Question

x is a normally distributed random variable with mean 70 and standard deviation 21. what is the probability that x is between 7 and 91? 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 $\mu = 70$ is the mean and $\sigma = 21$ is the standard deviation.
For $x = 7$, $z_1=\frac{7 - 70}{21}=\frac{- 63}{21}=-3$.
For $x = 91$, $z_2=\frac{91 - 70}{21}=\frac{21}{21}=1$.

Step2: Apply the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that about 68% of the data lies within 1 standard deviation of the mean, about 95% lies within 2 standard deviations, and about 99.7% lies within 3 standard deviations.
The area to the left of $z=-3$ is approximately 0.0015 (since the area within 3 standard deviations of the mean is 99.7%, so the area in the two tails outside of 3 standard - deviations is $1 - 0.997=0.003$, and the area in the left - tail is $\frac{0.003}{2}=0.0015$).
The area to the left of $z = 1$ is approximately $0.5+0.34 = 0.84$ (since the area between $z = 0$ and $z = 1$ is 34% or 0.34 and the area to the left of $z = 0$ is 0.5).

Step3: Calculate the probability

The probability $P(7$P(7

Answer:

0.839