QUESTION IMAGE
Question
using the given calculator, find the iqr, standard deviation, and any outliers. 52, 42, 49, 48, 34, 42, 49, 54, 41 multiple-choice question what is the standard deviation? 5.98 1.75 2.56 6.34
Step 1: Find the mean ($\mu$)
First, we calculate the mean of the data set. The data set is \( 52, 42, 49, 48, 34, 42, 49, 54, 41 \). The sum of the data points is:
\[
52 + 42 + 49 + 48 + 34 + 42 + 49 + 54 + 41 = 411
\]
There are \( n = 9 \) data points. So the mean is:
\[
\mu=\frac{411}{9} = 45.666\ldots\approx45.67
\]
Step 2: Calculate the squared differences from the mean
For each data point \( x \), we calculate \( (x - \mu)^2 \):
- For \( 52 \): \( (52 - 45.67)^2=(6.33)^2 = 40.0689 \)
- For \( 42 \): \( (42 - 45.67)^2=(- 3.67)^2 = 13.4689 \)
- For \( 49 \): \( (49 - 45.67)^2=(3.33)^2 = 11.0889 \)
- For \( 48 \): \( (48 - 45.67)^2=(2.33)^2 = 5.4289 \)
- For \( 34 \): \( (34 - 45.67)^2=(-11.67)^2 = 136.1889 \)
- For \( 42 \): \( (42 - 45.67)^2=(-3.67)^2 = 13.4689 \)
- For \( 49 \): \( (49 - 45.67)^2=(3.33)^2 = 11.0889 \)
- For \( 54 \): \( (54 - 45.67)^2=(8.33)^2 = 69.4089 \)
- For \( 41 \): \( (41 - 45.67)^2=(-4.67)^2 = 21.8089 \)
Step 3: Find the sum of squared differences
Now we sum up all these squared differences:
\[
\]
Step 4: Calculate the variance ($\sigma^2$)
The variance is the average of the squared differences. Since this is a population (we are not dealing with a sample), the variance \( \sigma^2=\frac{\sum(x - \mu)^2}{n} \)
\[
\sigma^2=\frac{322.0201}{9}\approx35.78
\]
Step 5: Calculate the standard deviation ($\sigma$)
The standard deviation is the square root of the variance:
\[
\sigma=\sqrt{35.78}\approx5.98
\]
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5.98