Question

1.The trend of thinner beauty pageant winners has generated charges that the contest encourages unhealthy diet habits among young women. Listed below are body mass indexes​ (BMI) for beauty pageant winners from two different time periods. Find the coefficient of variation for each of the two sets of​ data, then compare the variation.

BMI​ (from the 1920s and​ 1930s):	20.4	21.9	22.1	22.2	20.3	18.7	18.9	19.4	18.5	19.1
BMI​ (from recent​ winners):	19.5	20.3	19.7	20.1	17.7	17.9	19.1	18.7	17.7	16.8

a) The coefficient of variation for the​ BMI's of beauty pageant winners from the 1920s and 1930s is-------% ?

b)The coefficient of variation for the​ BMI's of recent beauty pageant winners is-----%?

c) Is there a difference in variation between the two data sets ?

2. Because the mean is very sensitive to extreme​ values, it is not a resistant measure of center. The trimmed mean is more resistant. To find the​ 10% trimmed mean for a data​ set, first arrange the data in​ order, then delete the bottom​ 10% of the values and the top​ 10% of the​ values, then calculate the mean of the remaining values. For the following​ credit-rating scores.

706	715	784	809	799
795	716	680	767	609
695	832	770	329	653
564	737	789	704	755

a. The mean is......?

(b) the​ 10% trimmed​ mean is.....?

​ (c) the​ 20% trimmed mean is...?

d) How do the results​ compare?

Answer 1

1)

A)

X	(X - X̄)²
20.4	0.06
21.9	3.06
22.1	3.80
22.2	4.20
20.3	0.02
18.7	2.10
18.9	1.563
19.4	0.563
18.5	2.723
19.1	1.102

	X	(X - X̄)²
total sum	201.5	19.21
n	10	10

mean =    ΣX/n =    201.500   /   10   =   20.15
                            
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (19.205/9)   =       1.46
coefficient of variation,CV=σ/µ=   0.07

The coefficient of variation for the​ BMI's of beauty pageant winners from the 1920s and 1930s is 7%

B)

X	(X - X̄)²
19.5	0.56
20.3	2.40
19.7	0.90
20.1	1.82
17.7	1.10
17.9	0.72
19.1	0.123
18.7	0.002
17.7	1.102
16.8	3.802

	X	(X - X̄)²
total sum	187.5	12.55
n	10	10

mean =    ΣX/n =    187.500   /   10   =   18.75
                            
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (12.545/9)   =       1.18
coefficient of variation,CV=σ/µ=   0.06

The coefficient of variation for the​ BMI's of recent beauty pageant winners iS 6%

c) Is there a difference in variation between the two data sets ?

difference in variation is of 1%

..............................

2)

a)

	X
total sum	14208
n	20

mean =    ΣX/n =    14208.000   /   20   =   710.40

b)

trimmed percent=   10.00%
n=   20
n*trimmed percent= 2

2 values will be removed from top and bottom

so now data set---------

10% TRIMMED
609
653
680
695
704
706
715
716
737
755
767
770
784
789
795
799

	X
total sum	11674
n	16

10% trimmed mean= 11674 / 16 = 729.63

c)

trimmed percent=   20.00%
n=   20
n*trimmed percent=   4
so 4 values will be remove from top and bottom

20% TRIMMED
680
695
704
706
715
716
737
755
767
770
784
789

	X
total sum	8818
n	12

20% trimmed mean = 8818 / 12 =734.833

d)

mean =   710.40
10% trimmed mean   = 729.63

20% trimmed mean =734.833

mean < 10% trimmed mean <20% trimmed mean

10% trimmed mean has been increased drastically due to removal of lower bound extreme value.