Chapter 12.4, Problem 44SR

To determine

To find: the mean, variance and standard deviation of the given set.

Answer to Problem 44SR

The mean is 65, the variance is 238.3 and the standard deviation is 15.4.

Explanation of Solution

Given information:

The given data:

  $76,47,59,47,72,89$

Calculation:

To find the mean, add the number and then divide by the number of values in the data set.

  $\begin{array}{l}\overline{\text{X}}=\frac{76+47+59+47+72+89}{6}\\ =\frac{390}{6}\\ =65\end{array}$

To find the variance, square the difference between each number and the mean. Then add the squares and divide it by number of values.

  $\begin{array}{l}{\sigma }_{}^2=\frac{{\left(76-65\right)}^2+{\left(47-65\right)}^2+{\left(59-65\right)}^2+{\left(47-65\right)}^2+{\left(72-65\right)}^2+{\left(89-65\right)}^2}{6}\\ =\frac{{\left(11\right)}^2+{\left(-18\right)}^2+{\left(-6\right)}^2+{\left(-18\right)}^2+{\left(7\right)}^2+{\left(24\right)}^2}{6}\\ =\frac{121+324+36+324+49+576}{6}\\ =\frac{1430}{6}\\ =238.3\end{array}$

Standard deviation is the square of the variance. To find standard deviation,

  $\begin{array}{l}{\sigma }_{}^2=\frac{1430}{6}\\ \sqrt{{\sigma }_{}^2}=\sqrt{\frac{1430}{6}}\\ \sigma \approx 15.4\left[\text{Use a calculator}\text{.}\right]\end{array}$

Hence, the mean is 65, the variance is 238.3 and the standard deviation is about 15.4.