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Answer :
Sure! Let's find the sample mean and standard deviation for the given data set. Here’s a step-by-step solution:
Given Data Set:
-8.00, -4.00, -7.00, -6.00, -8.00, -5.00, -7.00, -9.00, -2.00, 0.00
### Step 1: Calculate the Sample Mean
The sample mean is the average of all the data points. You can find it by summing up all the values and then dividing by the number of values in the sample.
1. Add all the numbers together:
[tex]\(-8.00 + (-4.00) + (-7.00) + (-6.00) + (-8.00) + (-5.00) + (-7.00) + (-9.00) + (-2.00) + 0.00 = -56.00\)[/tex]
2. Divide the sum by the number of data points. There are 10 data points:
[tex]\(\text{Sample Mean} = \frac{-56.00}{10} = -5.6\)[/tex]
### Step 2: Calculate the Sample Standard Deviation
The sample standard deviation is a measure of how spread out the numbers are around the mean. For a sample, we use Bessel's correction, where we divide by [tex]\(n-1\)[/tex] instead of [tex]\(n\)[/tex].
1. Find the deviation of each data point from the mean:
[tex]\[
\begin{array}{c}
-8.00 - (-5.6) = -2.4 \\
-4.00 - (-5.6) = 1.6 \\
-7.00 - (-5.6) = -1.4 \\
-6.00 - (-5.6) = -0.4 \\
-8.00 - (-5.6) = -2.4 \\
-5.00 - (-5.6) = 0.6 \\
-7.00 - (-5.6) = -1.4 \\
-9.00 - (-5.6) = -3.4 \\
-2.00 - (-5.6) = 3.6 \\
0.00 - (-5.6) = 5.6 \\
\end{array}
\][/tex]
2. Square each deviation:
[tex]\[
\begin{array}{c}
(-2.4)^2 = 5.76 \\
(1.6)^2 = 2.56 \\
(-1.4)^2 = 1.96 \\
(-0.4)^2 = 0.16 \\
(-2.4)^2 = 5.76 \\
(0.6)^2 = 0.36 \\
(-1.4)^2 = 1.96 \\
(-3.4)^2 = 11.56 \\
(3.6)^2 = 12.96 \\
(5.6)^2 = 31.36 \\
\end{array}
\][/tex]
3. Find the sum of these squared deviations:
[tex]\[
5.76 + 2.56 + 1.96 + 0.16 + 5.76 + 0.36 + 1.96 + 11.56 + 12.96 + 31.36 = 74.0625
\][/tex]
4. Divide by [tex]\(n-1\)[/tex] (in this case [tex]\(n=10, n-1=9\)[/tex]) to find the variance:
[tex]\(\text{Variance} = \frac{74.0625}{9} \approx 8.25\)[/tex]
5. The standard deviation is the square root of the variance:
[tex]\(\text{Standard Deviation} \approx \sqrt{8.25} \approx 2.875\)[/tex]
### Final Results:
- Sample Mean: [tex]\(-5.6\)[/tex]
- Sample Standard Deviation: [tex]\(2.875\)[/tex]
These values give us a measure of the average value of the sample and how much the values typically differ from this average.
Given Data Set:
-8.00, -4.00, -7.00, -6.00, -8.00, -5.00, -7.00, -9.00, -2.00, 0.00
### Step 1: Calculate the Sample Mean
The sample mean is the average of all the data points. You can find it by summing up all the values and then dividing by the number of values in the sample.
1. Add all the numbers together:
[tex]\(-8.00 + (-4.00) + (-7.00) + (-6.00) + (-8.00) + (-5.00) + (-7.00) + (-9.00) + (-2.00) + 0.00 = -56.00\)[/tex]
2. Divide the sum by the number of data points. There are 10 data points:
[tex]\(\text{Sample Mean} = \frac{-56.00}{10} = -5.6\)[/tex]
### Step 2: Calculate the Sample Standard Deviation
The sample standard deviation is a measure of how spread out the numbers are around the mean. For a sample, we use Bessel's correction, where we divide by [tex]\(n-1\)[/tex] instead of [tex]\(n\)[/tex].
1. Find the deviation of each data point from the mean:
[tex]\[
\begin{array}{c}
-8.00 - (-5.6) = -2.4 \\
-4.00 - (-5.6) = 1.6 \\
-7.00 - (-5.6) = -1.4 \\
-6.00 - (-5.6) = -0.4 \\
-8.00 - (-5.6) = -2.4 \\
-5.00 - (-5.6) = 0.6 \\
-7.00 - (-5.6) = -1.4 \\
-9.00 - (-5.6) = -3.4 \\
-2.00 - (-5.6) = 3.6 \\
0.00 - (-5.6) = 5.6 \\
\end{array}
\][/tex]
2. Square each deviation:
[tex]\[
\begin{array}{c}
(-2.4)^2 = 5.76 \\
(1.6)^2 = 2.56 \\
(-1.4)^2 = 1.96 \\
(-0.4)^2 = 0.16 \\
(-2.4)^2 = 5.76 \\
(0.6)^2 = 0.36 \\
(-1.4)^2 = 1.96 \\
(-3.4)^2 = 11.56 \\
(3.6)^2 = 12.96 \\
(5.6)^2 = 31.36 \\
\end{array}
\][/tex]
3. Find the sum of these squared deviations:
[tex]\[
5.76 + 2.56 + 1.96 + 0.16 + 5.76 + 0.36 + 1.96 + 11.56 + 12.96 + 31.36 = 74.0625
\][/tex]
4. Divide by [tex]\(n-1\)[/tex] (in this case [tex]\(n=10, n-1=9\)[/tex]) to find the variance:
[tex]\(\text{Variance} = \frac{74.0625}{9} \approx 8.25\)[/tex]
5. The standard deviation is the square root of the variance:
[tex]\(\text{Standard Deviation} \approx \sqrt{8.25} \approx 2.875\)[/tex]
### Final Results:
- Sample Mean: [tex]\(-5.6\)[/tex]
- Sample Standard Deviation: [tex]\(2.875\)[/tex]
These values give us a measure of the average value of the sample and how much the values typically differ from this average.
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