Question

x is a normally distributed random variable with mean 17 and standard deviation 22. what is the probability that x is between 61 and 83? 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 for 61

The z-score formula is $z=\frac{X-\mu}{\sigma}$, where $\mu=17$, $\sigma=22$, $X=61$.
$z_1=\frac{61-17}{22}=\frac{44}{22}=2$

Step2: Calculate z-score for 83

Use the same z-score formula with $X=83$.
$z_2=\frac{83-17}{22}=\frac{66}{22}=3$

Step3: Apply 0.68-0.95-0.997 rule

The rule states 99.7% of data is within $z=\pm3$ (probability 0.997), and 95% within $z=\pm2$ (probability 0.95). The area between $z=2$ and $z=3$ is half the difference of these two values (since we look at the upper tail between 2 and 3).
$\text{Probability}=\frac{0.997-0.95}{2}=\frac{0.047}{2}=0.0235$

Step4: Round to nearest thousandth

Round 0.0235 to three decimal places.
$0.0235\approx0.024$

Answer:

0.024