Question

x is a normally distributed random variable with mean 42 and standard deviation 19. what is the probability that x is less than 61? 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

$z = \frac{X - \mu}{\sigma} = \frac{61 - 42}{19} = 1$

Step2: Apply the 68-95-997 rule

The rule states 68% of data lies within $\mu \pm 1\sigma$, so the proportion of data less than $\mu + 1\sigma$ is $\frac{1 + 0.68}{2} = 0.84$

Answer:

0.840