Davison John
08/20/2023 · Senior High School
he length (in pages) of math research projects is given elow. Using this information, calculate the range, ariance, and standard deviation. \[ 19,16,34,11,29,24,16,299,21 \] ange \( = \) ariance \( = \) tandard deviation =
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To calculate the range, variance, and standard deviation of the given data set, we will follow these steps:
1. **Calculate the Range**:
The range is the difference between the maximum and minimum values in the data set.
Given data: \( 19, 16, 34, 11, 29, 24, 16, 299, 21 \)
- Maximum value = 299
- Minimum value = 11
\[
\text{Range} = \text{Maximum} - \text{Minimum} = 299 - 11 = 288
\]
2. **Calculate the Variance**:
Variance is calculated using the formula:
\[
\text{Variance} = \frac{\sum (x_i - \mu)^2}{N}
\]
where \( \mu \) is the mean of the data set, \( x_i \) are the individual data points, and \( N \) is the number of data points.
- First, calculate the mean (\( \mu \)):
\[
\mu = \frac{\sum x_i}{N} = \frac{19 + 16 + 34 + 11 + 29 + 24 + 16 + 299 + 21}{9} = \frac{ 19 + 16 + 34 + 11 + 29 + 24 + 16 + 299 + 21 }{9} = \frac{ 449 }{9} \approx 49.89
\]
- Next, calculate the squared differences from the mean:
\[
\begin{align*}
(19 - 49.89)^2 & \approx 952.5921 \\
(16 - 49.89)^2 & \approx 1135.5921 \\
(34 - 49.89)^2 & \approx 252.5921 \\
(11 - 49.89)^2 & \approx 1495.5921 \\
(29 - 49.89)^2 & \approx 436.5921 \\
(24 - 49.89)^2 & \approx 653.5921 \\
(16 - 49.89)^2 & \approx 1135.5921 \\
(299 - 49.89)^2 & \approx 61756.5921 \\
(21 - 49.89)^2 & \approx 826.5921 \\
\end{align*}
\]
- Now sum these squared differences:
\[
\sum (x_i - \mu)^2 \approx 952.5921 + 1135.5921 + 252.5921 + 1495.5921 + 436.5921 + 653.5921 + 1135.5921 + 61756.5921 + 826.5921 \approx 65192.0
\]
- Finally, calculate the variance:
\[
\text{Variance} = \frac{65192.0}{9} \approx 7243.56
\]
3. **Calculate the Standard Deviation**:
The standard deviation is the square root of the variance.
\[
\text{Standard Deviation} = \sqrt{\text{Variance}} \approx \sqrt{7243.56} \approx 85.0
\]
### Summary of Results:
- Range \( = 288 \)
- Variance \( \approx 7243.56 \)
- Standard Deviation \( \approx 85.0 \)
Please note that the calculations for variance and standard deviation are based on the assumption that the data represents a population. If it represents a sample, the denominator in the variance calculation would be \( N-1 \) instead of \( N \).
Quick Answer
Range = 288
Variance ≈ 7243.56
Standard Deviation ≈ 85.0
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