Question

According to survey​ data, the distribution of arm spans for males is approximately Normal with a mean of 71.4 inches and a standard deviation of 3.1 inches. a. What percentage of men have arm spans between 66 and 76 ​inches? b. A particular professional basketball player has an arm span of almost 89 inches. Find the​ z-score for this​ person's arm span. What percentage of males have an arm span at least as long as this​ person?

Answer 1

Answer:

a) 89.97% of men have arm spans between 66 and 76 ​inches.

b) The z-score for this​ person's arm span is 5.68. 0% of males have an arm span at least as long as this​ person

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 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.

Mean of 71.4 inches and a standard deviation of 3.1 inches.

This means that $\mu = 71.4, \sigma = 3.1$

a. What percentage of men have arm spans between 66 and 76 ​inches?

The proportion is the pvalue of Z when X = 76 subtracted by the pvalue of Z  when X = 66. The percentage is the proportion multiplied by 100.

X = 76

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

$Z = \frac{76 - 71.4}{3.1}$

$Z = 1.48$

$Z = 1.48$ has a pvalue of 0.9306

X = 66

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

$Z = \frac{66 - 71.4}{3.1}$

$Z = -1.74$

$Z = -1.74$ has a pvalue of 0.0409

0.9306 - 0.0409 = 0.8997

0.8997*100% = 89.97%

89.97% of men have arm spans between 66 and 76 ​inches.

b. A particular professional basketball player has an arm span of almost 89 inches. Find the​ z-score for this​ person's arm span. What percentage of males have an arm span at least as long as this​ person?

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

$Z = \frac{89 - 71.4}{3.1}$

$Z = 5.68$

The z-score for this​ person's arm span is 5.68.

Z = 5.68 has a pvalue of 1

1 - 1 = 0

0% of males have an arm span at least as long as this​ person