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?
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To calculate the probability that the car will be repaired during the 38 to 55 minute timeframe, we can use the concept of the standard normal distribution.
Step 1: Convert the given time values to the corresponding z-scores.
The z-score is calculated using the formula:
z = (x - μ) / σ
where x is the given value, μ is the mean, and σ is the standard deviation.
For the lower bound, 38 minutes:
z_lower = (38 - 45) / 6 = -1.17
For the upper bound, 55 minutes:
z_upper = (55 - 45) / 6 = 1.67
Step 2: Find the cumulative probability of the z-scores.
Using a standard normal distribution table or a statistical calculator, we can find the cumulative probability associated with each z-score.
For the lower bound:
P(z ≤ -1.17) ≈ 0.121 (from the table or calculator)
For the upper bound:
P(z ≤ 1.67) ≈ 0.952 (from the table or calculator)
Step 3: Calculate the probability within the given range.
To find the probability of the car being repaired within the 38 to 55 minutes timeframe, we subtract the cumulative probability of the lower bound from the cumulative probability of the upper bound.
P(38 ≤ x ≤ 55) = P(z ≤ 1.67) - P(z ≤ -1.17)
= 0.952 - 0.121
≈ 0.831
Therefore, the probability that the car will be repaired during the 38 to 55 minute timeframe is approximately 0.831, or 83.1%.