Rewrite the interval in terms of the standard deviations.
81.5 %
Practice makes perfect
We are told that the upper-arm length of adult males in the U.S. is normally distributed with a mean of μ=39.4cm and a standard deviation of σ=2.3cm. We will find the percent of adult males that have an upper-arm length between 34.8 cm and 41.7cm.
34.8 cmNext, we will divide the difference by the standard deviation, 2.3cm.
4.6cm/2.3cm=2
This means that 34.8 cm is 2 standard deviations below the mean. We will do the same thing for the upper limit. This time, we will subtract the mean 39.4cm from the upper limit 41.7cm.
41.7cm- 39.4cm= 2.3cm
Then, once again, we will divide the difference by the standard deviation.
2.3cm/2.3cm=1
The upper limit is 1 standard deviation above the mean. As a result, we will shade the region between 2 standard deviations below and 1 standard deviation above the mean.
Finally, we can calculate the percentage in this region by adding the percents of the individual regions.
