Video: Determining an Outlier and Then Excluding It to Find the Mean of a Data Set Represented on a Bar Graph

Bethani Gasparine

The bar graph shows the prices of six different kitchen gadgets. Identify the outlier, and then find the mean of the data if the outlier is not included.

02:37

Video Transcript

The bar graph shows the prices of six different kitchen gadgets. Identify the outlier and then find the mean of the data if the outlier is not included.

So an outlier will be a piece of the data that just doesn’t quite seem to fit. So just looking at the graph, more than likely it’s this 22. It’s not quite as high as all of the other ones. It’s very different. But we need to actually calculate to find the outlier just to be safe.

So to find the outlier, we need to begin by putting the price of the gadgets in ascending order — so from smallest to largest. So we have 22, 151, 160, 178, 193, and 197. So there are six numbers here. So we don’t have to worry about finding the median — the middle. But we do need the upper and lower quartile. So here we have a lower half and the lower quartile will be the middle of that. So our lower quartile will be 151. And then, the middle of the upper half will be our upper quartile. So what we need to do is find the interquartile range by subtracting these. So the interquartile range will be 42 IQR.

And to find where the outlier must be located outside of, we need to take the interquartile range and multiply it by 1.5. And we get 63. So we need to subtract 63 from the lower quartile and add 63 to the upper quartile. If we take the lower quartile of 151 and subtract 63, we get 88. And if we take 193, the upper quartile, and add 63, we get 256. So If there are any numbers that are less than the 88 or greater than the 256, these will be considered outliers. And the only number that does that is 22, just like we thought.

So it also says to find the mean of the data if the outlier is not Included. So we need to find the mean or the average of all of these numbers. So there are five of them. So we need to add them all up and divide by five. And when adding them together, we get 879. And now, we need to divide by five and we get a mean of 175.8.

So our final answer will be that the outlier will be the 22 dollars. And the mean of the data if the outlier was not included would be 175 dollars and 80 cents.

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