Question

x is a normally distributed random variable with mean 57 and standard deviation 12. what is the probability that x is between 33 and 69? 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-score for 33

$z_1 = \frac{33 - 57}{12} = -2$

Step2: Calculate z-score for 69

$z_2 = \frac{69 - 57}{12} = 1$

Step3: Apply empirical rule values

• 95% of data lies within $z=\pm2$, so $\frac{0.95}{2}=0.475$ for $z=-2$ to $\mu$
• 68% of data lies within $z=\pm1$, so $\frac{0.68}{2}=0.34$ for $\mu$ to $z=1$

Step4: Sum the two probabilities

$0.475 + 0.34 = 0.815$

Answer:

0.815