Question

x is a normally distributed random variable with mean 60 and standard deviation 6.
what is the probability that x is less than 42?
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

$z = \frac{X - \mu}{\sigma} = \frac{42 - 60}{6} = -3$

Step2: Apply 0.68-0.95-0.997 rule

The rule states 99.7% of data lies within $\mu \pm 3\sigma$. So the remaining data outside is $1 - 0.997 = 0.003$. Since we want the probability for values less than $\mu - 3\sigma$, we take half of this remaining area: $\frac{0.003}{2} = 0.0015$

Step3: Round to nearest thousandth

$0.0015 \approx 0.002$

Answer:

0.002