### Video Transcript

A quiz was completed by 92 students and their scores were recorded in the following frequency table. Find the standard deviation to two decimal places.

The data for the scores achieved by the 92 students in the quiz has been presented in a grouped frequency table. We know that there are 26 students who achieved a score of more than zero and no more than 20, but we donβt know the exact scores for any of the students. We are asked to find the standard deviation of the scores, which is a measure of how dispersed they are about their mean value. But as we donβt know any of the exact values in the dataset, our answer will be an estimate.

We recall that for a dataset with a mean value of π and data values π₯ one, π₯ two up to π₯ π with frequencies π one, π two, up to π π, the standard deviation is given by Ξ£ π₯ is equal to the square root of the sum from π equals one to π of π₯ π minus π squared multiplied by π π over the sum from π equals one to π of π π. Letβs just think about what each of these pieces of notation mean. In the denominator of the fraction under the square root, we have the sum from π equals one to π of π π. Thatβs the sum of the frequencies or in other words the total frequency. In the numerator, we have π₯ π minus π all squared, so we subtract the mean from each data value and then square this. We multiply by the frequency for that data value and then find the sum.

We also need to recall that to calculate the mean of a set of data presented in a frequency table, we use the formula the sum from π equals one to π of π₯ π multiplied by π π over the sum from π equals one to π of π π. Now, the difference here is that instead of having a single π₯-value for each data point, weβre given intervals. And as a result, we canβt directly apply the formulae weβve written down. Instead, we will use the midpoint of each interval as the π₯ π value for that interval as this is our best guess of what the π₯-values are with the least error on average.

To find the midpoint of each interval, we calculate the mean of the endpoints of that interval, so zero plus 20 over two, which gives a midpoint of 10, 20 plus 40 over two, which gives a midpoint of 30, and so on to give the final midpoints 50, 70, and 90. Now, the frequencies π π are given to us in the second row of the table. We need to find the mean π before we can find the standard deviation, so weβll add another row to our table, in which we calculate π₯ π multiplied by π π. Thatβs the midpoint of each interval multiplied by its frequency. That gives the values 260, 300, 1200, 350, and 2430. The sum of these five values is 4540.

We already know that the total frequency the sum form π equals one to five of π sub π is 92. And so we can now calculate the mean of the dataset. Itβs 4540 over 92. In its simplest fractional form, thatβs 1135 over 23 and as a decimal 49.34 continuing.

So, weβve calculated an estimate for the mean of the dataset, and now we need to work through the process of finding the standard deviation. Weβre going to need to add various rows to our table: one for subtracting the mean from each of the midpoints, one for squaring each of these values, and the final row for multiplying by the corresponding frequency. Weβre going to use exact values throughout. So, to subtract the mean from each midpoint, weβre going to subtract the fractional value of 1135 over 23. That gives the values negative 39.3478 continuing, negative 19.3478 continuing, 0.6521 continuing, 20.6521 continuing, and 40.6521 continuing.

Weβre then going to square each of these values, and in fact you may find it more efficient to square each value immediately after you calculated it. So, we populate this row in the table. Next, we need to multiply each of these values by the frequency for that class. And again, you may find it more efficient to do this as you calculate each value. So, you may prefer to work down each column rather than across each row.

So, weβve now completed the table. To find the standard deviation, we need to calculate the sum of the five values in the bottom row of the table, divide by the total frequency, which we know to be 92, and then find the square root. That gives the square root of 90760.8695 continuing over 92. As a decimal, that is 31.4090 continuing. And then we round to two decimal places as required to give 31.41. We found then that the standard deviation of the scores, which remember is an estimate because the data were grouped, is 31.41 to two decimal places.