To determine the correct numerator for the calculation of variance and standard deviation, let's follow these steps:
1. Calculate the mean of the areas:
The areas given are:
[tex]\( 2400, 1750, 1900, 2500, 2250, 2100 \)[/tex]
The mean [tex]\(\mu\)[/tex] is calculated as:
[tex]\[
\mu = \frac{\sum_{i=1}^{n} x_i}{n}
\][/tex]
Where [tex]\( \sum_{i=1}^{n} x_i \)[/tex] is the sum of all areas, and [tex]\( n \)[/tex] is the number of areas.
Sum of areas:
[tex]\[
2400 + 1750 + 1900 + 2500 + 2250 + 2100 = 12900
\][/tex]
Number of areas [tex]\( n \)[/tex]:
[tex]\[
n = 6
\][/tex]
Mean [tex]\(\mu\)[/tex]:
[tex]\[
\mu = \frac{12900}{6} = 2150
\][/tex]
2. Calculate the squared differences from the mean (each area [tex]\( x_i \)[/tex] minus the mean [tex]\(\mu\)[/tex]), and then sum these squared differences to get the numerator for the variance calculation:
For each area, calculate the difference from the mean, square it, and then add up these squared differences:
[tex]\[
\begin{align*}
(2400 - 2150)^2 & = (250)^2 = 62500 \\
(1750 - 2150)^2 & = (-400)^2 = 160000 \\
(1900 - 2150)^2 & = (-250)^2 = 62500 \\
(2500 - 2150)^2 & = (350)^2 = 122500 \\
(2250 - 2150)^2 & = (100)^2 = 10000 \\
(2100 - 2150)^2 & = (-50)^2 = 2500 \\
\end{align*}
\][/tex]
Sum of the squared differences (numerator):
[tex]\[
62500 + 160000 + 62500 + 122500 + 10000 + 2500 = 420000
\][/tex]
Therefore, the correct representation of the numerator in the calculation of variance and standard deviation is:
[tex]\[
(250)^2 + (-400)^2 + (-250)^2 + (350)^2 + (100)^2 + (-50)^2 = 420,000
\][/tex]
Therefore, the correct option is:
[tex]\[
(250)^2+(-400)^2+(-250)^2+(350)^2+(100)^2+(-50)^2=420,000
\][/tex]
