Question

What is the standard deviation of 90, 96, 147, 1371, and 49258?

Answer 1

SOLUTION

The sample standard deviation is given by the formula

$\begin{gathered} s=\sqrt[]{\frac{\sum ^n_{i\mathop=1}(x_i-\mu)^2}{n}} \\ \text{Where n is the number of values, }x_i,i=1\ldots.n\text{ are the values } \\ \mu\text{ is the mean of the values } \end{gathered}$

The mean of the data is

$\begin{gathered} \mu=\frac{90+96+147+1371+49258}{5} \\ \mu=\frac{509625}{5} \end{gathered}$

Also

$\begin{gathered} \text{The summation of }(x_i-\mu)^2\text{ becomes } \\ (90-\frac{509625}{5})^2+(96-\frac{509625}{5})^2+(147-\frac{509625}{5})^2 \\ +(1371-\frac{509625}{5})^2+(49258-\frac{509625}{5})^2 \\ =\frac{9544220206}{5} \end{gathered}$

Hence

$\begin{gathered} \frac{\sum ^n_{i\mathop{=}1}(x_i-\mu)^2}{n}=\frac{\frac{9544220206}{5}}{5}=\frac{\frac{9544220206}{5}}{5} \\ =381768808.2 \end{gathered}$

So we have

$\begin{gathered} s=\sqrt[]{\frac{\sum ^n_{i\mathop{=}1}(x_i-\mu)^2}{n}}=\sqrt[]{381768808.2}=19538.90499 \\ =19538.90\text{ to the nearest hundredth} \end{gathered}$

Hence the Answer is 19538.90 to the nearest hundredth