Question

approximate the mean of the frequency distribution for the populations (in thousands) of the counties in a certain state.
population (in thousands) frequency
0 - 19 47
20 - 39 2
40 - 59 1
60 - 79 1
80 - 99 1
100 - 119 5
120 - 139 0
140 - 159 1
the approximate mean population is thousand. (round to one decimal place as needed)

Explanation:

Step1: Find mid - points of each class

For 0 - 19, mid - point $x_1=\frac{0 + 19}{2}=9.5$; for 20 - 39, $x_2=\frac{20+39}{2}=29.5$; for 40 - 59, $x_3=\frac{40 + 59}{2}=49.5$; for 60 - 79, $x_4=\frac{60+79}{2}=69.5$; for 80 - 99, $x_5=\frac{80 + 99}{2}=89.5$; for 100 - 119, $x_6=\frac{100+119}{2}=109.5$; for 120 - 139, $x_7=\frac{120+139}{2}=129.5$; for 140 - 159, $x_8=\frac{140+159}{2}=149.5$.

Step2: Calculate the product of mid - points and frequencies

$f_1x_1=47\times9.5 = 446.5$; $f_2x_2=2\times29.5 = 59$; $f_3x_3=1\times49.5 = 49.5$; $f_4x_4=1\times69.5 = 69.5$; $f_5x_5=1\times89.5 = 89.5$; $f_6x_6=5\times109.5 = 547.5$; $f_7x_7=0\times129.5 = 0$; $f_8x_8=1\times149.5 = 149.5$.

Step3: Find the sum of products $\sum_{i = 1}^{8}f_ix_i$

$\sum_{i = 1}^{8}f_ix_i=446.5+59+49.5+69.5+89.5+547.5+0+149.5 = 1411$.

Step4: Find the sum of frequencies $\sum_{i = 1}^{8}f_i$

$\sum_{i = 1}^{8}f_i=47 + 2+1+1+1+5+0+1=58$.

Step5: Calculate the mean

$\bar{x}=\frac{\sum_{i = 1}^{8}f_ix_i}{\sum_{i = 1}^{8}f_i}=\frac{1411}{58}\approx24.3$.

Answer:

24.3