Probability of Missed Deadline for Transmission Service
The amount of time required for routine automobile transmission service is normally distributed with the mean 45 minutes and the standard deviation 8.0 minutes.
The service manager plans to have work begin on the transmission of a customer’s car 10 min after the car is dropped off, and the customer is told that the car will be ready within 1 hour total time (i.e. after the car is dropped off).
What is the probability that the service manager will be wrong?
Solution
\[X \sim N(\mu,\sigma^2) \]
Z-Score Computation and Probability Rules
Compute the Z-Score
\[ z_o = \frac{x_o - \mu}{\sigma} \]
Probability Rules
- \(P(X \leq x_o) = P(Z \leq z_o)\)
- \(P(Z \leq -z_o) = P(Z \geq z_o)\)
Scenario
- The time to complete transmission service is normally
distributed:
\(X \sim N(45, 8^2)\) - The service starts 10 minutes after drop-off, and the car must be ready within 60 minutes total
- Therefore, the actual service must be completed within 50 minutes
Goal
Find the probability that the service time exceeds 50 minutes, i.e.,
\[ P(X > 50) \]
Step 1: Standardize Using Z-Score
\[ z = \frac{x - \mu}{\sigma} = \frac{50 - 45}{8} = \frac{5}{8} = 0.625 \]
Step 2: Find the Corresponding Probability
Use standard normal distribution tables or a calculator:
\[ P(X > 50) = P(Z > 0.625) \]
From tables:
\[ P(Z < 0.625) \approx 0.7340 \Rightarrow P(Z > 0.625) = 1 - 0.7340 = \mathbf{0.2660} \]
Final Answer
There is approximately a 26.6% chance that the service manager will be wrong and miss the 60-minute deadline.