Question

Heights of fourth graders are Normally distributed with a mean of 52 inches and a standard deviation of 3.5 inches. What percentage of fourth graders are between 50 inches and 56 inches? A. 68.13% B. 27.42% C. 59.01% D. 95.27% E. 86.43%

Answer 1

Answer:

C. 59.01%

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 52, \sigma = 3.5$

What percentage of fourth graders are between 50 inches and 56 inches?

This is the pvalue of Z when X = 56 subtracted by the pvalue of Z when X = 50.

X = 56

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

$Z = \frac{56 - 52}{3.5}$

$Z = 1.14$

$Z = 1.14$ has a pvalue of 0.8729

X = 50

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

$Z = \frac{50 - 52}{3.5}$

$Z = -0.57$

$Z = -0.57$ has a pvalue of 0.2843.

0.8729 - 0.2843 = 0.5886, which is close to 59.01%.

So the correct answer is:

C. 59.01%