Video Transcript
The table shows the heights in
meters of the tallest buildings in a city. If there are any outliers in the
data, find their values.
Because we’re just given a table of
data and we want to find out if there are any outliers, we can use the 1.5-times-IQR
rule. An outlier 𝑥 would be less than 𝑄
one minus 1.5 times the IQR or, or the outlier would be greater than 𝑄 three plus
1.5 times IQR.
Our first step here is to calculate
the interquartile range and find these boundaries. And to do that, the first thing we
do is put the data in size order. We also know that each quartile is
25 percent of the data. That would be one-fourth of the
data. Since we have 12 building heights,
we can divide 12 by four, which is three. And that means our first quartile
will occur after the third data point, our second quartile after the sixth data
point, and our third quartile after the ninth data point.
Since quartile one is between the
third and fourth data point, we need to average the third and fourth data points to
find its value. 𝑄 one equals 561 plus 607 divided
by two, which is 584. We need to do the same thing for 𝑄
three. We average the ninth and 10th data
points, 714 plus 725 divided by two, which is 719.5. The interquartile range is 𝑄 three
minus 𝑄 one, for us, 719.5 minus 584, which equals 135.5.
Let’s make a list of what we
know. 𝑄 one is 584. 𝑄 three is 719.5. And our IQR is 135.5. We’re now ready to go back and use
these rules to calculate the upper and lower bounds for outliers. The lower bound for outliers will
be 𝑥 less than 𝑄 one minus 1.5 times the IQR. And the upper bound for outliers
will be such that 𝑥 is greater than 𝑄 three plus 1.5 times the IQR. Let’s plug in the values we have,
𝑄 one, 584, and the IQR, 135.5. When we do that calculation, we get
380.75.
And that means in order for there
to be an outlier on the low end, it needs to be less than 380.75. Our smallest data point is 502. And that means we don’t have an
outlier on the lower end. We’ll check the upper end. Plug in the values for 𝑄 three and
the IQR. And we find that the upper bound
for outliers is 922.75. In order for there to be an outlier
on the upper end, it would need to be greater than 922.75. Our largest data point is 901,
which is less than this value. And since none of our data values
are less than the lower bound for the outliers or greater than the upper bound for
outliers, there are no outliers in this data set.
