$ 0.13. This gives, -1.13 is the Z score corresponding to the value Y0. This gives, -1.13 = (Y0-19) / 1. Solving the equation for "Y0" we get, Y0 = 19 - 1.13 = 17.87  $\approx$ 18.

The 13th percentile for incubation time is 18 days.

(b). The incubation times that will make up the 95% are given by, P(Y1 < Y < Y2) = 0.95 such that, P(Y < Y1) = 0.025 and P(Y < Y2) = 0.975. from the Z table we get, P(Z < -1.96) = 0.025 and P(Z < 1.96) =0.975.

This gives, -1.96 and 1.96 are the Z scores corresponding to the values Y1 and Y2. This gives, -1.96 = (Y1-19) / 1 and 1.96 = (Y2-19) / 1. Solving the equation for "Y1" and "Y2" we get, Y1 = 19 - 1.96 = 17.04  $\approx$ 17 and Y2 = 19 + 1.96 = 20.96 $\approx$ 21.

The incubation times that will make up the 95% are 17 to 21 days.