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 106. 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
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 491\), \(\mu=500\), and \(\sigma = 10.6\).

$$z=\frac{491 - 500}{10.6}=\frac{-9}{10.6}\approx - 0.85$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt491)\), which is equivalent to \(P(Z\lt - 0.85)\) (since \(X\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\) and \(Z=\frac{X - \mu}{\sigma}\)).
Looking up the value of \(P(Z\lt - 0.85)\) in the standard - normal distribution table (or using a calculator with a normal - distribution function, e.g., in Excel: NORM.S.DIST(-0.85, TRUE)), we get \(P(Z\lt - 0.85)=0.1977\)

Answer:

\(0.1977\)