Exercise 29 Page P3

We want to find the mean, median, mode, range, and standard deviation of the given data set, as well as whether there are any outliers.

Mean

The mean of a data set is the sum of the values divided by the total number of values in the set. Let's start by calculating the sum of the given values.

85 + 100 + 92 + 36 + 74 + 88 + 92 + 86 + 88 + 82 + 98 + 70 + 78 + 92 + 84 + 100 = 1345

There are

16

values in our set, so we have to divide the sum by

16 .

Mean : \frac{1 3 4 5}{1 6} = 84.0625

Median

When the data are arranged in numerical order, the median is the middle value — or the mean of the two middle values — in a set of data. Let's arrange the given values and find the median.

36, 70, 74, 78, 82, 84, 85, 86 ∣ 88, 88, 92, 92, 92, 98, 100, 100

Since there are

16

values, there is no one middle value. Thus, for this exercise, the median is the mean of the two middle values.

Median : \frac{8 6 + 8 8}{2} = 87

Mode

The mode is the value or values that appear most often in a set of data. Let's find the mode of the given values.

85, 100, 92, 36, 74, 88, 92, 86 88, 82, 98, 70, 78, 92, 84, 100

The value that appears most often is

92 .

Mode : 92

Range

The range is the difference between the greatest and least values in a set of data. For our exercise, the greatest value is

100

and the least value is

36 .

Range : 100 - 36 = 64

Standard Deviation

The standard deviation of a set of data is the average amount by which each individual value deviates or differs from the mean.

\underline{Standard Deviation} \frac{( x _{1} - μ ) ^{2} + ( x _{2} - μ ) ^{2} + \dots + ( x _{n} - μ ) ^{2}}{n}

In the above formula,

x_{1}, \dots, x_{n}

are the values of the set of data,

μ

is the mean, and

n

is the number of values. For this exercise, the mean is the sum of the values

1345

divided by the number of values

16 .

μ = \frac{1 3 4 5}{1 6} = 84.0625

Let's use this value and apply the formula to each value in the set.

$x_{n}$	$x_{n} - μ$	$(x_{n} - μ)^{2}$
$85$	$85 - 84.0625 = 0.9375$	$0.937 5^{2} \approx 0.9$
$100$	$100 - 84.0625 = 15.9375$	$15.937 5^{2} \approx 254$
$92$	$92 - 84.0625 = 7.9375$	$7.937 5^{2} \approx 63$
$36$	$36 - 84.0625 = - 48.0625$	$(- 48.0625)^{2} \approx 2310$
$74$	$74 - 84.0625 = - 10.0625$	$(- 10.0625)^{2} \approx 101.3$
$88$	$88 - 84.0625 = 3.9375$	$3.937 5^{2} \approx 15.5$
$92$	$92 - 84.0625 = 7.9375$	$7.937 5^{2} \approx 63$
$86$	$86 - 84.0625 = 1.9375$	$1.937 5^{2} \approx 3.8$
$88$	$88 - 84.0625 = 3.9375$	$3.937 5^{2} \approx 15.5$
$82$	$82 - 84.0625 = - 2.0625$	$(- 2.0625)^{2} \approx 4.3$
$98$	$98 - 84.0625 = 13.9375$	$83.937 5^{2} \approx 194.3$
$70$	$70 - 84.0625 = - 14.0625$	$(- 14.0625)^{2} \approx 197.8$
$78$	$78 - 84.0625 = - 6.0625$	$(- 6.0625)^{2} \approx 36.8$
$92$	$92 - 84.0625 = 7.9375$	$7.937 5^{2} \approx 63$
$84$	$84 - 84.0625 = - 0.0625$	$(- 0.0625)^{2} \approx 0.004$
$100$	$100 - 84.0625 = 15.9375$	$15.937 5^{2} \approx 254$
Sum of Values		$\approx 3577.204$

Finally, we need to divide by

16

and then calculate the square root.

Standard Deviation : \frac{3 5 7 7 . 2 0 4}{1 6} \approx 14.95

Outliers

To find the outliers, let's start by defining some important characteristics of data sets.

Lower Quartile $(Q_{1})$ is the median of the lower half of the data set.
Upper Quartile $(Q_{3})$ is the median of the upper half of the data set.
Interquartile Range is the difference between the upper and lower quartiles $(Q_{3} - Q_{1}) .$

Let's find the upper quartile, the lower quartile, and the interquartile range for the given data set. Do not forget to write the values in numerical order!

36, 70, 74, 78, 82, 84, 85, 86 ∣ 88, 88, 92, 92, 92, 98, 100, 100

We have two middle values for each half. Thus, we need to calculate the mean of those middle values.

Upper Quartile : Lower Quartile : Interquartile Range : \frac{9 2 + 9 2}{2} = 92 \frac{7 8 + 8 2}{2} = 80 92 - 80 = 12

An outlier is an extremely high or extremely low value when compared with the rest of the values in the set. To check for them, we look for data values that are beyond the upper or lower quartiles by more than

1.5

times the interquartile range.

Minimum = Maximum = Q_{1} - I Q R \times 1.5 Q_{2} + I Q R \times 1.5

Let's find the minimum and maximum for a value not to be an outlier.

Minimum 80 - 1.5 \times 12 = 62 Maximum 92 + 1.5 \times 12 = 110

Any value between

62

and

110

is not an outlier. Thus, the data set has only one outlier,

36 .