Question

x is a normally distributed random variable with mean 53 and standard deviation 22. what is the probability that x is between 31 and 97? 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 - scores

For $x_1 = 31$, $z_1=\frac{31 - 53}{22}=\frac{- 22}{22}=-1$. For $x_2 = 97$, $z_2=\frac{97 - 53}{22}=\frac{44}{22}=2$.

Step2: Apply the 0.68 - 0.95 - 0.997 rule

The 0.68 - 0.95 - 0.997 rule states that the probability within 1 standard deviation of the mean is 0.68, within 2 standard deviations is 0.95 and within 3 standard deviations is 0.997. The probability $P(-1

Answer:

0.815