Question

x is a normally distributed random variable with mean 83 and standard deviation 14. what is the probability that x is between 41 and 55? 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 $\mu$ is the mean and $\sigma$ is the standard deviation.
For $x = 41$, $z_1=\frac{41 - 83}{14}=\frac{-42}{14}=- 3$.
For $x = 55$, $z_2=\frac{55 - 83}{14}=\frac{-28}{14}=-2$.

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that about 95% of the data lies within 2 standard deviations of the mean and about 99.7% of the data lies within 3 standard deviations of the mean.
The proportion of data between $z=-3$ and $z = - 2$ is $\frac{0.997 - 0.95}{2}$.
$0.997$ is the proportion of data within 3 standard - deviations ($z=-3$ to $z = 3$) and $0.95$ is the proportion of data within 2 standard - deviations ($z=-2$ to $z = 2$).
$\frac{0.997 - 0.95}{2}=\frac{0.047}{2}=0.0235\approx0.024$.

Answer:

$0.024$