Question

good cholesterol: high - density lipoprotein cholesterol (hdl) is sometimes called \good cholesterol,\ because it carries exce is removed from the body. high hdl levels are desirable. following are hdl measurements, in milligrams per deciliter, for san between the ages of 18 and 29.
men 30 41 32 45 46 47 62 61 52 55
women 40 37 43 54 51 41 82 83 80 69
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part: 0 / 3
part 1 of 3
(a) find the sample standard deviation of hdl levels for men. round the answer to one decimal place as needed.
the sample standard deviation of hdl levels for men is

Explanation:

Step1: Calculate the mean

Let the data for men be $x_1 = 30,x_2 = 41,x_3 = 32,x_4 = 45,x_5 = 46,x_6 = 47,x_7 = 62,x_8 = 61,x_9 = 52,x_{10}=55$.
The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$, where $n = 10$.
$\sum_{i=1}^{10}x_i=30 + 41+32 + 45+46+47+62+61+52+55=471$.
$\bar{x}=\frac{471}{10}=47.1$.

Step2: Calculate the squared - differences

$(x_1-\bar{x})^2=(30 - 47.1)^2=(-17.1)^2 = 292.41$.
$(x_2-\bar{x})^2=(41 - 47.1)^2=(-6.1)^2 = 37.21$.
$(x_3-\bar{x})^2=(32 - 47.1)^2=(-15.1)^2 = 228.01$.
$(x_4-\bar{x})^2=(45 - 47.1)^2=(-2.1)^2 = 4.41$.
$(x_5-\bar{x})^2=(46 - 47.1)^2=(-1.1)^2 = 1.21$.
$(x_6-\bar{x})^2=(47 - 47.1)^2=(-0.1)^2 = 0.01$.
$(x_7-\bar{x})^2=(62 - 47.1)^2=(14.9)^2 = 222.01$.
$(x_8-\bar{x})^2=(61 - 47.1)^2=(13.9)^2 = 193.21$.
$(x_9-\bar{x})^2=(52 - 47.1)^2=(4.9)^2 = 24.01$.
$(x_{10}-\bar{x})^2=(55 - 47.1)^2=(7.9)^2 = 62.41$.
The sum of squared - differences $\sum_{i = 1}^{n}(x_i-\bar{x})^2=292.41+37.21+228.01+4.41+1.21+0.01+222.01+193.21+24.01+62.41 = 1064.9$.

Step3: Calculate the sample standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$.
Here, $n = 10$, so $s=\sqrt{\frac{1064.9}{9}}\approx\sqrt{118.322}\approx10.9$.

Answer:

$10.9$