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(a) Approximate the population mean and standard deviation of age for males And For females.

A Approximate The Population Mean And Standard Deviation Of Age For Males And For Females class=

Sagot :

[tex]\begin{gathered} \mu=38.68 \\ \sigma=452.67 \end{gathered}[/tex]

1) Since we have a table for grouped data, we need to place into that table another column with the middle point of each interval to get the mean.

2) Setting that table with another column we've got:

Age Middlepoint Male Female Male*freq Female*fr

0-9 (0+9)/2 =4.5 10 9 10*4.5=45 9*4.5=40.5

10-19 (10+19)/2=14.5 11 5 159.5 72.5

20-29 (20+29)/2= 24.5 12 13 294 318.5

30-39 (30+39)/2= 34.5 16 19 552 655.5

40-49 (40+49)/2= 44.5 25 21 1112.5 934.5

50-59 (50+59)/2= 54.5 20 24 1090 1308

60-69 (60+69)/2=64.5 18 18 1161 1161

70-79 (70+79)/2= 74.5 15 14 117.5 1043

Now, we can pick the absolute frequency of males and females and multiply by the middle point.

Now we can add the number of males multiplied by the frequency and divide them by the sum of the frequencies, this way:

[tex]\mu=\frac{45+159.5+294+552+1112.5+1090+1161+117.5}{10+11+12+16+25+20+18+15}\approx38.68[/tex]

3) Now to find the standard deviation of this population, we can write out the following:

[tex]\begin{gathered} \sigma=\frac{\sqrt[]{(45-38.68)^2+(159.5-38.68)^2+(294-38.68)^2+(552-38.68)^2+(1112.5-38.68)^2+(1090-38.68)^2+(1161-38.68)^2+(117.5-38.68)^2}}{8} \\ \\ \sigma=452.6694 \end{gathered}[/tex]