Question

scores on a test are normally distributed with a mean of 63.2 and a standard deviation of 11.7. find $p_{81}$, which separates the bottom 81% from the top 19%. round to two decimal places.
a. 0.29
b. 66.60
c. 0.88
d. 73.47

Explanation:

Step1: Find the z - score

We know that we want to find the z - score corresponding to a cumulative probability of 0.81. Using a standard normal distribution table (z - table) or a calculator with a normal - distribution function (e.g., invNorm on a TI - 84 Plus), the z - score $z$ such that $P(Z

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value we want to find. We know that $\mu = 63.2$, $\sigma=11.7$, and $z = 0.88$. Rearranging the formula for $x$ gives $x=\mu + z\sigma$.
Substitute the values: $x=63.2+0.88\times11.7$.
First, calculate $0.88\times11.7 = 10.296$.
Then, $x=63.2 + 10.296=73.496\approx73.47$.

Answer:

D. 73.47