approximate the mean of the frequency distribution for the ages of the residents of a town.\nage…

approximate the mean of the frequency distribution for the ages of the residents of a town.\nage frequency\n0 - 9 28\n10 - 19 20\n20 - 29 16\n30 - 39 30\n40 - 49 29\n50 - 59 52\n60 - 69 49\n70 - 79 11\n80 - 89 3\nthe approximate mean age is years. (round to one decimal place as needed.)

approximate the mean of the frequency distribution for the ages of the residents of a town.\nage frequency\n0 - 9 28\n10 - 19 20\n20 - 29 16\n30 - 39 30\n40 - 49 29\n50 - 59 52\n60 - 69 49\n70 - 79 11\n80 - 89 3\nthe approximate mean age is years. (round to one decimal place as needed.)

Answer

Explanation:

Step1: Find mid - points of age intervals

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

Step2: Calculate the product of mid - points and frequencies

$(4.5\times28)+(14.5\times20)+(24.5\times16)+(34.5\times30)+(44.5\times29)+(54.5\times52)+(64.5\times49)+(74.5\times11)+(84.5\times3)$ $=126+290+392+1035+1290.5+2834+3160.5+819.5+253.5$ $=10001$

Step3: Calculate the total frequency

$28 + 20+16+30+29+52+49+11+3=238$

Step4: Calculate the mean

$\bar{x}=\frac{10001}{238}\approx42.0$

Answer:

$42.0$