Question

Ch 03: Mastery Assignment - Descriptive Statistics: Numerical Measures < Back to Assignment Attempts: Average: 75 6. Measures of variability Consider a data set containing the following values: 88 96 72 84 95 100 92 90 The smallest value is and the largest value is Therefore, the range is The first quartile is The third quartile is Therefore, the interquartile range (IQR) is The mean of the preceding values is 89.625. The deviations from the mean have been calculated: -1.625 6.375 -17.625 -5.625 5.375 10.375 2.375 0.375 o ir E

Answer 1

x	x-mean
1	72	-17.625	310.641
2	84	-5.625	31.641
3	88	-1.625	2.641
4	90	0.375	0.141
5	92	2.375	5.641
6	95	5.375	28.891
7	96	6.375	40.641
8	100	10.375	107.641
Total	717	0	527.875
Mean	89.625
pop Var	65.984	Sample var	75.411
Pop SD	8.123	Sample SD	8.684

1)

We first arrange the data in ascending order. This would make further calculations easier.

Smallest = 72

Largest = 100

Range = Max - min = 100 -72

Range = 28

The quartile are calculated using the following formula

$Q_{i}=(\frac{i(N+1)}{4})^{th}value$

Quartile	th value
Q1	2.25 (2nd +0.5(3rd -2nd)	85
Q3	6.75	95.75

IQR = Q3 - Q1

IQR = 10.75

For sample we use 'n-1' and for population we use 'n'

Therefore sample values

$SD=\sqrt{Var}=\sqrt{\frac{\sum (x_{i}-\bar{x})^{2}}{n-1}}$

Var = 527.875 / 7

Var = 75.411

SD = 8.684

Coefficient of variation = $\frac{S}{\bar{x}}*100\%$

=8.684 / 89.625

Coefficient of variation = 9.689%

Therefore population values

$SD=\sqrt{Var}=\sqrt{\frac{\sum (x_{i}-\bar{x})^{2}}{n}}$

Var = 527.875 / 8

Var = 65.984

SD = 8.123

Coefficient of variation = $\frac{\sigma}{\bar{x}}*100\%$

=8.123 / 89.625

Coefficient of variation = 9.063%

Measures of variation help us to know if the values are far from each or are more similar. Higher these values, more different values from each other and vice versa.

If the minimum value is increased from 72 to 80 then

• The range will decrease. this is because now the value is closer to the maximum value proof : 100-80=20

• The IQR will not change this is because as seen in the formula IQR does not use the minimum value at all so it is unaffected by the outliers or extreme values

• The variance and SD would decrease. same reason as range. Now that value is 80, the deviation from mean will reduce from 17.625 to 9.625.