In Exercises 5 and 6, construct the dotplot.

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Question

In Exercises 5 and 6, construct the dotplot.

Pulse Rates Listed below are pulse rates (beats per minute) of females selected from Data Set 1 “Body Data” in Appendix B. All of those pulse rates are even numbers. Is there a pulse rate that appears to be an outlier? What is its value?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says, in a health study, the resting pulse rates, beats per minute, of a group of adult males were recorded. The recorded pulse rates are given below, and we have 62, 70, 72, 80, 74, 66, 78, 84, 88, 92, 70, 78, 66, 72, and 80. There are two parts to this question. For Part A, construct a dot plot for the given data, and for Part B, is there a pulse rate that appears to be an outlier? And the second part to this question is what is the outlier's value? Well, the problem we're given an empty graph on which to create our dot plot. Now, for Part A, we need to construct a dot plot. So the first thing we want to do is order our data, starting from the smallest value and going to the largest. So, if I look at the data, the smallest value is going to be 62. The next value is going to be 66, but that appears twice, so we'll have another value of 66. Then we'll have 70. And that actually appears twice, so our next value is gonna be 70 again. Then the next value is going to be 72. And it also appears twice, it will have another value of 72. Then we will have a value of 74. Then we'll have 78, but it appeared twice, so we'll have another value of 78. Then our next value is going to be 80. And it also appears twice, so we'll have another value of 80. Then the next value is going to be 84. The next value is going to be 88. And our final value is going to be 92. So, now we're going to create a dot plot. So, what we're going to do is place a dot for each of the values as it occurs in our sequence here. So, for 62, that appears once, so it will just have one dot. The next value is going to be 66, and it actually appears twice, it will have two dots. The next value is 70, and it appears twice, so it will have two dots. Next value is gonna be 72. It also appears twice, so it will have two dots. Then we'll have the value of 74. It appears once, so it will have 1 dot. The next value is 78, and it appears twice, so we'll have two dots for that value. The next value is going to be 80. It also appears twice. We'll have two dots on our plot. The next value is 84, it appears once, so we'll just have one dot. The next value is 88, it appears once, so we'll just have one dot, and the final value is going to be 92, and it appears once, and so we'll just have one dot. So, the plot that we have created is actually the answer for part A. Now, for Part B, we need to find if there is an outlier, and for this we're going to use our interquartile range in order to find outliers. So, the first thing we need to do is determine our 1st and 3rd quartiles. So, recall that the 1st quartile is going to be The median of the first half of our data. So, we're gonna be looking at the 1st 7 points. And if I look at the median value, that's going to be the 4th value in the sequence, and that turns out to be 70. So, the value of our first quartile, which we'll label as Q1, is going to be equal to 70. Now, to find the value of our 3rd quartile, we'll do the same thing with the last 7 values. And here we're again going to find the median. And so this again is going to be the 4th value in the sequence, and that turns out to be AD. So, Q3, which is going to be our third quartile, is going to be equal to 80. So now we just need to calculate our interquartile range. Which will label as I RQ or IQR, sorry. And that is going to be equal to 23. Minus 21, which is going to be equal to 80 minus 70, which is equal to 10. So now that we have our interquartile range, we can find the upper bound and lower bound. Which will help us to determine if we have any outliers. So for our lower bound, which we'll label as LV, That is going to be equal to, if you recall. First quartile Q1 minus. 1.5, multiplied by our interquartile range, IQR. And this is going to be equal to 70 minus 1.5, multiplied by 10. And this is going to be equal to 55 when we evaluate that expression. Now, we can also look for upper bound, which we'll label as UV. And that's going to be equal to our third quartile, Q3, plus 1.5 multiplied by our interquartile range, IQR. And so, this is going to be equal to for Q3, we had 80. Then we'll have plus 1.5, multiplied by 10, and evaluating this expression, this is going to be equal to 95. So, if we look at our values, we see that all of our values fall within the range of 62 to 92, and that is actually in the range of 55 to 95. So, since all of our data points fall within our lower bound and upper bound, that means that we have no outliers. So, the answer For Part B is that we have no outliers. So that is the answer we're looking for for that part. So, I hope that this explanation has been useful, and I'll see you in the next video.

In Exercises 5 and 6, construct the dotplot.

Pulse Rates Listed below are pulse rates (beats per minute) of females selected from Data Set 1 “Body Data” in Appendix B. All of those pulse rates are even numbers. Is there a pulse rate that appears to be an outlier? What is its value?

Key Concepts

Dot Plot

Outlier

Pulse Rate

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