Question

2. Find the grouped means for the ages of male and females listed belovw Male 12 10 Female Age <15 years 15-24 years 25-54 years 55-64 years 65+ years 10 31 6 3. Find the unweighted and weighted mean centers for the following settlements Settlement (x, y) Population 1,000 50,000 2,500 3,000 8,000 (7.8, 2.3) (9.4, 3.3) (8.8, 3.7)

Answer 1

2) solution :

Given data:

Ages	Male	Female
< 15 years	12	10
15-24 years	10	9
24-54 years	33	31
55-64 years	8	8
65+years	6	8

Now, we have to fin dout the:

Grouped means for both male and females :

We already know the formula for the  Grouped mean is,

  $\mu =\frac{\sum mf}{ n}$

Here m = mid point

f = frequency

n = total

Now, Grouped means for males:

Ages	Male (f)	mid point ( m)	mf
< 15 years	12	15/2 = 7.5	7.5 *12= 90
15-24 years	10	(15+24) /2 = 19.5	19.5*10= 195
25-54 years	33	( 25+54 )/2 =39.5	39.5*33 = 1303.5
55-64 years	8	(55+64)/2 = 59.5	59.5*8 = 476
65+years	6	(65+100)/2 = 82.5	82.5*6 = 495
Total	69		$\sum$mf = 2559.5

Now, Grouped mean   $\mu =\frac{\sum mf}{ n}$

= 2559.5 / 69

=37.094

$\therefore$ Grouped mean for males  = 37.094

Now, we fin dout the:

Grouped mean for Females :

Ages	Female (f)	mid point ( m)	mf
< 15 years	10	15/2 = 7.5	7.5*10= 75
15-24 years	9	(15+24) /2 = 19.5	19.5*9 =175.5
24-54 years	31	( 25+54 )/2 =39.5	39.5*31= 1224.5
55-64 years	8	(55+64)/2 = 59.5	59.5* 8 = 476
65+years	8	(65+100)/2 = 82.5	82.5*8 = 660
Total	66		$\sum$mf = 2611

Now, Grouped mean $\mu =\frac{\sum mf}{ n}$

= 2611 / 66

= 39.5606

$\therefore$ Grouped mean for Females  =39.5606

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