Question

x is a normally distributed random variable with mean 79 and standard deviation 5. what is the probability that x is between 72 and 86? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate z - scores for 72 and 86

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the random variable.
For $x = 72$, $\mu=79$, $\sigma = 5$:
$z_1=\frac{72 - 79}{5}=\frac{- 7}{5}=- 1.4$
For $x = 86$, $\mu = 79$, $\sigma=5$:
$z_2=\frac{86 - 79}{5}=\frac{7}{5}=1.4$

Step2: Find the probabilities for the z - scores

We need to find $P(-1.4 < Z < 1.4)$, which is equal to $P(Z < 1.4)-P(Z < - 1.4)$.
Using the standard normal distribution table (or a calculator with normal distribution functions):
$P(Z < 1.4)\approx0.9192$
$P(Z < - 1.4)=1 - P(Z < 1.4)\approx1 - 0.9192 = 0.0808$

Step3: Calculate the probability between the two z - scores

$P(-1.4 < Z < 1.4)=P(Z < 1.4)-P(Z < - 1.4)=0.9192-0.0808 = 0.8384$

Answer:

0.838