Identify the outliers of the data set. Then determine if the outlier increases or decreases the value of the mean.176, 183, 186, 172, 177, 133, 181, 190, 179, 184, 188
133 is the outlier. Would including that score increase or decrease the mean?
To identify the outliers in a data set, you can use the following steps:
Step 1: Arrange the data set in ascending order.
133, 172, 176, 177, 179, 181, 183, 184, 186, 188, 190
Step 2: Calculate the interquartile range (IQR) of the dataset.
IQR = Q3 - Q1, where Q1 is the first quartile and Q3 is the third quartile.
Q1 = median of the lower half of the data = median of (133, 172, 176, 177, 179) = 176
Q3 = median of the upper half of the data = median of (183, 184, 186, 188, 190) = 186
IQR = 186 - 176 = 10
Step 3: Determine the lower and upper fence.
Lower fence = Q1 - (1.5 * IQR)
Upper fence = Q3 + (1.5 * IQR)
Lower fence = 176 - (1.5 * 10) = 161
Upper fence = 186 + (1.5 * 10) = 201
Step 4: Identify the outliers.
Any data points below the lower fence or above the upper fence are considered outliers.
Outliers: 133 (below the lower fence), 190 (above the upper fence)
Step 5: Determine the effect of the outliers on the mean.
To determine if the outliers increase or decrease the mean, we need to calculate the mean of the original dataset (without the outliers) and compare it to the mean of the dataset including the outliers.
Mean (without outliers) = (176 + 177 + 172 + 179 + 181 + 183 + 184 + 186 + 188) / 9 = 180.11
Mean (with outliers) = (176 + 177 + 172 + 179 + 181 + 183 + 184 + 186 + 188 + 133 + 190) / 11 = 179.64
Comparing the means, we can see that the outlier values decrease the mean.
To identify the outliers in a data set, you need to calculate the mean and standard deviation of the data. The outliers will be values that are significantly higher or lower than the rest of the data.
Let's start by finding the mean of the data set. To do this, add up all the values and divide by the total number of values:
176 + 183 + 186 + 172 + 177 + 133 + 181 + 190 + 179 + 184 + 188 = 2049
Total number of values = 11
Mean = 2049 / 11 = 186.27
Next, calculate the standard deviation of the data set. The standard deviation tells you how spread out the data is from the mean.
To calculate the standard deviation, subtract the mean from each individual value, square the result, sum up the squared differences, divide by the total number of values, and take the square root of the result.
Here are the steps involved:
1. Subtract the mean from each value:
(176 - 186.27), (183 - 186.27), (186 - 186.27), (172 - 186.27), (177 - 186.27), (133 - 186.27), (181 - 186.27), (190 - 186.27), (179 - 186.27), (184 - 186.27), and (188 - 186.27)
2. Square each difference:
(-10.27)², (-3.27)², (-0.27)², (-14.27)², (-9.27)², (-53.27)², (-5.27)², (3.73)², (-7.27)², (-2.27)², and (1.73)²
3. Sum up the squared differences
(-10.27)² + (-3.27)² + (-0.27)² + (-14.27)² + (-9.27)² + (-53.27)² + (-5.27)² + (3.73)² + (-7.27)² + (-2.27)² + (1.73)² = 6089.3
4. Divide by the total number of values (11 in this case)
6089.3 / 11 = 553.57
5. Take the square root of the result
√553.57 = 23.53
Now that we have calculated the mean (186.27) and standard deviation (23.53), we can identify the outliers.
Typically, values that are more than 3 standard deviations away from the mean are considered outliers. In this case, we will use that criterion.
To find the outliers, we need to check if any of the values are more than 3 standard deviations away from the mean.
Mean - (3 * Standard Deviation) = 186.27 - (3 * 23.53) = 115.68
Mean + (3 * Standard Deviation) = 186.27 + (3 * 23.53) = 256.86
From the original data set, we can see that 133 is less than 115.68, which is more than 3 standard deviations away from the mean. Hence, 133 is an outlier.
Now, to determine if the outlier increases or decreases the value of the mean, we can compare the mean before and after removing the outlier.
Mean (with outlier) = 186.27
Mean (without outlier) = (2049 - 133) / 10 = 191.60
Comparing the two means, we see that removing the outlier increases the mean from 186.27 to 191.60.
Therefore, the outlier decreases the value of the mean.