Question

1. on a recent statewide math test, the raw score average was 56 points with a standard deviation of 18. if the scores were normally distributed and 24,000 students took the test, answer the following questions. (a) what percent of students scored below a 38 on the test? (b) how many students scored less than a 38? (c) if the state would like to scale the test so that a 90% would correspond to a raw score that is one and a half standard deviations above the mean, what raw score is needed for a 90%? (d) what should the raw passing score be set at so that no more than the 550 students fail?

Explanation:

Part (a)

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 38$, $\mu=56$, and $\sigma = 18$.
$z=\frac{38 - 56}{18}=\frac{- 18}{18}=-1$

Step2: Find the percentage from z - table

For a z - score of $z=-1$, we look up the value in the standard normal distribution table. The area to the left of $z = - 1$ (which represents the percentage of students with scores less than 38) is 0.1587 or 15.87%.

Step1: Use the percentage from part (a)

From part (a), we know that 15.87% of the students scored less than 38.

Step2: Calculate the number of students

The total number of students is $N = 24000$. The number of students with scores less than 38 is $0.1587\times24000$.
$0.1587\times24000 = 3808.8\approx3809$

Step1: Recall the formula for raw score

The formula for the raw score $x$ in a normal distribution is $x=\mu + z\sigma$. We know that $\mu = 56$, $z = 1.5$ (one and a half standard deviations above the mean), and $\sigma=18$.

Step2: Calculate the raw score

$x=56+1.5\times18$
$x = 56 + 27$
$x=83$

Answer:

15.87%

Part (b)