Select the outlier in the data set. $\newline$ $4, 68, 76, 77, 83, 85, 87, 95, 98$ $\newline$ If the outlier were removed from the data set, would the mean increase or decrease? $\newline$ Choices: $\newline$ (A)increase $\newline$ (B)decrease

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Algebra 1

Identify an outlier and describe the effect of removing it

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Q. Select the outlier in the data set.

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4, 68, 76, 77, 83, 85, 87, 95, 98

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If the outlier were removed from the data set, would the mean increase or decrease?

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Choices:

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(A)increase

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(B)decrease

Identify Outlier: Identify the outlier in the data set. $\newline$ To find the outlier, we can look for a number that is significantly different from the rest of the numbers in the set. In this case, the number $4$ stands out as it is much lower than all other numbers.
Determine Outlier Using IQR: Determine if $4$ is an outlier using the interquartile range (IQR) method. First, we need to find the quartiles ( $Q1$ , $Q2$ , and $Q3$ ) of the data set. Since the data set is already ordered, we can easily find the median ( $Q2$ ), which is $83$ . The lower half of the data set is $4$ , $68$ , $76$ , $77$ , and the upper half is $Q1$ $0$ , $Q1$ $1$ , $Q1$ $2$ , $Q1$ $3$ . The median of the lower half is $Q1$ $4$ (average of $68$ and $76$ ), which is $Q1$ , and the median of the upper half is $Q1$ $8$ (average of $Q1$ $1$ and $Q1$ $2$ ), which is $Q3$ . Next, calculate the IQR: $Q2$ $2$ . Now, calculate the lower bound for outliers: $Q2$ $3$ . Since $4$ is less than $Q2$ $5$ , it is considered an outlier.
Calculate Mean with Outlier: Calculate the mean of the data set with and without the outlier. $\newline$ First, calculate the mean with the outlier: $(4 + 68 + 76 + 77 + 83 + 85 + 87 + 95 + 98) / 9 = 673 / 9 \approx 74.78$ . $\newline$ Next, calculate the mean without the outlier: $(68 + 76 + 77 + 83 + 85 + 87 + 95 + 98) / 8 = 669 / 8 \approx 83.63$ .
Determine Mean Change: Determine if the mean would increase or decrease after removing the outlier. Comparing the two means calculated in Step $3$ , we see that the mean without the outlier ( $83.63$ ) is higher than the mean with the outlier ( $74.78$ ). Therefore, removing the outlier would increase the mean.