Question

Find the measures of center for following.

Data	Frequency
40 - 44	10
45 - 49	23
50 - 54	12
55 - 59	10
60 - 64	5
65 - 69	4
70 - 74	2
75 - 79	0
80 - 84	1

mode =

median =

mean =  (round to 4 decimal places)

Answer 1

solution:

Class (1)	Frequency (f) (2)	Mid value (x) (3)	d=x-Ah=x-625 A=62,h=5 (4)	f⋅d (5)=(2)×(4)	cf (7)
40 - 44	10	42	-4	-40	10
45 - 49	23	47	-3	-69	33
50 - 54	12	52	-2	-24	45
55 - 59	10	57	-1	-10	55
60 - 64	5	62=A	0	0	60
65 - 69	4	67	1	4	64
70 - 74	2	72	2	4	66
75 - 79	0	77	3	0	66
80 - 84	1	82	4	4	67
---	---	---	---	---	---
	n=67	-----	-----	∑f⋅d=-131

Mean =A+ ∑fd/ n⋅h

=62+-131/67⋅5

=62+-1.9552⋅5

=62+-9.7761

=52.2239

Median
= value of (n/2)^th observation

= value of (67/2)^th observation

= value of 33rd observation

From the column of cumulative frequency cf, we find that the 33rd observation lies in the class 50-54.

∴ The median class is 49.5-54.5.

Now,
∴L=lower boundary point of median class =49.5

∴n=Total frequency =67

∴cf=Cumulative frequency of the class preceding the median class =33

∴f=Frequency of the median class =12

∴c=class length of median class =5

Median M=L+n/2-cf/f⋅c

=49.5+33-33/12⋅5

=49.5+012⋅5

=49.5+0

=49.7083

Mode
Here, maximum frequency is 23.

∴ The mode class is 44.5-49.5.

∴L=lower boundary point of mode class =44.5

∴f1= frequency of the mode class =23

∴f0= frequency of the preceding class =10

∴f2= frequency of the succedding class =12

∴c= class length of mode class =5

Z=L+(f₁-f₀ / 2⋅f₁-f₀-f₂)⋅c

=44.5+(23-10 / 2⋅23-10-12)⋅5

=44.5+(13/24)⋅5

=44.5+2.7083

=47.2083