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
difference between each number and mean,
[tex]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[/tex]
[tex]\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[/tex]
= 25 + 36 + 256 + 49 + 196 + 9 + 16 + 1 + 100 +16 + 81
= 785
[tex]\rm{ Standard\ Deviation = \sqrt{\dfrac{\sum{(X-\mu)^2}} {n}[/tex]
[tex]\sigma= \sqrt{\dfrac{785}{11}}\\\\ \sigma = 8.4477[/tex]
Variance = [tex]\sigma ^2[/tex]
= 71.40
Hence, the standard deviation is 8.4477 and the variance is 71.40.
Learn more about Standard Deviation:
https://brainly.com/question/12402189
Answer:
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 =
71.36
Find the standard deviation of her grades rounded to the nearest hundredth.
σ =
8.45
Step-by-step explanation:
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