Question

To qualify as a contestant in a​ race, a runner has to be in the fastest​ 16% of all applicants. The running times are normally​ distributed, with a mean of 63 min and a standard deviation of 5 min. To the nearest​ minute, what is the qualifying time for the​ race?

Answer 1

To find the qualifying time for the race, we need to determine the cutoff time which corresponds to the fastest 16% of runners. Since the running times are normally distributed, we can use the properties of the normal distribution.

1. **Identify the mean and standard deviation:**
- Mean (\(\mu\)) = 63 minutes
- Standard deviation (\(\sigma\)) = 5 minutes

2. **Find the Z-score that corresponds to the top 16% of the distribution:**
The fastest 16% corresponds to the 84th percentile of the normal distribution (100% - 16% = 84%). We can use a Z-table or a calculator to find the Z-score for 0.84.

The Z-score that corresponds to the 84th percentile is approximately **0.995**.

3. **Use the Z-score formula to find the qualifying time:**
The Z-score formula is given by:

\[
Z = \frac{X - \mu}{\sigma}
\]

Rearranging this to solve for \(X\) (the qualifying time):

\[
X = \mu + Z \cdot \sigma
\]

Substituting in the known values:

\[
X = 63 + (0.995 \cdot 5)
\]

\[
X = 63 + 4.975
\]

\[
X \approx 67.975
\]

4. **Round to the nearest minute:**
The qualifying time for the race is approximately **68 minutes**.

Thus, the qualifying time for the race is **68 minutes**.