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What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United States gave the following information about a prehistoric extended family group of 88 members on what is now a Native American reservation. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint. (Round your answers to one decimal place.)

col1 Age range (years) 1-10 11-20 21-30 31 and over
col2 Number of individuals 40 24 15 9

Respuesta :

Answer:

mean=14.7045

sample variance=103.96

standard deviation=10.196

Step-by-step explanation:

Class interval midpoints(x) frequency(f)  fx

1-10                   5.5                  40               220

11-20                 15.5                24                372

21-30                 25.5               15                 382.5

31 and over      35.5                 9                  319.5

mean=sumfx/sumf

sum fx=1294

sum f=n=40+24+15+9=88

mean=1294/88=14.7045

[tex]Variance={\frac{sumfx^2-\frac{(sumfx)^2}{n}}{n-1} }[/tex]

sumfx²=40(5.5)²+24(15.5)²+15(25.5)²+9(35.5)²=28072

[tex]Variance={\frac{28072-\frac{(1294)^2}{88}}{87} }[/tex]

Variance=103.9577=103.96

[tex]Standard deviation=\sqrt{\frac{sumfx^2-\frac{(sumfx)^2}{n}}{n-1} }[/tex]

[tex]Standard deviation=\sqrt{\frac{28072-\frac{(1294)^2}{88}}{87} }[/tex]

[tex]standard deviation=\sqrt{103.9577}[/tex]

standard deviation=10.196