Question

x is a normally distributed random variable with mean 97 and standard deviation 21. what is the probability that x is greater than 76? use the 0.68-0.95-0.997 rule and write your answer as a decimal. round to the nearest thousandth if necessary.

Explanation:

Step1: Calculate z-score

$z = \frac{X - \mu}{\sigma} = \frac{76 - 97}{21} = -1$

Step2: Apply 0.68-0.95-0.997 rule

The rule states 68% of data lies within $\mu \pm 1\sigma$. So the proportion of data below $\mu - 1\sigma$ is $\frac{1 - 0.68}{2} = 0.16$.

Step3: Find upper tail probability

$P(X > 76) = 1 - 0.16 = 0.84$

Answer:

0.84