To solve the problem of finding the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes, we can use the concepts of normal distribution and z-scores. Here's a step-by-step approach to solving the problem:
### Step 1: Convert Times to Minutes
First, let's convert the given times from hours and minutes to minutes:
- Mean time ([tex]\(\mu\)[/tex]): 3 hours and 50 minutes
[tex]\[
\mu = 3 \times 60 + 50 = 230 \text{ minutes}
\][/tex]
- Runner's time ([tex]\(X\)[/tex]): 3 hours and 20 minutes
[tex]\[
X = 3 \times 60 + 20 = 200 \text{ minutes}
\][/tex]
### Step 2: Identify the Standard Deviation
The given standard deviation ([tex]\(\sigma\)[/tex]) is 30 minutes.
### Step 3: Calculate the Z-score
The z-score represents the number of standard deviations a data point is from the mean. It's calculated by the formula:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
Plugging in the values:
[tex]\[
z = \frac{200 - 230}{30} = \frac{-30}{30} = -1
\][/tex]
### Step 4: Use the Standard Normal Table
The standard normal table provides the cumulative probability associated with a given z-score. For [tex]\(z = -1\)[/tex]:
From the table:
[tex]\[
z = 1.00 \quad \text{corresponds to a probability of} \quad 0.8413
\][/tex]
Since the normal distribution is symmetric:
[tex]\[
P(Z \leq -1) = 1 - P(Z < 1) = 1 - 0.8413 = 0.1587
\][/tex]
However, since the table commonly provides the cumulative probability [tex]\(P(Z \leq z)\)[/tex], for [tex]\(z = -1\)[/tex]:
[tex]\[
P(Z \leq -1) = 0.1587
\][/tex]
Given [tex]\(z = -1\)[/tex] provides probability [tex]\(0.1587 + 0.8413 = 1\)[/tex]
### Step 5: Converting to Percentage
To find the percentage:
[tex]\[
\text{Probability} = 0.8413 \times 100 = 84.13\%
\][/tex]
### Conclusion
The probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is approximately [tex]\(84.13\%\)[/tex]. Among the given options, the closest match is:
[tex]\[
\boxed{84\%}
\][/tex]
