Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
4. Normal Distributions
Continue to next subchapter

Exercise 6 Page 341

Express the interval using the standard deviation.

99.7 %

Practice makes perfect
We have been told that the masses of pennies minted in the U.S. after 1982 are normally distributed with a mean of 2.50g and a standard deviation of 0.02g. We will find the percentages of pennies that have a mass between 2.44g and 2.56g. 2.44< x < 2.56 First, let's find the difference between the mean, 2.50g, and the lower limit, 2.44g, in terms of standard deviations. To do that, we will subtract the lower limit from the mean. 2.50g- 2.44g= 0.06 g

Next, we will divide the difference by the standard deviation, 0.02g. 0.06 g/0.02g=3 This means that 2.44 is 3 standard deviations below the mean. We will do the same thing for the upper limit. This time, we will subtract the mean, 2.50g, from the upper limit, 2.56g. 2.56g- 2.50g= 0.06 g Then, we will divide the difference by the standard deviation, 0.02g. 0.06 g/0.02g=3 The upper limit is also 3 standard deviations above the mean. As a result, the masses of the pennies between 2.44g and 2.56g fall within 3 standard deviations of the mean.

Normally distributed data always contains approximately 99.7 % of the data within 3 standard deviations on both sides.