Question

x is a normally distributed random variable with mean 73 and standard deviation 8. what is the probability that x is between 49 and 89? 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 $x = 49$, $z_1=\frac{49 - 73}{8}=\frac{- 24}{8}=-3$.
For $x = 89$, $z_2=\frac{89 - 73}{8}=\frac{16}{8}=2$.

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that:

• Approximately 68% of the data lies within 1 standard deviation of the mean ($z=\pm1$).
• Approximately 95% of the data lies within 2 standard deviations of the mean ($z = \pm2$).
• Approximately 99.7% of the data lies within 3 standard deviations of the mean ($z=\pm3$).

The probability that $z$ is between - 3 and 2 is the probability that $z$ is between - 3 and 3 minus the probability that $z$ is between 2 and 3.
The probability that $z$ is between - 3 and 3 is 0.997.
The probability that $z$ is between - 2 and 2 is 0.95. So the probability that $z$ is between 2 and 3 is $\frac{0.997 - 0.95}{2}=0.0235$.
The probability that $z$ is between - 3 and 2 is $0.997-0.0235 = 0.9735\approx0.974$.

Answer:

0.974