Question

x is a normally distributed random variable with mean 30 and standard deviation 11. what is the probability that x is greater than 8? 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 the z-score

The z-score formula is $z = \frac{X - \mu}{\sigma}$, where $\mu=30$, $\sigma=11$, $X=8$.
$z = \frac{8 - 30}{11} = \frac{-22}{11} = -2$

Step2: Apply the 0.68-0.95-0.997 rule

The rule states 95% of data lies within $\mu\pm2\sigma$. So the remaining data outside this range is $1 - 0.95 = 0.05$. This is split equally in the two tails, so the left tail (values less than $\mu-2\sigma$) has $\frac{0.05}{2}=0.025$.

Step3: Find $P(X>8)$

$P(X>8) = 1 - P(X<8) = 1 - 0.025$

Answer:

0.975