Question

At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 40 minutes and a standard deviation of 2 minutes. What percentage of customers have to wait between 37 minutes and 40 minutes, to the nearest tenth?

Answer 1

To find the percentage of customers who wait between 37 minutes and 40 minutes for their food, we can use the properties of the normal distribution.

1. **Determine the parameters**:
- Mean (\( \mu \)) = 40 minutes
- Standard Deviation (\( \sigma \)) = 2 minutes

2. **Convert the waiting times to Z-scores** using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
where \( X \) is the value for which we are calculating the Z-score.

3. **Calculate the Z-score for 37 minutes**:
\[
Z_{37} = \frac{(37 - 40)}{2} = \frac{-3}{2} = -1.5
\]

4. **Calculate the Z-score for 40 minutes**:
\[
Z_{40} = \frac{(40 - 40)}{2} = \frac{0}{2} = 0
\]

5. **Look up the Z-scores in the standard normal distribution table** (or use a calculator):
- For \( Z = -1.5 \), the cumulative probability \( P(Z < -1.5) \) is approximately \( 0.0668 \).
- For \( Z = 0 \), the cumulative probability \( P(Z < 0) \) is \( 0.5 \).

6. **Calculate the percentage of customers waiting between 37 and 40 minutes**:
\[
P(37 < X < 40) = P(Z < 0) - P(Z < -1.5) = 0.5 - 0.0668 = 0.4332
\]

7. **Convert to percentage**:
\[
0.4332 \times 100 \approx 43.32\%
\]

8. **Round to the nearest tenth**:
\[
\text{Final answer} = 43.3\%
\]

Therefore, approximately **43.3%** of customers have to wait between 37 minutes and 40 minutes.