Question

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal​ distribution, with a mean of 15 minutes and a standard deviation of 3.5 minutes. ​(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take​ longer, the customer will receive the service for​ half-price. What percent of customers receive the service for​ half-price? ​(b) If the automotive center does not want to give the discount to more than 3​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer 1

Answer:

a) Therefore 92.64% of customers receive the service for​ half-price.

b) Therefore it would take 8.52 minutes

Step-by-step explanation:

Given that ime required for an automotive center to complete an oil change service on an automobile approximately follows a normal​ distribution with:

mean (μ) = 15 minutes

Standard deviation (σ) = 3.5 minutes

a) The z score is given by the equation:

$z=\frac{x-\mu}{\sigma}$

Since the service will take no longer than 20 minutes,

$z=\frac{x-\mu}{\sigma}=\frac{20-15}{3.5} = 1.43$

Using the probability table to get the probability of the z score

percent of customers receive the service for​ half-price = P(X < 20) = P(z < 1.43) = 0.9236 = 92.64%

Therefore 92.64% of customers receive the service for​ half-price.

b) Since the probability is 3% = 0.03, this corresponds with a z score of -1.88

Therefore:

$z=\frac{x-\mu}{\sigma}=\frac{x-15}{3.5} =-1.88$

$\frac{x-15}{3.5} =-1.88$

${x-15} =-6.58\\x=-6.58+15=8.52\\x=8.52 minutes$

Therefore it would take 8.52 minutes