Let x denote the time taken to run a road race. Suppose x is approximately normally distributed with a mean of 190 minutes a a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete road race in 224 to 236 minutes? Round your answer to four decimal places.

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**Educational Resource on Normal Distribution**

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**Probability Calculation for Time to Complete a Road Race**

Let \( x \) denote the time taken to run a road race. Suppose \( x \) is approximately normally distributed with a mean (\( \mu \)) of 190 minutes and a standard deviation (\( \sigma \)) of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race in 224 to 236 minutes?

**Steps to Solve:**

1. **Standardize the Values**: Convert the given times (224 and 236 minutes) into their respective z-scores using the formula:
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   where:
   - \( x \) is the value from the dataset,
   - \( \mu \) is the mean,
   - \( \sigma \) is the standard deviation.

2. **Calculate Z-scores**:
   - For \( x = 224 \) minutes:
     \[
     z = \frac{224 - 190}{21} \approx 1.62
     \]
   - For \( x = 236 \) minutes:
     \[
     z = \frac{236 - 190}{21} \approx 2.19
     \]

3. **Find the Probability**:
   - Use a standard normal distribution table or a calculator to find the probabilities corresponding to \( z = 1.62 \) and \( z = 2.19 \).
   - Let \( P(Z < 1.62) \) and \( P(Z < 2.19) \) be these probabilities.

4. **Compute the Desired Probability**:
   - The probability that the runner completes the race between 224 and 236 minutes is:
     \[
     P(224 < x < 236) = P(Z < 2.19) - P(Z < 1.62)
     \]

**Note:** Round your answer to four decimal places.

**Reference Table for Z-scores:**

| Z-Score   | Probability |
|-----------|--------------|
| 1.62      | To be looked from statistical table/calculator |
| 2.19      | To be looked from statistical table/calculator |

**Interactive Element:**

- An input box is provided for students to enter their calculated probability value:
Transcribed Image Text:**Educational Resource on Normal Distribution** --- **Probability Calculation for Time to Complete a Road Race** Let \( x \) denote the time taken to run a road race. Suppose \( x \) is approximately normally distributed with a mean (\( \mu \)) of 190 minutes and a standard deviation (\( \sigma \)) of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race in 224 to 236 minutes? **Steps to Solve:** 1. **Standardize the Values**: Convert the given times (224 and 236 minutes) into their respective z-scores using the formula: \[ z = \frac{x - \mu}{\sigma} \] where: - \( x \) is the value from the dataset, - \( \mu \) is the mean, - \( \sigma \) is the standard deviation. 2. **Calculate Z-scores**: - For \( x = 224 \) minutes: \[ z = \frac{224 - 190}{21} \approx 1.62 \] - For \( x = 236 \) minutes: \[ z = \frac{236 - 190}{21} \approx 2.19 \] 3. **Find the Probability**: - Use a standard normal distribution table or a calculator to find the probabilities corresponding to \( z = 1.62 \) and \( z = 2.19 \). - Let \( P(Z < 1.62) \) and \( P(Z < 2.19) \) be these probabilities. 4. **Compute the Desired Probability**: - The probability that the runner completes the race between 224 and 236 minutes is: \[ P(224 < x < 236) = P(Z < 2.19) - P(Z < 1.62) \] **Note:** Round your answer to four decimal places. **Reference Table for Z-scores:** | Z-Score | Probability | |-----------|--------------| | 1.62 | To be looked from statistical table/calculator | | 2.19 | To be looked from statistical table/calculator | **Interactive Element:** - An input box is provided for students to enter their calculated probability value:
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