Question

in a recent year, the total scores for a certain standardized test were normally distributed, with a mean of 500 and a standard deviation of 10.6. answer parts (a)-(d) below.
(a) find the probability that a randomly selected medical student who took the test had a total score that was less than 491.
the probability that a randomly selected medical student who took the test had a total score that was less than 491 is 0.1977
(round to four decimal places as needed.)
(b) find the probability that a randomly selected medical student who took the test had a total score that was between 500 and 512.
the probability that a randomly selected medical student who took the test had a total score that was between 500 and 512 is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score for \(x = 512\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 500\) (mean) and \(\sigma=10.6\) (standard deviation).
For \(x = 512\), \(z=\frac{512 - 500}{10.6}=\frac{12}{10.6}\approx1.13\)

Step2: Find the probability using the standard normal distribution

The probability that \(z<1.13\) is \(P(Z < 1.13)\). Looking up in the standard - normal table, \(P(Z < 1.13)=0.8708\)
The probability that \(z < 0\) (since for \(x = 500\), \(z=\frac{500 - 500}{10.6}=0\)) is \(P(Z < 0)=0.5\)
The probability that \(0\(P(0 < Z < 1.13)=0.8708 - 0.5\)

Answer:

\(0.3708\)