Griffin Wheeler
08/17/2023 · Primary School
The numbers of regular season wins for 10 football teams in a given season are given below. Determine the range, mean, variance, and standard deviation of the population data set. \( 2,8,15,3,14,10,13,9,4,7 \)
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Step-by-step Solution
To calculate the range, mean, variance, and standard deviation of the given data set, we will follow these steps:
### Given Data
The numbers of regular season wins are:
\[ 2, 8, 15, 3, 14, 10, 13, 9, 4, 7 \]
### Step 1: Calculate the Range
The range is calculated as:
\[
\text{Range} = \text{Maximum} - \text{Minimum}
\]
- Maximum = 15
- Minimum = 2
\[
\text{Range} = 15 - 2 = 13
\]
### Step 2: Calculate the Mean
The mean is calculated as:
\[
\text{Mean} = \frac{\sum x_i}{N}
\]
where \( N \) is the number of data points.
Calculating the sum:
\[
\sum x_i = 2 + 8 + 15 + 3 + 14 + 10 + 13 + 9 + 4 + 7 = 91
\]
Number of data points \( N = 10 \).
Now, calculate the mean:
\[
\text{Mean} = \frac{91}{10} = 9.1
\]
### Step 3: Calculate the Variance
The variance is calculated using the formula:
\[
\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}
\]
where \( \mu \) is the mean.
First, we calculate \( (x_i - \mu)^2 \) for each data point:
- For \( 2 \): \( (2 - 9.1)^2 = (-7.1)^2 = 50.41 \)
- For \( 8 \): \( (8 - 9.1)^2 = (-1.1)^2 = 1.21 \)
- For \( 15 \): \( (15 - 9.1)^2 = (5.9)^2 = 34.81 \)
- For \( 3 \): \( (3 - 9.1)^2 = (-6.1)^2 = 37.21 \)
- For \( 14 \): \( (14 - 9.1)^2 = (4.9)^2 = 24.01 \)
- For \( 10 \): \( (10 - 9.1)^2 = (0.9)^2 = 0.81 \)
- For \( 13 \): \( (13 - 9.1)^2 = (3.9)^2 = 15.21 \)
- For \( 9 \): \( (9 - 9.1)^2 = (-0.1)^2 = 0.01 \)
- For \( 4 \): \( (4 - 9.1)^2 = (-5.1)^2 = 26.01 \)
- For \( 7 \): \( (7 - 9.1)^2 = (-2.1)^2 = 4.41 \)
Now, sum these squared differences:
\[
\sum (x_i - \mu)^2 = 50.41 + 1.21 + 34.81 + 37.21 + 24.01 + 0.81 + 15.21 + 0.01 + 26.01 + 4.41 = 194.1
\]
Now, calculate the variance:
\[
\sigma^2 = \frac{194.1}{10} = 19.41
\]
### Step 4: Calculate the Standard Deviation
The standard deviation is the square root of the variance:
\[
\sigma = \sqrt{\sigma^2} = \sqrt{19.41} \approx 4.41
\]
### Summary of Results
- **Range**: 13
- **Mean**: 9.1
- **Variance**: 19.41
- **Standard Deviation**: 4.41
These are the calculated statistics for the given data set.
Quick Answer
- **Range**: 13
- **Mean**: 9.1
- **Variance**: 19.41
- **Standard Deviation**: 4.41
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