You are the owner of an auto repair service. History tells you it takes on average 45 minutes to complete a repair job. You have determined the standard deviation for a job is 6 minutes. A women comes into your shop and tells you she must leave her car for repair, and that she will be shopping at the mall across the street. She says she will be back no earlier than 38 minutes, but she absolutely must leave no later than 55 minutes to pick up her child from school. What is the probability her car will be repaired during the 38 to 55 minute timeframe?

Question

You are the owner of an auto repair service. History tells you it takes on average 45 minutes to complete a repair job. You have determined the standard deviation for a job is 6 minutes. A women comes into your shop and tells you she must leave her car for repair, and that she will be shopping at the mall across the street. She says she will be back no earlier than 38 minutes, but she absolutely must leave no later than 55 minutes to pick up her child from school. What is the probability her car will be repaired during the 38 to 55 minute timeframe?

Thanks!

Answer 1

Calculate Z scores for both values.

Z = (x-μ)/SD, where x = score, μ = mean, and SD = standard deviation.

In table on back of your statistics book called something like "areas under normal distribution," use Z scores to find areas between scores and mean. Add the two values.

I hope this helps. Thanks for asking.

Answer 2

To find the probability of the car being repaired within the 38 to 55 minute timeframe, we need to calculate the z-scores for both time limits and then look up the corresponding probabilities in the standard normal distribution table.

First, let's calculate the z-score for 38 minutes:
Z = (X - μ) / σ
Z = (38 - 45) / 6
Z = -7 / 6
Z = -1.17 (rounded to two decimal places)

Next, let's calculate the z-score for 55 minutes:
Z = (X - μ) / σ
Z = (55 - 45) / 6
Z = 10 / 6
Z = 1.67 (rounded to two decimal places)

Now, we can use the z-scores to find the probability of the car being repaired within these time limits.

To find the probability for the lower limit (38 minutes), we need to look up the z-score -1.17 in the standard normal distribution table. The corresponding probability is 0.1210.

To find the probability for the upper limit (55 minutes), we need to look up the z-score 1.67 in the standard normal distribution table. The corresponding probability is 0.9525.

Now, we subtract the lower probability from the higher probability to find the probability that the car will be repaired within the 38 to 55 minute timeframe:
0.9525 - 0.1210 = 0.8315

Therefore, the probability that the car will be repaired during the 38 to 55 minute timeframe is approximately 0.8315, or 83.15%.