Question

approximate the mean of the frequency distribution for the ages of the residents of a town.
the approximate mean age is
(round to one decimal place as needed.)
age
frequency
0 - 9
24
10 - 19
38
20 - 29
15
30 - 39
30
40 - 49
20
50 - 59
48
60 - 69
35
70 - 79
11
80 - 89
9

Explanation:

Step1: Find mid - points

For age group \(0 - 9\), mid - point \(x_1=\frac{0 + 9}{2}=4.5\).
For age group \(10 - 19\), mid - point \(x_2=\frac{10+19}{2}=14.5\).
For age group \(20 - 29\), mid - point \(x_3=\frac{20 + 29}{2}=24.5\).
For age group \(30 - 39\), mid - point \(x_4=\frac{30+39}{2}=34.5\).
For age group \(40 - 49\), mid - point \(x_5=\frac{40 + 49}{2}=44.5\).
For age group \(50 - 59\), mid - point \(x_6=\frac{50+59}{2}=54.5\).
For age group \(60 - 69\), mid - point \(x_7=\frac{60 + 69}{2}=64.5\).
For age group \(70 - 79\), mid - point \(x_8=\frac{70+79}{2}=74.5\).
For age group \(80 - 89\), mid - point \(x_9=\frac{80 + 89}{2}=84.5\).

Step2: Calculate \(\sum(f\times x)\)

\(f_1 = 24,x_1 = 4.5,f_1x_1=24\times4.5 = 108\).
\(f_2 = 38,x_2 = 14.5,f_2x_2=38\times14.5=551\).
\(f_3 = 15,x_3 = 24.5,f_3x_3=15\times24.5 = 367.5\).
\(f_4 = 30,x_4 = 34.5,f_4x_4=30\times34.5=1035\).
\(f_5 = 20,x_5 = 44.5,f_5x_5=20\times44.5 = 890\).
\(f_6 = 48,x_6 = 54.5,f_6x_6=48\times54.5=2616\).
\(f_7 = 35,x_7 = 64.5,f_7x_7=35\times64.5 = 2257.5\).
\(f_8 = 11,x_8 = 74.5,f_8x_8=11\times74.5=819.5\).
\(f_9 = 9,x_9 = 84.5,f_9x_9=9\times84.5 = 760.5\).
\(\sum(f\times x)=108 + 551+367.5+1035+890+2616+2257.5+819.5+760.5=9405\).

Step3: Calculate \(\sum f\)

\(\sum f=24 + 38+15+30+20+48+35+11+9=230\).

Step4: Calculate the mean

The formula for the mean of a frequency distribution is \(\bar{x}=\frac{\sum(f\times x)}{\sum f}\).
\(\bar{x}=\frac{9405}{230}\approx40.9\).

Answer:

\(40.9\)