Answer:
(1) Variance = 4.5 and Standard deviation = 2.121.
(2) Variance = 4.5 and Standard deviation = 2.121.
(3) The effect on the measure of dispersion if each value is changed uniformly is that it remains unchanged.
Step-by-step explanation:
We are given with the following set of data below;
X [tex]X-\bar X[/tex] [tex](X-\bar X)^{2}[/tex]
5 5 - 8 = -3 9
5 5 - 8 = -3 9
8 8 - 8 = 0 0
10 10 - 8 = 2 4
10 10 - 8 = 2 4
10 10 - 8 = 2 4
9 9 - 8 = 1 1
9 9 - 8 = 1 1
6 6 - 8 = -2 4
Total 72 36
Firstly, the mean of the above data is given by;
Mean, [tex]\bar X[/tex] = [tex]\frac{\sum X}{n}[/tex]
= [tex]\frac{72}{9}[/tex] = 8
(1)Now, the variance of the given data is;
Variance = [tex]\frac{\sum (X-\bar X)^{2} }{n-1}[/tex]
= [tex]\frac{36}{9-1}[/tex] = 4.5
So, the standard deviation, (S.D.) = [tex]\sqrt{\text{Variance}}[/tex]
= [tex]\sqrt{4.5}[/tex] = 2.12
(2) Now, each value is added by 2; so the new data set is given by;
X [tex]X-\bar X[/tex] [tex](X-\bar X)^{2}[/tex]
7 7 - 10 = -3 9
7 7 - 10 = -3 9
10 10 - 10 = 0 0
12 12 - 10 = 2 4
12 12 - 10 = 2 4
12 12 - 10 = 2 4
11 11 - 10 = 1 1
11 11 - 10 = 1 1
8 8 - 10 = -2 4
Total 90 36
Firstly, the mean of the above data is given by;
Mean, [tex]\bar X[/tex] = [tex]\frac{\sum X}{n}[/tex]
= [tex]\frac{90}{9}[/tex] = 10
(1)Now, the variance of the given data is;
Variance = [tex]\frac{\sum (X-\bar X)^{2} }{n-1}[/tex]
= [tex]\frac{36}{9-1}[/tex] = 4.5
So, the new standard deviation, (S.D.) = [tex]\sqrt{\text{Variance}}[/tex]
= [tex]\sqrt{4.5}[/tex] = 2.12
(3) The effect on the measure of dispersion if each value is changed uniformly is that it remains unchanged as we see in the case of variance or standard deviation.
