Question

Assume the upper arm length of males over 20 years old in the United States is approximately Normal with mean 39.2 centimeters (cm) and standard deviation 2.2 cm. Use the 68–95–99.7 rule to answer the given questions. (a) What range of lengths covers almost all, 99.7% , of this distribution? Enter your answers rounded to one decimal place.

Answer 1

Answer:

The interval [32.6 cm, 45.8 cm]

Step-by-step explanation:

According with the 68–95–99.7 rule for the Normal distribution:  If $\large \bar x$  is the mean of the distribution and s the standard deviation, around 68% of the data must fall in the interval

$\large [\bar x - s, \bar x +s]$

around 95% of the data must fall in the interval

$\large [\bar x -2s, \bar x +2s]$ 

around 99.7% of the data must fall in the interval

$\large [\bar x -3s, \bar x +3s]$

So, the range of lengths that covers almost all the data (99.7%) is the interval

[39.2 - 3*2.2, 39.2 + 3*2.2] = [32.6, 45.8]

This means that if we measure the upper arm length of a male over 20 years old in the United States, the probability that the length is between 32.6 cm and 45.8 cm is 99.7%