Suppose the waiting times for patients needing emergency service are normally distributed with a mean of $\mu =12$ minutes and a standard deviation of $\sigma =3$ minutes. What proportion of patients wait $9$ minutes or less?

• A. $0.9772$
• B. $0.8413$
• C. $0.1587$
• D. $0.5$

Answer: (B)

Verified step by step guidance

1. Identify the given parameters: the waiting times are normally distributed with a mean \( \mu = 12 \) minutes and a standard deviation \( \sigma = 3 \) minutes.
2. Define the problem: we want to find the proportion of patients who wait 9 minutes or less. This means we are looking for \( P(X \leq 9) \), where \( X \) is the waiting time.
3. Convert the raw score (9 minutes) to a standard normal score (z-score) using the formula: \[ z = \frac{X - \mu}{\sigma} \] Substitute the values to get: \[ z = \frac{9 - 12}{3} \]
4. Use the standard normal distribution table (z-table) or a calculator to find the cumulative probability corresponding to the calculated z-score. This gives the proportion of patients waiting 9 minutes or less.
5. Interpret the result: the value obtained from the z-table is the proportion of patients whose waiting time is at most 9 minutes.