Video Transcript
The frequency table below gives the
heights of a group of students. Which of the following histograms
represents the given data?
In this question, we’re thinking
about histograms. When we have a histogram, on the
𝑦-axis, we have frequency density, rather than frequency, which we would have in a
bar chart. This is why if we have a quick look
at our options, on our first bar, the height doesn’t go up to 60 as this would be
simply representing the frequency. The frequency in a histogram is
found by calculating the areas of the bars. To create a histogram from the
given frequency table, the first thing we need to do is calculate the frequency
density for each class. We can find frequency density by
finding the frequency and dividing by the class width. And what exactly do we mean by
class width?
Well, if we take a look at our
first class, the height ℎ of this group of students, in the first interval, we have
the inequality that 150 is less than ℎ is less than or equal to 153 centimeters. In this interval, we would
therefore have students who are 151 centimeters, 152 centimeters, or 153
centimeters. We couldn’t have students here
exactly equal to 150 centimetres because of this inequality which doesn’t include
the equals to. The class width in this case would
be three. So to find the frequency density,
we would take our frequency of 60 and divide by the class width of three. And 60 divided by three is 20.
In the second class, we can find
our class width by having 156 subtract 153. In this case, the class width would
be three. The values in this interval would
be 154, 155, and 156. Easing the formula to find the
frequency density, we would take our frequency of 45, divide by the class width of
three. And we’d have a value of 15 for
this frequency density. In our next column, the class width
is a little bit more interesting. In this case, we can’t include the
value of 156 centimeters, since our inequality says that we can’t have an equals to
this value. This interval will go up to and
including the value of 157. So we just have one possible
value. Frequency density here is the
frequency of 30 divided by the class width of one, giving us a frequency density of
30.
The next class will have a class
width of five, since we have the values 158, 159, 160, 161, and 162. The frequency density is the
frequency of 100 divided by the class width of five, which is 20. In our final class interval, we’ll
have a frequency density of 15, as 45 divided by the class width of three gives us
15.
So let’s take a look at the
histogram given in option (A). Our first interval goes from 150 to
153. The frequency density is 20. The first histogram shows this bar
correctly. The interval of 153 to 156 has a
correct frequency density of 15. The third interval from 156 to 157,
however, in option (A) is not given correctly. Here, the frequency density would
be given as 15. We could assume that the person who
drew this histogram incorrectly worked out the class width as two. So we can rule out option (A) as
the histogram.
If we look at options (B) and (C),
both of these have the same correct two bars to begin with. If we take a close look at option
(B), the interval 156 to 157 we can see on the histogram is incorrectly given as the
interval from 156 to 159. So this bar is wrong. And we can rule out option (B) as
the histogram. We need to check that the other
bars in option (C) are completely correct. We can see that our third bar
correctly has a frequency density of 30. The fourth bar, which structures
correctly along from 157 to 162, does have a correct frequency density of 20. The final bar does correctly go
from 162 to 165 on our 𝑥-axis. And the frequency density is
correct at 15. We can therefore say that option
(C) is the histogram to represent the data in the frequency table.
