Chapter 13.2, Problem 41E

To determine

To find: mean, variance and standard deviation of the given data. Then find what percent of data lies within one standard deviation of the mean.

Answer to Problem 41E

Mean $\left(\overline{x}\right)\approx 743.05$ , standard deviation $\left(\sigma \right)\approx 472.34$ , variance $\left({\sigma }^2\right)\approx 223106.85$ and 75% of the data lies within one standard deviation of the mean.

Explanation of Solution

Given information:

The data are 389, 332, 295, 280, 280, 272, 311, 365, 411, 446, 606, 699, 874, 975, 1227, 1572, 1673, 1415, 1269 and 1170.

Concept used:

Steps to find Mean $\left(\overline{x}\right)$ using graphing calculator:

1. For returning home press 2^nd “MODE (QUIT)”
2. Select 2^nd “STAT (LIST)”. Go to the right for “MATH”
3. choose the third option “mean(”
4. Enter the data in L1 and press enter.

Steps to find Standard Deviation $\left(\sigma x\right)$ using graphing calculator:

1. For returning home press 2^nd “MODE (QUIT)”
2. Select 2^nd “STAT (LIST)”. Go to the right for “MATH”
3. choose the seventh option “stdDev(”
4. Enter the data in L1 and press enter.

Also, $\text{Variance}={\left(\text{Standard}\text{deviation}\right)}^2$

Calculation:

Arranging the observations are 272, 280, 280, 295, 311, 332, 365, 389, 411, 446, 606, 699, 874, 975, 1170, 1227, 1269, 1415, 1572, 1673

There are 20 observations.

Using the graphing calculator

Mean $\left(\overline{x}\right)=743.05$

Standard deviation $\left(\sigma \right)=472.3418757$

Variance $\left({\sigma }^2\right)=223106.84753979$

Approximately mean $\left(\overline{x}\right)\approx 743.05$

Standard deviation $\left(\sigma \right)\approx 472.34$

Variance $\left({\sigma }^2\right)\approx 223106.85$

Values within one standard deviation of mean can be calculated as:

Mean $-$ Standard deviation $=743.05-472.3418757=270.7081243\approx 270.71$

Mean $+$ Standard deviation $=743.05+472.3418757=1215.3918757\approx 1215.39$

From the given data there are 15 data points which lies between 270.71 and 1215.39 so the required percentage can be calculated as

  $\frac{15}{20}\times 100=75\%$

Hence 75% of the data lies within one standard deviation of the mean.