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Petroleum pollution in oceans stimulates the growth of certain bacteria. An assessment of this growth has been made by counting the bacteria in each of 6 randomly chosen specimens of ocean water (of a fixed size). The 6 counts obtained were as follows.66, 45, 65, 70, 58, 62Find the standard deviation of this sample of numbers. Round your answer to two decimal places.

Petroleum pollution in oceans stimulates the growth of certain bacteria. An assessment-example-1
User Algenis
by
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1 Answer

20 votes
20 votes

Answer:

8.81

Explanation:

The standard deviation for a sample is calculated using the formula:


s=\sqrt{\frac{\sum^{}_{}(x-\text{Mean)}^2}{n-1}}

First, find the mean of the data set.


\text{Mean}=(66+45+65+70+58+62)/(6)=(366)/(6)=61

Next, we find the square of the mean deviations for each x.


\begin{gathered} \sum (x-\text{Mean)}^2=(66-61)^2+(45-61)^2+\mleft(65-61\mright)^2 \\ +\mleft(70-61\mright)^2+(58-61)^2+(62-61)^2 \\ =(5)^2+(-16)^2+(4)^2+(9)^2+(-3)^2+(1)^2 \\ =25+256+16+81+9+1 \\ \sum (x-\text{Mean)}^2=388 \end{gathered}

Therefore, the standard deviation of this sample of numbers will be:


s=\sqrt[]{\frac{\sum^{}_{}(x-\text{Mean)}^2}{n-1}}=\sqrt[]{(388)/(6-1)}=\sqrt[]{(388)/(5)}=8.81

The standard deviation is 8.81 (correct to 2 decimal places).

User Mahamutha M
by
2.7k points
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