in a recent year, the scores for the reading portion of a test were normally distributed, with a mean of…

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 0.7498 (round to four decimal places as needed.) (c) find the probability that a randomly selected high school student who took the reading portion of the test has a score that is more than 36.2. the probability of a student scoring more than 36.2 is (round to four decimal places as needed.)

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 0.7498 (round to four decimal places as needed.) (c) find the probability that a randomly selected high school student who took the reading portion of the test has a score that is more than 36.2. the probability of a student scoring more than 36.2 is (round to four decimal places as needed.)

Answer

Explanation:

Step1: Calculate the z - score formula

The z - score is calculated as $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 22.5$ and $\sigma=6.6$.

Step2: Calculate the z - score for $x = 36.2$

$z=\frac{36.2 - 22.5}{6.6}=\frac{13.7}{6.6}\approx2.08$

Step3: Find the probability using the standard normal distribution table

We want $P(X>36.2)$, which is equivalent to $P(Z > 2.08)$ in the standard - normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$. From the standard - normal table, $P(Z\leq2.08) = 0.9812$. So $P(Z > 2.08)=1 - 0.9812=0.0188$

Answer:

$0.0188$