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?
the qualifying time for the race is \\(\square\\) min.
(round to the nearest minute as needed.)

Explanation:

Step1: Identify the problem type

This is a normal distribution problem where we need to find the value corresponding to the 84th percentile (since the fastest 16% means we want the value such that 84% of the data is above it, or equivalently, the 84th percentile from the left? Wait, no: fastest 16% means the smallest 16% of times? Wait, no: running times, faster means lower time. So to be in the fastest 16%, the time must be less than or equal to the value such that 16% of the times are below it. Wait, no: "fastest 16% of all applicants" – so the 16% with the lowest running times. So we need to find the value \( x \) such that \( P(X \leq x) = 0.16 \), where \( X \) is normally distributed with mean \( \mu = 63 \) min and standard deviation \( \sigma = 5 \) min.

Step2: Use the z-score formula

The z-score corresponding to a cumulative probability of 0.16. We look up the z-score in the standard normal table (or use a calculator) for \( P(Z \leq z) = 0.16 \). Looking up in the z-table, the z-score for which the area to the left is 0.16. Let's find it: the z-table values. For example, \( z = -0.99 \) (since \( P(Z \leq -0.99) \approx 0.1611 \), which is close to 0.16).

Step3: Calculate the x-value

Using the z-score formula \( z = \frac{x - \mu}{\sigma} \), we solve for \( x \):

\( x = \mu + z \cdot \sigma \)

Substituting \( \mu = 63 \), \( z = -0.99 \), \( \sigma = 5 \):

\( x = 63 + (-0.99) \cdot 5 = 63 - 4.95 = 58.05 \)

Wait, but let's check the z-score more accurately. Using a calculator or more precise table: the z-score for \( P(Z \leq z) = 0.16 \). Let's use the inverse of the standard normal CDF. Using a calculator, the z-score for 0.16 is approximately -0.9945.

Then \( x = 63 + (-0.9945) \cdot 5 = 63 - 4.9725 = 58.0275 \), which rounds to 58 minutes.

Wait, but let's confirm: if we use a more accurate z-score. Let's use the formula for the inverse normal. Using a calculator (like TI-84: invNorm(0.16, 63, 5)). Let's compute that:

invNorm(0.16, 63, 5) = 63 + invNorm(0.16, 0, 1)5. invNorm(0.16) is approximately -0.9945, so 63 - 0.99455 = 63 - 4.9725 = 58.0275, which rounds to 58 minutes.

Answer:

58