Question

Eight trials are simulated. The results are shown in the table.Simulation109105112110115106108109What is the estimated margin of error, using standard deviation?Enter your answer, rounded to two decimal places, in the box.

Answer 1

The standard deviation of a sample of data can be computed using the equation

`\sigma=\sqrt[]{(1)/(N)\sum^N_(n\mathop=1)(x_n-\mu)^2}`

where μ stands for the mean of the data and N as the number of samples.

The first step in solving the standard deviation of the given sample is to find the mean. The mean of the data given is

`\mu=(109+105+112+110+115+106+108+109)/(8)=109.25`

The second step is to subtract each number to the mean of the data then get the square of the difference. We have the following

`\begin{gathered} (x_1-\mu)^2=(109-109.25)^2=0.0625 \\ (x_2-\mu)^2=(105-109.25)^2=18.0625 \\ (x_3-\mu)^2=(112-109.25)^2=7.5625 \\ (x_4-\mu)^2=(110-109.25)^2=0.5625 \\ (x_5-\mu)^2=(115-109.25)^2=33.0625 \\ (x_6-\mu)^2=(106-109.25)^2=10.5625 \\ (x_7-\mu)^2=(1098109.25)^2=1.5625 \\ (x_8-\mu)^2=(109-109.25)^2=0.0625 \end{gathered}`

The third step is to find the sum of these squared differences and get the mean. We get

`(1)/(N)\sum ^N_{n\mathop{=}1}(x_n-\mu)^2=(0.0625+18.0625+7.5625+0.5625+33.0625+10.5625+1.5625+0.0625)/(8)=8.933`

Finally, for the last step, get the square root of the result in step 3, we have

`\sqrt[]{(1)/(N)\sum^N_{n\mathop{=}1}(x_n-\mu)^2}=\sqrt[]{8.933}=\pm2.99`

Therefore, the standard deviation of the given data is equal to ± 2.99.