Question

adimas found the mean of her 11 math test scores for the first semester.
overline { x } = \frac { ( 76 + 87 + 65 + 88 + 67 + 84 + 77 + 82 + 91 + 85 + 90 ) } { 11 } = \frac { 892 } { 11 } approx 81
using 81 as the mean, find the variance of her grades rounded to the nearest hundredth.
sigma ^ { 2 } =
find the standard deviation of her grades rounded to the nearest hundredth.
sigma =

Explanation:

Step1: Calculate \((x_i - \bar{x})^2\) for each score

For \(x_1 = 76\): \((76 - 81)^2=(- 5)^2 = 25\)
For \(x_2 = 87\): \((87 - 81)^2=(6)^2 = 36\)
For \(x_3 = 65\): \((65 - 81)^2=(-16)^2 = 256\)
For \(x_4 = 88\): \((88 - 81)^2=(7)^2 = 49\)
For \(x_5 = 67\): \((67 - 81)^2=(-14)^2 = 196\)
For \(x_6 = 84\): \((84 - 81)^2=(3)^2 = 9\)
For \(x_7 = 77\): \((77 - 81)^2=(-4)^2 = 16\)
For \(x_8 = 82\): \((82 - 81)^2=(1)^2 = 1\)
For \(x_9 = 91\): \((91 - 81)^2=(10)^2 = 100\)
For \(x_{10}=85\): \((85 - 81)^2=(4)^2 = 16\)
For \(x_{11}=90\): \((90 - 81)^2=(9)^2 = 81\)

Step2: Calculate the variance \(\sigma^{2}\)

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Step3: Calculate the standard deviation \(\sigma\)

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Answer:

The variance \(\sigma^{2}\approx80.45\) and the standard deviation \(\sigma\approx8.97\)