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Adimas found the mean of her 11 math test scores for the first semester.
[tex]$
\bar{x}=\frac{(76+87+65+88+67+84+77+82+91+85+90)}{11}=\frac{892}{11} \approx 81
$[/tex]

Using 81 as the mean, find the variance of her grades rounded to the nearest hundredth.
[tex]$
\sigma^2=
$[/tex]

Find the standard deviation of her grades rounded to the nearest hundredth.
[tex]$
\sigma=
$[/tex]

Sagot :

GinnyAnswer
To address this problem, let's break it down systematically and thoroughly to ensure precision in our calculations.

### Calculate the Mean

First, you correctly calculated the mean of Adimas' math test scores:

[tex]\[ \bar{x} = \frac{(76 + 87 + 65 + 88 + 67 + 84 + 77 + 82 + 91 + 85 + 90)}{11} = \frac{892}{11} \approx 81 \][/tex]

### Calculate the Variance

Next, we need to find the variance, which measures how much the scores are spread out from the mean.

To find the variance, we'll first calculate each individual deviation from the mean, square those deviations, sum them up, and then find the average of those squared deviations.

Step-by-step, let's calculate each deviation from the mean (81) and square it:

1. [tex]\((76 - 81)^2 = (-5)^2 = 25\)[/tex]
2. [tex]\((87 - 81)^2 = 6^2 = 36\)[/tex]
3. [tex]\((65 - 81)^2 = (-16)^2 = 256\)[/tex]
4. [tex]\((88 - 81)^2 = 7^2 = 49\)[/tex]
5. [tex]\((67 - 81)^2 = (-14)^2 = 196\)[/tex]
6. [tex]\((84 - 81)^2 = 3^2 = 9\)[/tex]
7. [tex]\((77 - 81)^2 = (-4)^2 = 16\)[/tex]
8. [tex]\((82 - 81)^2 = 1^2 = 1\)[/tex]
9. [tex]\((91 - 81)^2 = 10^2 = 100\)[/tex]
10. [tex]\((85 - 81)^2 = 4^2 = 16\)[/tex]
11. [tex]\((90 - 81)^2 = 9^2 = 81\)[/tex]

Next, sum these squared deviations:

[tex]\[ 25 + 36 + 256 + 49 + 196 + 9 + 16 + 1 + 100 + 16 + 81 = 785 \][/tex]

To find the variance, we take the average of these squared deviations:

[tex]\[ \sigma^2 = \frac{785}{11} \approx 71.36 \][/tex]

So, the variance of her grades, rounded to the nearest hundredth, is:

[tex]\[ \sigma^2 \approx 71.36 \][/tex]

### Calculate the Standard Deviation

Finally, the standard deviation is the square root of the variance, which helps us understand the spread of the scores in the same units as the original scores.

[tex]\[ \sigma = \sqrt{71.36} \approx 8.45 \][/tex]

So, the standard deviation of her grades, rounded to the nearest hundredth, is:

[tex]\[ \sigma \approx 8.45 \][/tex]

### Summary

The variance of her grades rounded to the nearest hundredth is:

[tex]\[ \sigma^2 \approx 71.36 \][/tex]

The standard deviation of her grades rounded to the nearest hundredth is:

[tex]\[ \sigma \approx 8.45 \][/tex]
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