Question

suppose a normal distribution has a mean of 79 and a standard deviation of 7. what is ( p(x leq 72) )?

choice

a. 0.84
b. 0.16
c. 0.025
d. 0.975

Explanation:

Step1: Calculate the z - score

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 dataset.
Given $\mu = 79$, $\sigma=7$, and $x = 72$.
Substitute these values into the z - score formula: $z=\frac{72 - 79}{7}=\frac{- 7}{7}=- 1$.

Step2: Use the empirical rule for normal distribution

The empirical rule (68 - 95 - 99.7 rule) states that for a normal distribution:

• Approximately 68% of the data lies within $\mu\pm\sigma$ (i.e., $z=\pm1$), so the area outside of $z=- 1$ to $z = 1$ is $1 - 0.68=0.32$. Since the normal distribution is symmetric about the mean, the area to the left of $z=-1$ (which is $P(X\leq72)=P(Z\leq - 1)$) is half of 0.32, so $\frac{0.32}{2}=0.16$.

Answer:

B. 0.16