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Find the standard deviation ofthe given data rounded to thenearest hundredth.12, 53, 141, 219, 500

Sagot :

[tex]173.014[/tex]

1) Let's find the standard deviation for this data set:

[tex]12,53,141,219,500[/tex]

2) So, let's apply the formula for standard deviation:

[tex]\begin{gathered} S\left(X\right)=\sqrt{\frac{\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}{n}} \\ \end{gathered}[/tex]

3) Let's find the mean and compute the variance:

[tex]\bar{x}=\sum_{i=1}^na_i=\frac{12+53+141+219+500}{5}=\frac{925}{5}=185[/tex]

The sum of all entries is divided by the number of data points.

Now, for the variance:

[tex]\begin{gathered} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2=\frac{\left(12-185\right)^2+\left(53-185\right)^2+\left(141-185\right)^2+\left(219-185\right)^2+\left(500-185\right)^2}{5} \\ \frac{149670}{5}=29934 \end{gathered}[/tex]

Finally, we can take the square root of that variance to get the standard deviation:

[tex]\sigma\left(X\right)=\sqrt{\sum_{i=1}^n\frac{\left(x_i-\bar{x}\right)^2}{n}}=\sqrt{29934}=173.014[/tex]