What are the mean and standard deviation of {34, 98, 20, -1200, -90}?

Trevor Rush
2022-08-14
Answered

What are the mean and standard deviation of {34, 98, 20, -1200, -90}?

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Jaxson White

Answered 2022-08-15
Author has **15** answers

Explanation:

Calculate the mean as the sum of the numbers divided by the number of observations

$Mean=\frac{34+98+20-1200-90}{5}=227.6$

Calculate the standard deviation as the square root of the sum of the squared difference each observation and the mean divided by the number of observations.

Standard deviation

$=\sqrt{\frac{(34-227.6{)}^{2}+(98-227.6{)}^{2}+(20-227.6{)}^{2}+(-1200-227.6{)}^{2}+(-90-227.6{)}^{2}}{5}}=489.9492$

Calculate the mean as the sum of the numbers divided by the number of observations

$Mean=\frac{34+98+20-1200-90}{5}=227.6$

Calculate the standard deviation as the square root of the sum of the squared difference each observation and the mean divided by the number of observations.

Standard deviation

$=\sqrt{\frac{(34-227.6{)}^{2}+(98-227.6{)}^{2}+(20-227.6{)}^{2}+(-1200-227.6{)}^{2}+(-90-227.6{)}^{2}}{5}}=489.9492$

balafiavatv

Answered 2022-08-16
Author has **2** answers

Data: S={34,98,20,-1200,-90}

Mean: $\sum \frac{s}{5}=-227.6$

Variance square differences are $(34-(-227.6){)}^{2}=68434.56$

$(98-(-227.6){)}^{2}=106015.36,(20-(-227.6){)}^{2}=61305.76,$

$(-1200-(-227.6){)}^{2}=945561.76,(-90-(-227.6){)}^{2}=18933.76$

Average variance square differences is

${\sigma}^{2}=\frac{1200251.2}{5}=240050.24$

Standard deviation is $\sqrt{{\sigma}^{2}}=489.9492$

Mean: $\sum \frac{s}{5}=-227.6$

Variance square differences are $(34-(-227.6){)}^{2}=68434.56$

$(98-(-227.6){)}^{2}=106015.36,(20-(-227.6){)}^{2}=61305.76,$

$(-1200-(-227.6){)}^{2}=945561.76,(-90-(-227.6){)}^{2}=18933.76$

Average variance square differences is

${\sigma}^{2}=\frac{1200251.2}{5}=240050.24$

Standard deviation is $\sqrt{{\sigma}^{2}}=489.9492$

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