The following table shows sample
data organized in a frequency distribution. If the estimated mean is 10, find
the missing frequency in the table.
We’re given data organized in a
frequency distribution. And we’re told that the estimated
mean is 10. This is only an estimate of the
mean because we don’t know any of the exact data values themselves. We only know how many values lie in
each class of the distribution.
We recall that the estimated mean
for a frequency distribution is found by multiplying the midpoint of each class by
its frequency, adding all these values together, and dividing by the total
frequency. This gives an estimate of the total
of all the values in the distribution divided by how many values there are.
We can’t yet work out either the
numerator or denominator of this fraction, because one of the frequencies in the
table is missing. Let’s call this missing frequency
𝑛 and work through the process of calculating the mean.
The first step is to calculate the
midpoint of each group, which we often denote as 𝑥. This is found by taking the average
of the class boundaries. In this distribution, the first
class, which is written as zero dash, contains all the values greater than or equal
to zero and strictly less than four, because that’s the lower boundary for the next
class. The midpoint is therefore equal to
zero plus four over two, which is two. Applying the same method, the next
three midpoints are six, 10, and 14.
For the final class, we have to
make an assumption about its width, because there is no class above it. We assume it has the same width as
the previous class. In fact all classes in this
distribution have a width of four. So we assume the upper boundary for
the final class is 20, and hence the midpoint is 18. Next, we multiply the midpoint of
each class by the frequency, which gives an estimate of the total of the data values
in each class. This gives 10, six 𝑛, 100, 56, and
72. We then need to find the total of
these values. This gives the simplified
expression six 𝑛 plus 238, which will be the numerator of the fraction to estimate
To find the total frequency, we sum
the values in the second row of the table, which gives the expression 𝑛 plus
23. We’re now ready to substitute these
expressions and the value of the estimated mean into the formula, which will give an
equation in 𝑛. Doing so gives 10 equals six 𝑛
plus 238 over 𝑛 plus 23. To solve this equation for 𝑛, we
first multiply both sides by the expression 𝑛 plus 23, which gives 10 multiplied by
𝑛 plus 23 equals six 𝑛 plus 238.
Moving to the top of the screen, we
can distribute the parentheses on the left-hand side to give 10𝑛 plus 230 equals
six 𝑛 plus 238. Then, we can collect like terms on
the same side of the equation by subtracting both six 𝑛 and 230 from each side,
giving four 𝑛 equals eight. Finally, dividing both sides of the
equation by four gives 𝑛 equals two.
At this point, we should feel
reassured that our answer is an integer, because it represents a frequency. So we can be reasonably confident
that we are correct. We’ve found that the missing
frequency in the table is two.