Question

Complete the 2nd table below by supplementing the data from the first table. Create 13 classes/groups using
7 as the class interval. The lowest value of the first class/group is the lowest value of the given set of data.
Thus the class mark of the first class is 10. The last class, class 13th, is from 91 to 97.

This data is a sample. The variance of this sample grouped data is 764.4   

Find the following: Mean using Short Method, Mode, SKP, SKB, SKK, Five number summary (Lowest and
highest value from the ungrouped data and Quartiles of the grouped data), normality and the outliers.




















X Values
25 38 44 94
65 61 55 92
56 65 21 78
22 60 88 87
18 44 41 8
96 10 32 8
42 84 8 44
96 54 21 63
7 63 25 40
47 87 39 80

Answer 1

SOLUTION:-

Class Marks	Class	Obs.
10	7-13	8,8,10,8,7
17	14-20	18
24	21-27	25,21,22,21,25
31	28-34	32
38	35-41	38,41,40,39
45	42-48	44,44,42,44,47
52	49-55	55,54
59	56-62	61,56,60
66	63-69	65,65,63,63
73	70-76
80	77-83	78,80
87	84-90	88,87,84,87
94	91-97	94,92,96,96

> x
[1] 25 38 44 94 65 61 55 92 56 65 21 78 22 60 88 87 18 44 41 8 96 10 32 8 42 84 8 44 96 54 21 63 7 63 25 40 47 87 39 80
> mean(x)
[1] 50.2
> median(x)
[1] 45.5
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
7.00 25.00 45.50 50.20 68.25 96.00
> table(x)
x
7 8 10 18 21 22 25 32 38 39 40 41 42 44 47 54 55 56 60 61 63 65 78 80 84 87 88 92 94 96
1 3 1 1 2 1 2 1 1 1 1 1 1 3 1 1 1 1 1 1 2 2 1 1 1 2 1 1 1 2
> mode=8 ##observation with high frequency
> skp=3*(mean(x)-median(x))/sd(x);skp
[1] 0.5044433
> qqnorm(x)
> qqline(x)

> s=shapiro.test(x);s

Shapiro-Wilk normality test

data: x
W = 0.9444, p-value = 0.04876

> ifelse(s$p.value<0.05,"sample is not normal","sample is normally distributed")
[1] "sample is not normal"

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