Question

in a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 22.5 and a standard deviation of 6.6. complete parts (a) through (d) below.
(a) find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 16.
the probability of a student scoring less than 16 is 0.1623. (round to four decimal places as needed.)
(b) find the probability that a randomly selected high school student who took the reading portion of the test has a score that is between 14.9 and 30.1.
the probability of a student scoring between 14.9 and 30.1 is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 22.5$ is the mean and $\sigma = 6.6$ is the standard deviation.
For $x = 14.9$, $z_1=\frac{14.9 - 22.5}{6.6}=\frac{-7.6}{6.6}\approx - 1.15$.
For $x = 30.1$, $z_2=\frac{30.1 - 22.5}{6.6}=\frac{7.6}{6.6}\approx1.15$.

Step2: Use the standard normal distribution table

We want $P(14.9Since $P(-1.15 < Z < 1.15)=P(Z < 1.15)-P(Z < - 1.15)$.
From the standard - normal table, $P(Z < 1.15)=0.8749$ and $P(Z < - 1.15)=1 - P(Z < 1.15)=1 - 0.8749 = 0.1251$.
So $P(-1.15 < Z < 1.15)=0.8749-0.1251 = 0.7498$.

Answer:

$0.7498$