Standard Deviation
Find the sample standard deviation for this set of values: 1, 3, 7, 6, 8.
Expand Hint
The sample standard deviation is:
$$$s =\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (X_{1}-\mu )^{2}}$$$
where
$$n$$
is the number of items or observations,
$$X$$
is the value from the set, and
$$\mu$$
is the mean.
Hint 2
To find the mean:
$$$\mu=\frac{sum\:of\:terms}{number\:of\:terms}$$$
First, let’s find the mean of the values:
$$$\mu=\frac{sum\:of\:terms}{number\:of\:terms}=\frac{1+3+7+6+8}{5}=\frac{25}{5}=5$$$
The sample standard deviation is:
$$$s =\sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (X_{1}-\mu )^{2}}$$$
where
$$n$$
is the number of items or observations,
$$X$$
is the value from the set, and
$$\mu$$
is the mean. Do not confuse this with a population standard deviation. Thus,
$$$s=\sqrt{\frac{1}{5-1}\cdot[(X_1-\mu)^2+(X_2-\mu)^2+(X_3-\mu)^2+(X_4-\mu)^2+(X_5-\mu)^2]}$$$
$$$=\sqrt{\frac{1}{4}\times[(1-5)^2+(3-5)^2+(7-5)^2+(6-5)^2+(8-5)^2]}$$$
$$$=\sqrt{\frac{1}{4}\times[(-4)^2+(-2)^2+(2)^2+(1)^2+(3)^2]}$$$
$$$=\sqrt{\frac{1}{4}\times[16+4+4+1+9]}=\sqrt{\frac{34}{4}}=\sqrt{8.5}=2.9$$$
2.9
Time Analysis
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