Question

x is a normally distributed random variable with mean 84 and standard deviation 24. what is the probability that x is between 12 and 60? 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

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 84$ is the mean and $\sigma = 24$ is the standard deviation.
For $x = 12$: $z_1=\frac{12 - 84}{24}=\frac{-72}{24}=-3$.
For $x = 60$: $z_2=\frac{60 - 84}{24}=\frac{-24}{24}=-1$.

Step2: Apply the 0.68 - 0.95 - 0.997 rule

The 0.68 - 0.95 - 0.997 rule states that about 68% of the data lies within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations.
The area between $z=-3$ and $z = - 1$ can be found as follows:
The area between $z=-3$ and $z = 3$ is 0.997. The area between $z=-1$ and $z = 1$ is 0.68.
The area between $z=-3$ and $z=-1$ is $\frac{0.997-0.68}{2}=0.1585$.

Answer:

0.159