Chapter B.2, Problem 32E

To determine

To find: The mean, variance and standard deviation of the data using graphing utility and the percent of data lie within one standard deviation of the mean.

Answer to Problem 32E

Mean = 753.7917

Variance = 227009.7

Standard Deviation = 476.4554

The percent of data lie within one standard deviation of the mean is 75%.

Explanation of Solution

Given information: Average prices of gold (in dollars per ounce) for the years 1994 through 2017.

Formula used: Mean ( $\overline{x_i}$ ) $=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}$

Standard deviation ( $\sigma$ ) $=\sqrt{\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x_i}\right)}^2}}{n-1}}$

Graph:

  

Interpretation:

Mean = 753.7917

Variance = 227009.7

Standard Deviation = 476.4554

The interval that contain all numbers that lie within one standard deviation of the mean is

  $\begin{array}{l}\left[753.7917-1\left(476.4554\right),753.7917+1\left(476.4554\right)\right]\\ =\left[277.3363,1230.2471\right]\end{array}$

From the given data, 18 numbers lie in this interval.

Thus, the percent of data lie within one standard deviation of the mean is 75%.