Suppose x is a normally distributed random variable with a mean of 5 and a variance of 4. The probability that x is greater than 10.52 is _____.
A) .0029
B) .0838
C) .4971
D) .9971

Question

Quizplus · Accepted Answer

To find the probability that x is greater than 10.52, we first calculate the z-score using the formula $z = \frac{x - \mu}{\sigma}$, where $\mu$ is the mean and $\sigma$ is the standard deviation. Given $\mu = 5$ and variance $= 4$, the standard deviation $\sigma = \sqrt{4} = 2$. So, $z = \frac{10.52 - 5}{2} = \frac{5.52}{2} = 2.76$. Looking up 2.76 in the standard normal distribution table or using a calculator gives a probability in the right tail of approximately .0029.

Suppose X Is a Normally Distributed Random Variable with a Mean