Questions: If X is normal with μ=100 and σ=10, what is the probability that X exceeds 115?

Solution Steps

We are tasked with finding the probability that a normally distributed random variable \( X \) with mean \( \mu = 100 \) and standard deviation \( \sigma = 10 \) exceeds the value \( 115 \). Mathematically, we want to calculate \( P(X > 115) \).

To find this probability, we first convert the value \( 115 \) into a Z-score using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

Substituting the values:

\[ Z = \frac{115 - 100}{10} = \frac{15}{10} = 1.5 \]

The probability \( P(X > 115) \) can be expressed in terms of the cumulative distribution function \( \Phi \) of the standard normal distribution:

\[ P(X > 115) = 1 - P(X \leq 115) = 1 - \Phi(1.5) \]

From the output, we find:

\[ P(X > 115) = \Phi(\infty) - \Phi(1.5) = 1 - \Phi(1.5) \]

Given that \( \Phi(1.5) \approx 0.9332 \), we can calculate:

\[ P(X > 115) = 1 - 0.9332 = 0.0668 \]

Final Answer