Answer to Question #185399 in Statistics and Probability for sean

Question

Sixteen smokers whose ages were between 50 and 65 years were randomly selected. Blood samples were taken from them. their cholesterol levels were determined and are presented below.181 287 240 256257 179 209 296211 231 171 248189 212 189 187Compute the following:a. Sample meanb. Sample variancec. Sample standard deviation

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Answer 1

a. Sample mean is

$\displaystyle\overline{x}=\dfrac{1}{16}\sum\limits_{i=1}^{16}x_i=$ $\dfrac{1}{16}\cdot(181+ 287+ 240+ 256+257+ 179+ 209+ 296+211 +231+ 171+ 248+189+ 212+ 189+ 187)=$

$\dfrac{3543}{16}=221.4375$

b. Sample variance is

$\displaystyle \sigma^2=\dfrac{1}{16}\sum\limits_{i=1}^{16}x_i^2-\overline{x}^2=$

$\dfrac{1}{16}\cdot(181^2+ 287^2+ 240^2+ 256^2+257^2+ 179^2+ 209^2+ 296^2+211^2 +231^2+ 171^2+ 248^2+189^2+ 212^2+ 189^2+ 187^2)-\left(\dfrac{3543}{16}\right)^2=$

$1442.621$

c. Sample standard deviation is

$\sigma=\sqrt{\sigma^2}=\sqrt{1442.621}\approx37.98$