Question

approximate the mean of the frequency distribution for the ages of the residents of a town.
age frequency
0-9 31
10-19 36
20-29 12
30-39 24
40-49 15
50-59 51
60-69 38
70-79 19
80-89 5
the approximate mean age is \\(\square\\) years.
(round to one decimal place as needed.)

Explanation:

Step1: Find Midpoints

For each age group, calculate the midpoint ($x_i$).

• 0 - 9: $\frac{0 + 9}{2} = 4.5$
• 10 - 19: $\frac{10 + 19}{2} = 14.5$
• 20 - 29: $\frac{20 + 29}{2} = 24.5$
• 30 - 39: $\frac{30 + 39}{2} = 34.5$
• 40 - 49: $\frac{40 + 49}{2} = 44.5$
• 50 - 59: $\frac{50 + 59}{2} = 54.5$
• 60 - 69: $\frac{60 + 69}{2} = 64.5$
• 70 - 79: $\frac{70 + 79}{2} = 74.5$
• 80 - 89: $\frac{80 + 89}{2} = 84.5$

Step2: Calculate $f_i x_i$

Multiply each midpoint by its frequency ($f_i$):

• 4.5 × 31 = 139.5
• 14.5 × 36 = 522
• 24.5 × 12 = 294
• 34.5 × 24 = 828
• 44.5 × 15 = 667.5
• 54.5 × 51 = 2779.5
• 64.5 × 38 = 2451
• 74.5 × 19 = 1415.5
• 84.5 × 5 = 422.5

Step3: Sum $f_i$ and $\sum f_i x_i$

Sum of frequencies:
$31 + 36 + 12 + 24 + 15 + 51 + 38 + 19 + 5 = 231$

Sum of $f_i x_i$:
$139.5 + 522 + 294 + 828 + 667.5 + 2779.5 + 2451 + 1415.5 + 422.5 = 9519.5$

Step4: Compute Mean

Mean ($\bar{x}$) = $\frac{\sum f_i x_i}{\sum f_i}$ = $\frac{9519.5}{231} \approx 41.2$

Answer:

41.2