Question

Raw scores on a certain standardized test one year were normally distributed, with a mean of 156 and a standard deviation of 23. If
48,592 students took the test, about how many of the students scored less than 96?

Answer 1

Answer:

About 220 of the students scored less than 96

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 156 and a standard deviation of 23.

This means that $\mu = 156, \sigma = 23$

Proportion that scored less than 96:

p-value of Z when X = 96. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{96 - 156}{23}$

$Z = -2.61$

$Z = -2.61$ has a p-value of 0.00453.

About how many of the students scored less than 96?

0.00453 out of 48592.

0.00453*48592 = 220.1.

Rounding to the closest integer:

About 220 of the students scored less than 96