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The standard deviation is approximately??​

The standard deviation is approximately??​-example-1

1 Answer

5 votes

Answer:


SD = \$ 239

Explanation:

Given


\$277.36, \$269.29, \$307.98, \$361.63, \$407.97,\\ \$442.99, \$601.71, \$693.64, \$870.21, \$970.60

Required

Determine the standard deviation

First, we calculate the mean.


Mean = (\$277.36+ \$269.29+ \$307.98+ \$361.63+ \$407.97+ \$442.99+ \$601.71+ \$693.64+ \$870.21+ \$970.60)/(10)


Mean = (\$5203.38)/(10)


Mean = \$520.338


Mean = \$520.34 --- approximately

Next, subtract the mean from each data


\$277.36 - \$520.34 =-\$242.98


\$269.29 - \$520.34 = -\$251.05


\$307.98 - \$520.34 = -\$212.36


\$361.63 - \$520.34 = -\$158.71


\$407.97 - \$520.34 = -\$112.37


\$442.99 - \$520.34 = -\$77.35


\$601.71 - \$520.34 = \$81.37


\$693.64 - \$520.34 = \$173.3


\$870.21 - \$520.34 =\$349.87


\$970.60 - \$520.34 =\$450.26

Square the results above


(-\$242.98)^2 = \$59039.2804


(-\$251.05)^2 =\$63026.1025


(-\$212.36)^2 =\$45096.7696


(-\$158.71)^2 =\$25188.8641


(-\$112.37)^2 =\$12627.0169


(-\$77.35)^2 =\$5983.0225


(-\$81.37)^2 =\$6621.0769


(-\$173.3)^2 =\$30032.89


(-\$349.87)^2 =\$122409.0169


(-\$450.26)^2 =\$202734.0676

Add the squared results


\$59039.2804 + \$63026.1025 + \$45096.7696 + \$25188.8641 + \$12627.0169 + \$5983.0225 +\$6621.0769 + \$30032.89 + \$122409.0169 + \$202734.0676


= \$572758.1074

Divide by number of data, to get the variance


Variance = (\$572758.1074)/(10)


Variance = \$57275.81074

Square the above, to get the standard deviation


SD = √(\$57275.81074)


SD = \$ 239.323652697


SD = \$ 239 --- approximated

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