Question

approximate the mean of the frequency distribution for the ages of the residents of a town.
age \tfrequency
0-9 \t32
10-19 \t26
20-29 \t17
30-39 \t23
40-49 \t30
50-59 \t46
60-69 \t31
70-79 \t11
80-89 \t9
the approximate mean age is \boxed{} years.
(round to one decimal place as needed.)

Explanation:

Step1: Find midpoints of each age group

For a group \(a - b\), midpoint \(x=\frac{a + b}{2}\).

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

Step2: Calculate \(f_i\times x_i\) for each group

• \(f_1 = 32\), \(f_1x_1=32\times4.5 = 144\)
• \(f_2 = 26\), \(f_2x_2=26\times14.5 = 377\)
• \(f_3 = 17\), \(f_3x_3=17\times24.5 = 416.5\)
• \(f_4 = 23\), \(f_4x_4=23\times34.5 = 793.5\)
• \(f_5 = 30\), \(f_5x_5=30\times44.5 = 1335\)
• \(f_6 = 46\), \(f_6x_6=46\times54.5 = 2507\)
• \(f_7 = 31\), \(f_7x_7=31\times64.5 = 1999.5\)
• \(f_8 = 11\), \(f_8x_8=11\times74.5 = 819.5\)
• \(f_9 = 9\), \(f_9x_9=9\times84.5 = 760.5\)

Step3: Find total frequency \(N=\sum f_i\)

\(N=32 + 26+17 + 23+30 + 46+31 + 11+9\)
\(N = 32+26 = 58; 58 + 17=75; 75+23 = 98; 98+30 = 128; 128+46 = 174; 174+31 = 205; 205+11 = 216; 216+9 = 225\)

Step4: Find total of \(f_i\times x_i\), \(\sum f_ix_i\)

\(\sum f_ix_i=144+377 + 416.5+793.5+1335+2507+1999.5+819.5+760.5\)
Calculate step by step:
\(144+377 = 521\); \(521+416.5 = 937.5\); \(937.5+793.5 = 1731\); \(1731+1335 = 3066\); \(3066+2507 = 5573\); \(5573+1999.5 = 7572.5\); \(7572.5+819.5 = 8392\); \(8392+760.5 = 9152.5\)

Step5: Calculate mean \(\bar{x}=\frac{\sum f_ix_i}{N}\)

\(\bar{x}=\frac{9152.5}{225}\approx40.7\) (rounded to one decimal place)

Answer:

\(40.7\)