Question

x is a normally distributed random variable with mean 71 and standard deviation 21. what is the probability that x is between 8 and 92? 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 8 and 92

The z-score formula is $z = \frac{X - \mu}{\sigma}$, where $\mu=71$, $\sigma=21$.
For $X=8$: $z_1 = \frac{8 - 71}{21} = \frac{-63}{21} = -3$
For $X=92$: $z_2 = \frac{92 - 71}{21} = \frac{21}{21} = 1$

Step2: Apply 0.68-0.95-0.997 rule

• The rule states 99.7% of data is within $z=\pm3$, so $\frac{0.997}{2}=0.4985$ of data is between $z=-3$ and $\mu$.
• The rule states 68% of data is within $z=\pm1$, so $\frac{0.68}{2}=0.34$ of data is between $\mu$ and $z=1$.

Step3: Sum the two probabilities

Add the probabilities for the two intervals: $0.4985 + 0.34 = 0.8385$

Step4: Round to nearest thousandth

$0.8385 \approx 0.839$

Answer:

0.839