2.8 Measuring the spread of the data (modified r. bloom) (Page 3/6)

Collaborative statistics: custom Page 3 / 6

At an elementary school, a teacher was interested in the average age and the standard deviation of the ages of the students in the fifth grade. The following data are the ages for a SAMPLE of n=20 fifth grade students. The ages are rounded to the nearest half year:

9
9.5
9.5
10
10
10
10
10.5
10.5
10.5
10.5
11
11
11
11
11
11
11.5
11.5
11.5

\bar{x} = \frac{9 + 9.5 \times 2 + 10 \times 4 + 10.5 \times 4 + 11 \times 6 + 11.5 \times 3}{20} = 10.525

The sample average age is 10.53 years, rounded to 2 places.

The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating $s$ .

Data	Freq.	Deviations	${Deviations}^{2}$	(Freq.)( ${Deviations}^{2}$ )
$x$	$f$	$(x - x)$	${(x - \bar{x})}^{2}$	$(f) {(x - \bar{x})}^{2}$
$9$	$1$	$9 - 10.525 = - 1.525$	${(- 1.525)}^{2} = 2.325625$	$1 \times 2.325625 = 2.325625$
$9.5$	$2$	$9.5 - 10.525 = - 1.025$	${(- 1.025)}^{2} = 1.050625$	$2 \times 1.050625 = 2.101250$
$10$	$4$	$10 - 10.525 = - 0.525$	${(- 0.525)}^{2} = 0.275625$	$4 \times .275625 = 1.1025$
$10.5$	$4$	$10.5 - 10.525 = - 0.025$	${(- 0.025)}^{2} = 0.000625$	$4 \times .000625 = .0025$
$11$	$6$	$11 - 10.525 = 0.475$	${(0.475)}^{2} = 0.225625$	$6 \times .225625 = 1.35375$
$11.5$	$3$	$11.5 - 10.525 = 0.975$	${(0.975)}^{2} = 0.950625$	$3 \times .950625 = 2.851875$

The sample variance $s^{2}$ is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 - 1):

$s^{2} = \frac{9.7375}{20 - 1} = 0.5125$

The sample standard deviation $s$ is equal to the square root of the sample variance:

$s = \sqrt{0.5125} = 0.715891$ Rounded to two decimal places, $s = 0.72$

Typically, you do the calculation for the standard deviation on a calculator or computer. Also note that we only rounded at the end of the calculation. The intermediate results are not rounded; this is done for accuracy.

Verify the mean and standard deviation calculated above using a calculator or computer.

For the TI-83,83+,84+, enter data into the list editor.
Put the data values in list L1 and the frequencies in list L2.
STAT CALC 1-VarStats L1, L2
$x$ =10.525
Use Sx because this is sample data (not a population): Sx=.715891

For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation)
For a sample: $x$ = $x$ + (#ofSTDEVs)(s)
For a population: $x$ = $μ$ + (#ofSTDEVs)( $σ$ )
For this example, use $x$ = $x$ + (#ofSTDEVs)(s) because the data is from a sample

Find the value that is 1 standard deviation above the mean. Find $(\bar{x} + 1 s)$ .

$(\bar{x} + 1 s) = 10.53 + (1) (0.72) = 11.25$

Find the value that is two standard deviations below the mean. Find $(\bar{x} - 2 s)$ .

$(\bar{x} - 2 s) = 10.53 - (2) (0.72) = 9.09$

Find the values that are 1.5 standard deviations from (below and above) the mean.

$(\bar{x} - 1.5 s) = 10.53 - (1.5) (0.72) = 9.45$
$(\bar{x} + 1.5 s) = 10.53 + (1.5) (0.72) = 11.61$

Explanation of the standard deviation calculation shown in the table

The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11. The deviations 0.97 and 0.47 indicate that. A positive deviation occurs when the data value is greater than the mean. A negative deviation occurs when the data value is less than the mean; the deviation is -1.525 for the data value 9. If you add up all the deviations, the sum is always zero ; the positive and negative deviations offset each other. (For this example, there are n=20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, they become positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.

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Source: OpenStax, Collaborative statistics: custom version modified by r. bloom. OpenStax CNX. Nov 15, 2010 Download for free at http://legacy.cnx.org/content/col10617/1.4

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