Question

Arm span is the physical measurement of the length of an individual's arms from fingertip to fingertip. A man’s arm span is approximately normally distributed with mean of 70 inches with a standard deviation of 4.5 inches. Find length in inches of the 99th percentile for a man’s arm span. Round answer to 2 decimal places

Answer 1

Answer:

The 99th percentile for a man’s arm span is 80.47 inches.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

$\mu = 70, \sigma = 4.5$

99th percentile

X when Z has a pvalue of 0.99. So X when Z = 2.327.

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

$2.327 = \frac{X - 70}{4.5}$

$X - 70 = 2.327*4.5$

$X = 80.47$

The 99th percentile for a man’s arm span is 80.47 inches.