Modified Box-and-Whisker Plot In Exercises 59...

Question

Modified Box-and-Whisker Plot In Exercises 59–62, (a) identify any outliers and (b) draw a modified box-and-whisker plot that represents the data set. Use asterisks (*) to identify outliers.

75 78 80 75 62 72 74 75 80 95 76 72

Accepted Answer

Hello everyone. Let's take a look at the question together. So here we need to identify any outliers and draw a modified box and whisker plot that represents the following data set. We have our values or our data set, and our empty box plot. So, starting with the first part of the question, where we need to identify any outliers, the first step in solving this problem is to arrange the date in ascending order. So now we have our data set that is arranged in ascending order. Then the next step is to find Q1, Q2, and Q3. Well, we know that there are 12 values, so our Q2, which is the median, is calculated as the average of the 6th and 7th values, which looking at our data set, our Q2, or our median is calculated as 65 plus 67 divided by 2, giving us a Q2 value of 66. And using that Q2 value of 66, we know the lower half or the first numbers are the following values from our data set, which we can use to calculate our Q1 or the media of the first half. So our Q1 can be calculated as 62 plus 63, divided by 2, giving us our Q1 of 62.5. Then we have our upper half, which are the last 6 numbers, and are the following numbers from the data set, which we can use to find our Q or our median of this upper half. So then our Q3 is calculated as 70 + 72, divided by 2, giving us a 3 value of 71. Then using our Q1, Q2, and Q3 values, we can find IQ. Which we know IQR is calculated as Q3 minus Q1, so plugging in the values for both Q3 and Q1, our IQR is equal to 71 minus 62.5, giving us a final value of 8.5. Depending our calculated value so far, we can determine the outlier boundaries, starting with the lower fence, which is calculated as Q1 minus 1. multiplied by our IQR value and substituting the values into the formula, we get 62.5 minus 1.5. Applied by 8.5, so we have a lower fence value of 49.75, and then for the upper fence, we have Q3 plus 0.5 multiplied by IQR which substituting in the values, we have 71 plus 1.5, multiplied by 8.5, giving us an up value of 83.75. Therefore, any number below 49.75 and above 83.75 is an lier, which looking at our data set, using these boundaries, we can identify that the value of 90 is an outlier. Therefore, 90. is an outlier, and we can go ahead and draw a modified box and whisker plot that runs our data set. And so drawing the box plot, we first draw the box from 1 to Q3, then we make a line inside the box at Q2, which is our median. Then the left whisker connects the minimum value 58 to 1. And the right whisker connects Q3 to the maximum value of 75, as well as the outlier, which we know is 90, is shown as a dot, and following the steps gives us the following box of our data set. So we know that our outlier is 90, and we have the modified box and whisker plot that represents our data set. And looking at our answer choices can identify that answer choice D is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and

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Modified Box-and-Whisker Plot In Exercises 59–62, (a) identify any outliers and (b) draw a modified box-and-whisker plot that represents the data set. Use asterisks (*) to identify outliers.

75 78 80 75 62 72 74 75 80 95 76 72

Key Concepts

Outliers

Box-and-Whisker Plot

Interquartile Range (IQR)

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Modified Box-and-Whisker Plot In Exercises 59–62, (a) identify any outliers and (b) draw a modified box-and-whisker plot that represents the data set. Use asterisks (*) to identify outliers.75 78 80 75 62 72 74 75 80 95 76 72