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The test scores of a geometry class are given below.
[tex]\[ 70, 75, 72, 88, 85 \][/tex]

The teacher wants to find the variance for the class population. What is the value of the numerator of the calculation of the variance?

Variance: [tex]\[ \sigma^2=\frac{(x_1-\mu)^2 + (x_2-\mu)^2 + \ldots + (x_N-\mu)^2}{N} \][/tex]

A. -160
B. -6
C. 16
D. 258


Sagot :

To determine the variance for the class population, we need to perform several steps, focusing particularly on the numerator of the variance calculation formula. This can be broken down into the steps below:

1. List the test scores: The scores given are [tex]\(0, 75, 72, 88, 85\)[/tex].

2. Calculate the mean ([tex]\(\mu\)[/tex]) of the scores:
[tex]\[ \mu = \frac{\sum (scores)}{N} \][/tex]
where [tex]\(N\)[/tex] is the number of scores.

3. Subtract the mean from each score and square the result to calculate [tex]\((x_i - \mu)^2\)[/tex].

4. Sum these squared differences.

To proceed with the steps:

1. Calculate the mean:
[tex]\[ \mu = \frac{0 + 75 + 72 + 88 + 85}{5} = \frac{320}{5} = 64.0 \][/tex]

2. Calculate the squared deviations from the mean and sum them:
[tex]\[ (x_1 - \mu)^2 = (0 - 64.0)^2 = (-64.0)^2 = 4096.0 \][/tex]
[tex]\[ (x_2 - \mu)^2 = (75 - 64.0)^2 = (11.0)^2 = 121.0 \][/tex]
[tex]\[ (x_3 - \mu)^2 = (72 - 64.0)^2 = (8.0)^2 = 64.0 \][/tex]
[tex]\[ (x_4 - \mu)^2 = (88 - 64.0)^2 = (24.0)^2 = 576.0 \][/tex]
[tex]\[ (x_5 - \mu)^2 = (85 - 64.0)^2 = (21.0)^2 = 441.0 \][/tex]

Sum of the squared deviations:
[tex]\[ 4096.0 + 121.0 + 64.0 + 576.0 + 441.0 = 5298.0 \][/tex]

The numerator of the variance calculation is the sum of these squared differences, which is [tex]\(5298.0\)[/tex].

Therefore, the value of the numerator of the calculation of the variance is:
[tex]\[ 5298.0 \][/tex]
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