Questions: X follows a normal distribution with mean 3 and variance 25, the probability of X being larger than 10 is (A) 0.92 (B) 0.08 (C) 0.61 (D) 0.39

Solution Steps

Let \( X \) be a normally distributed random variable with mean \( \mu = 3 \) and variance \( \sigma^2 = 25 \). The standard deviation is given by:

\[ \sigma = \sqrt{25} = 5 \]

To find the probability that \( X \) is greater than 10, we first calculate the Z-score for \( X = 10 \):

\[ Z = \frac{X - \mu}{\sigma} = \frac{10 - 3}{5} = \frac{7}{5} = 1.4 \]

We need to find \( P(X > 10) \), which can be expressed in terms of the cumulative distribution function \( \Phi \):

\[ P(X > 10) = 1 - P(X \leq 10) = 1 - \Phi(1.4) \]

From the calculations, we have:

\[ P(X \leq 10) = \Phi(1.4) \approx 0.9192 \]

Thus, the probability that \( X \) is greater than 10 is:

\[ P(X > 10) = 1 - 0.9192 = 0.0808 \]

Final Answer