Question

Let X be a normally distributed random variable with mean \(\mu = 50\) and variance \(\sigma^2 = 9\). The probability that X is greater than \(\mu + \sigma\) is:

• A. 0.5000
• B. 0.8413
• C. 0.3413
• D. 0.1587

Hint:

Exam Tip:
For the standard normal distribution:

• \(P(Z \le 1) \approx 0.8413\).
• \(P(Z > 1) \approx 0.1587\).
• \(P(Z \le -1) \approx 0.1587\).
• \(P(Z > -1) \approx 0.8413\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find \(P(X > \mu + \sigma)\).
Given \(\mu = 50\), \(\sigma = \sqrt{9} = 3\).
So, \(\mu + \sigma = 53\).
We need \(P(X > 53)\).

Step 2: Key Formula or Approach:
Standardize using \(Z = \frac{X - \mu}{\sigma}\).
\(P(X > 53) = P\left(Z > \frac{53 - 50}{3}\right) = P(Z > 1)\).

Step 3: Detailed Explanation:
From the standard normal table, \(P(Z \le 1) = 0.8413\).
So, \(P(Z > 1) = 1 - 0.8413 = 0.1587\).
This is a standard result: the area to the right of \(z = 1\) is approximately 0.1587.

Step 4: Final Answer:
Therefore, option (D) is correct.