To find the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes, let's go through the problem step-by-step:
1. Convert the mean and target times from hours and minutes to total minutes:
- Mean time: [tex]\(3\)[/tex] hours [tex]\(50\)[/tex] minutes
[tex]\[
3 \text{ hours} \times 60 \text{ minutes/hour} + 50 \text{ minutes} = 230 \text{ minutes}
\][/tex]
- Target time: [tex]\(3\)[/tex] hours [tex]\(20\)[/tex] minutes
[tex]\[
3 \text{ hours} \times 60 \text{ minutes/hour} + 20 \text{ minutes} = 200 \text{ minutes}
\][/tex]
2. Calculate the z-score for the target time:
The z-score formula is:
[tex]\[
z = \frac{\text{Target time} - \text{Mean time}}{\text{Standard Deviation}}
\][/tex]
Substituting the given values:
[tex]\[
z = \frac{200 \text{ minutes} - 230 \text{ minutes}}{30 \text{ minutes}} = \frac{-30 \text{ minutes}}{30 \text{ minutes}} = -1.0
\][/tex]
3. Find the probability corresponding to the z-score:
Using the standard normal distribution table, we look up the probability for [tex]\(z = -1.0\)[/tex]. Since the standard normal table typically provides probabilities for positive z-values, we know that the symmetry of the normal distribution means:
For [tex]\( z = -1.0 \)[/tex], the probability is [tex]\(0.1587\)[/tex].
This means the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is approximately [tex]\(15.87\%\)[/tex].
4. Select the correct choice:
Given the options: [tex]$16 \%$[/tex], [tex]$32 \%$[/tex], [tex]$34 \%$[/tex], [tex]$84 \%$[/tex], the closest to [tex]\(15.87\%\)[/tex] is:
[tex]\[
16\%
\][/tex]
Therefore, the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is [tex]\(\boxed{16\%}\)[/tex].
