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?

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:

Since the service will take no longer than 20 minutes,

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:

Therefore it would take 8.52 minutes

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.