Video Transcript
The following table shows sample data organized in a frequency distribution. Estimate the value of the mean, giving your answer to one decimal place.
The data has been presented in a frequency distribution. We’re given classes, written as zero dash, three dash, six dash, nine dash, and 12
dash, and then the frequency for each class. We recall that, in general, the mean of a data set is found by calculating the sum of
all the data values and dividing this by how many values there are. Because we don’t know the exact values of the data, however, we can only estimate the
value of the mean.
We can work out the number of values, which is the total frequency, by summing the
values in the second row of the table. And it gives 30. We can only estimate the sum of all the data values. And to do this, we first need to find a single value that is representative of each
class. The best value to use is the midpoint of each class. This is found by taking the average of the class boundaries.
The first class, which is written as zero dash, contains all the values that are
greater than or equal to zero but strictly less than three, because that is the
lower boundary for the next class. We therefore take three to be the upper boundary for the first class. The midpoint is then equal to zero plus three over two, which is 1.5.
For the next class, the lower boundary is three and the upper boundary is six. The midpoint is therefore three plus six over two, which is 4.5. The midpoints for the next two classes are 7.5 and 10.5.
For the final class, we have to make an assumption about its width because there
isn’t a class that comes after it. We assume that this final class has the same width as the previous one. In fact, every class in this distribution has the same width of three. So we assume that the upper boundary for the final class is 15. The midpoint for the final class is therefore 12 plus 15 over two, which is 13.5.
We can then estimate the sum of the values in each class by multiplying the midpoint
by the frequency. For example, in the first class, there are five data values which are each
approximately equal to 1.5. So the total of the values in this class is approximately 1.5 multiplied by five. We can repeat this process for each of the remaining classes. The total of these five values, which is 237, gives our estimate for the sum of all
the data values. We can then tweak our formula for the mean, to say that the estimated mean is equal
to the estimated sum of all the data values divided by the number of values. That’s the sum of all the midpoints multiplied by the frequencies divided by the
total frequency.
Using the values we’ve calculated, this is equal to 237 over 30, which simplifies to
7.9. We’re asked to give our estimate to one decimal place, but as this value is exact to
one decimal place already, no rounding is required. And so we have that the estimated mean of the distribution is 7.9.
