Question number 9.

Suppose $X$ is a normal random variable with $\mu=40$ and $\sigma=20$. Find $P(X> -11.4)$.
0.9867
0.9938
0.9949
0.0062
0.0051
None of the above.

Step 1 ：Let's denote \(X\) as a normal random variable with mean \(\mu = 40\) and standard deviation \(\sigma = 20\). We are asked to find the probability \(P(X > -11.4)\).

Step 2 ：First, we calculate the z-score, which is a measure of how many standard deviations an element \(x\) is from the mean. The z-score is calculated as \(z = \frac{x - \mu}{\sigma}\).

Step 3 ：Substituting the given values, we get \(z = \frac{-11.4 - 40}{20} = -2.57\).

Step 4 ：We then find the probability \(P(X < -11.4)\) using the cumulative distribution function (CDF) of the normal distribution. The CDF at \(z\) is the probability that the random variable \(X\) will take a value less than or equal to \(z\).

Step 5 ：Using the z-score of -2.57, we find that \(P(X < -11.4) = 0.0051\).

Step 6 ：Finally, we find \(P(X > -11.4)\) by subtracting \(P(X < -11.4)\) from 1, as the total probability is 1. So, \(P(X > -11.4) = 1 - P(X < -11.4) = 1 - 0.0051 = 0.9949\).

Step 7 ：Final Answer: The probability that the random variable \(X\) is greater than -11.4 is approximately \(\boxed{0.9949}\).