Question

Adimas found the mean of her 11 math test scores for the first semester. X = StartFraction (76 87 65 88 67 84 77 82 91 85 90) Over 11 EndFraction = StartFraction 892 Over 11 EndFraction ≈ 81 Using 81 as the mean, find the variance of her grades rounded to the nearest hundredth. Σ2 = Find the standard deviation of her grades rounded to the nearest hundredth. Σ =.

Answer 1

To find the solution we will first find the difference between each number and mean, and then substitute the value in the formula of standard deviation.

The standard deviation is 8.4477 and the variance is 71.40.

Given to us

• Test scores of Adimas = (76, 87, 65, 88, 67, 84, 77, 82, 91, 85, 90)
• Mean = 81

Standard Deviation

difference between each number and mean,

$x_1 -\mu= 76-81 = -5\\ x_2-\mu = 87-81 = 6\\ x_3-\mu = 65-81 = -16\\ x_4-\mu = 88 -81 = 7\\ x_5-\mu = 67-81 = -14\\ x_6 -\mu= 84-81 = 3\\ x_7 -\mu= 77-81 = -4\\ x_8 -\mu= 81-81 = 1\\ x_9 -\mu= 91-81 = 10\\ x_{10}-\mu=85-81 = 4\\ x_{11} -\mu= 90-81 = 9$

$\sum(x-\mu)^2 = x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2+x_8^2+x_9^2+x_{10}^2+x_{11}^2$

                 =  25 + 36 + 256 + 49 + 196 + 9 + 16 + 1 + 100 +16 + 81

                 = 785

$\rm{ Standard\ Deviation = \sqrt{\dfrac{\sum{(X-\mu)^2}} {n}$

                             $\sigma= \sqrt{\dfrac{785}{11}}\\\\ \sigma = 8.4477$

Variance

Variance = $\sigma ^2$

              = 71.40

Hence, the standard deviation is 8.4477 and the variance is 71.40.

Learn more about Standard Deviation:

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