# Video: CBSE Class X • Pack 1 • 2018 • Question 26A

CBSE Class X • Pack 1 • 2018 • Question 26A

03:51

### Video Transcript

The mean of the following distribution is 18. Find the frequency, 𝑓, of the class 19 to 21.

So in this question, we’ve been given the mean of the data and asked to work backwards in order to calculate a missing frequency. Let’s recall the process for calculating the mean of a grouped frequency table.

First, we find the midpoint of each class by averaging out its endpoints. So, for example, for the first class, we add 11 and 13 together, divide by two, and it gives 12. The midpoints are therefore 12, 14, 16, 18, 20, 22, and 24.

As the data have been grouped, we don’t know the exact values of the original data. So these midpoints give an estimate of each value within the class. For each class, we then multiply its midpoint by its frequency. This gives 36, 84, 144, 234, 20𝑓, 110, and 96. Notice that, for the class of the missing frequency, the value of the midpoint multiplied by the frequency is in terms of 𝑓.

The next step is to find the sum of all of the midpoints multiplied by their frequencies, which gives an estimate of the sum of all the data values. Remember that this notation here, which is the symbol for the Greek letter Σ, means sum.

Now you may have to use perhaps a column addition method to add all of those values together, but it simplifies to 704 plus 20𝑓. So remember, this sum gives an estimate of the sum of all of the data values.

The next step in calculating the mean is to divide by the total frequency. So we have three plus six plus nine plus 13 plus 𝑓 plus five plus four. This simplifies to give the expression 40 plus 𝑓. So this is how the mean for this distribution would be calculated.

But remember, we already know that the mean of this distribution is 18. So we can set this expression equal to 18, and it gives us an equation. To find the value of 𝑓, we just need to solve this equation.

Now The first step is to eliminate the denominator on the left side of the equation. So we multiply both sides by 40 plus 𝑓, giving 704 plus 20𝑓 is equal to 18 multiplied by 40 plus 𝑓. To find 18 multiplied by 40, we can first find 18 multiplied by four by doubling 18 and then doubling it again. 18 multiplied by two is 36, and then multiplying by two gives 72. So if 18 multiplied by four is 72, then 18 multiplied by 40 is 720. And expanding the brackets, so multiplying the second term by 18 as well, gives 720 plus 18𝑓.

So now we have a relatively straightforward equation to solve, but it’s got 𝑓s on both sides. So the first step is gonna be to group all of the 𝑓s on the same side of the equation. We’re going to subtract 18𝑓 from each side. Doing so gives 704 plus two 𝑓 is equal to 720.

Right, we’re nearly there with solving this equation. The next step is to subtract 704 from each side, giving two 𝑓 is equal to 16. The final step is just to divide both sides of this equation by two. And this gives 𝑓 is equal to eight. So the frequency of the class 19 to 21 is eight.