Question

Suppose that you are an elementary school teacher and you are evaluating the reading levels of your students. You find an individual that reads 46.4 word per minute. You do some research and determine that the reading rates for their grade level are normally distributed with a mean of 90 words per minute and a standard deviation of 24 words per minute. At what percentile is the child's reading level (round final answer to one decimal place)?

Answer 1

Answer:

The child's reading level is at the 3.4th percentile.

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.

You do some research and determine that the reading rates for their grade level are normally distributed with a mean of 90 words per minute and a standard deviation of 24 words per minute.

This means that $\mu = 90, \sigma = 24$

You find an individual that reads 46.4 word per minute. At what percentile is the child's reading level?

The percentile is the p-value of Z when X = 46.4, multiplied by 100. So

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

$Z = \frac{46.4 - 90}{24}$

$Z = -1.82$

$Z = -1.82$ has a p-value of 0.034.

0.034*100 = 3.4

The child's reading level is at the 3.4th percentile.