FINAL ASSIGNMNET
Statistics
UU-MTH-2000
,1. a) Mode: The mode is 11 with the highest frequency of 12.
Median
n
−cf
2
x=L+ (w)
f
n 40
Median class position = = = 20
2 2
The median class = [16 – 18] and for L = 16, W= 18 – 16 = 2, N = 20
f = 40, cf = 5
40
−5
2
¿ 16+ ∗2
20
15
¿ 16+ ∗2
20
30
¿ 16+
20
= 17.5
Mean
∑ k 1 fiMi
i=¿¿
μ=
∑ k 1 fi
i=¿¿
f1M
¿
1 +f k Mk
f 1+ f k
5∗5+7∗8+ 12∗11+10∗14+ 6∗17
¿
5+7+ 12+ 10+ 6
455
=
40
=11.38
, 1 b) Range
For the grouped data, the range is the difference between the upper
boundary of the class having the maximum value and the lower
boundary of the class having the minimum value.
18+19
The upper boundary is: 20 = 18.5
4+ 3
The lower boundary is: 2 = 3.5
Therefore,
Range = 18.5 – 3.5 = 22
Statistics
UU-MTH-2000
,1. a) Mode: The mode is 11 with the highest frequency of 12.
Median
n
−cf
2
x=L+ (w)
f
n 40
Median class position = = = 20
2 2
The median class = [16 – 18] and for L = 16, W= 18 – 16 = 2, N = 20
f = 40, cf = 5
40
−5
2
¿ 16+ ∗2
20
15
¿ 16+ ∗2
20
30
¿ 16+
20
= 17.5
Mean
∑ k 1 fiMi
i=¿¿
μ=
∑ k 1 fi
i=¿¿
f1M
¿
1 +f k Mk
f 1+ f k
5∗5+7∗8+ 12∗11+10∗14+ 6∗17
¿
5+7+ 12+ 10+ 6
455
=
40
=11.38
, 1 b) Range
For the grouped data, the range is the difference between the upper
boundary of the class having the maximum value and the lower
boundary of the class having the minimum value.
18+19
The upper boundary is: 20 = 18.5
4+ 3
The lower boundary is: 2 = 3.5
Therefore,
Range = 18.5 – 3.5 = 22
