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A meteorologist is studying the monthly rainfall in a section of the Brazilian rainforest. She recorded the monthly rainfall, in inches, for last year. They were:

[tex]$
\begin{array}{l}
1.8, 2.5, 2.6, 4.4, 4.4, 7.3, 8.0, 9.5, 10.3, 10.4, 11.1, 11.7
\end{array}
$[/tex]

For this data set:

[tex]$
\begin{array}{l}
\mu=7, \quad N=12 \\
\sigma^2=\frac{\left(x_1-\mu\right)^2+\left(x_2-\mu\right)^2+\ldots+\left(x_N-\mu\right)^2}{N} \\
\sigma^2=\frac{(1.8-7)^2+(2.5-7)^2+\ldots+(11.7-7)^2}{12}
\end{array}
$[/tex]

Fill in the missing values in the formula. What is the variance?

A. 0
B. 3.217
C. 3.522
D. 12.405



Answer :

GinnyAnswer
To find the variance [tex]\(\sigma^2\)[/tex] of the given set of monthly rainfall data, we use the provided formula:
[tex]\[ \sigma^2 = \frac{(x_1 - \mu)^2 + (x_2 - \mu)^2 + \ldots + (x_{12} - \mu)^2}{12} \][/tex]

Here, the given data points are:
[tex]\[ 1.8, 2.5, 2.6, 4.4, 4.4, 7.3, 8.0, 9.5, 10.3, 10.4, 11.1, 11.7 \][/tex]

And we have:
[tex]\[ \mu = 7 \][/tex]
[tex]\[ N = 12 \][/tex]

Let's break it down into steps:

1. Subtract the mean [tex]\(\mu\)[/tex] from each data point [tex]\(x_i\)[/tex] and square the result:
[tex]\[ \begin{array}{l} (1.8 - 7)^2 = (-5.2)^2 = 27.04 \\ (2.5 - 7)^2 = (-4.5)^2 = 20.25 \\ (2.6 - 7)^2 = (-4.4)^2 = 19.36 \\ (4.4 - 7)^2 = (-2.6)^2 = 6.76 \\ (4.4 - 7)^2 = (-2.6)^2 = 6.76 \\ (7.3 - 7)^2 = (0.3)^2 = 0.09 \\ (8.0 - 7)^2 = (1.0)^2 = 1.00 \\ (9.5 - 7)^2 = (2.5)^2 = 6.25 \\ (10.3 - 7)^2 = (3.3)^2 = 10.89 \\ (10.4 - 7)^2 = (3.4)^2 = 11.56 \\ (11.1 - 7)^2 = (4.1)^2 = 16.81 \\ (11.7 - 7)^2 = (4.7)^2 = 22.09 \\ \end{array} \][/tex]

2. Sum these squared differences:
[tex]\[ 27.04 + 20.25 + 19.36 + 6.76 + 6.76 + 0.09 + 1.00 + 6.25 + 10.89 + 11.56 + 16.81 + 22.09 = 148.86 \][/tex]

3. Divide the sum by the number of data points [tex]\(N = 12\)[/tex]:
[tex]\[ \sigma^2 = \frac{148.86}{12} = 12.405 \][/tex]

Thus, the variance [tex]\(\sigma^2\)[/tex] is:
[tex]\[ 12.405 \][/tex]

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