# Question Video: Using the Mean of a Frequency Distribution to Calculate a Missing Frequency Mathematics

The following table shows sample data organized in a frequency distribution. If the estimated mean is 10, find the missing frequency in the table.

04:18

### Video Transcript

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 the mean.

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.