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Question Video: Determining the Standard Deviation of a Data Set Presented in a Frequency Table Mathematics • 9th Grade

The table shows the distribution of goals scored in the first half of a football season. Find the standard deviation of the number of goals scored. Give your answer to three decimal places.

04:55

Video Transcript

The table shows the distribution of goals scored in the first half of a football season. Find the standard deviation of the number of goals scored. Give your answer to three decimal places.

Looking at the table we’ve been given, we can see that the data has been presented in a frequency table. The first row gives the number of goals scored, and the second row gives the number of matches in which that number of goals were scored, or in other words, the frequencies. For example, there were five matches in which zero goals were scored. We could therefore add the labels π‘₯ sub 𝑖 to represent the data, which are the number of goals, and 𝑓 sub 𝑖 to represent the frequencies.

We are asked to find the standard deviation of the number of goals scored, which is a measure of how dispersed or spread out the data is around its mean value. We can recall that for a data set 𝑋 containing the values π‘₯ sub one, π‘₯ sub two, up to π‘₯ sub 𝑛 with corresponding frequencies 𝑓 sub one, 𝑓 sub two, up to 𝑓 sub 𝑛 and mean πœ‡, the standard deviation, which we denote as 𝜎 𝑋, is given by 𝜎 𝑋 equals the square root of the βˆ‘ from one to 𝑛 of π‘₯ 𝑖 minus πœ‡ squared multiplied by 𝑓 i over the βˆ‘ from one to 𝑛 of 𝑓 i.

We should also recall that the mean of a set of data presented in a frequency table, πœ‡, is given by the βˆ‘ from one to 𝑛 of π‘₯ 𝑖 multiplied by 𝑓 𝑖 over the βˆ‘ from one to 𝑛 of 𝑓 i.

Practically, what this means when we calculate the standard deviation of a set of data presented in a frequency table is we calculate the mean πœ‡ first. We then subtract this from each π‘₯-value and square. We multiplied by the frequency for that π‘₯-value and find the βˆ‘. We then divide this quantity by the βˆ‘ of the frequencies or the total frequency. And finally, we take the square root.

We can extend the table we’ve been given in order to work through the process. The first thing we need to find in order to calculate the mean is the product of each π‘₯-value with its corresponding frequency. We have zero times five, which is zero; one times two, which is two; three times seven, which is 21; four times seven, which is 28; and six times four, which is 24. The βˆ‘ of these values, and that’s the βˆ‘ for 𝑖 from one to five because there are five π‘₯-values in the data set, is 75.

We also need to calculate the total frequency by summing the values in the second row of the table, five plus two plus seven plus seven plus four, which is equal to 25.

To find the mean, we divide the βˆ‘ of each π‘₯-value multiplied by its frequency, which is 75, by the total frequency of 25 giving πœ‡ is equal to three. So, we’ve calculated the mean, and now we need to calculate the standard deviation. In the next row of our table, we’re going to subtract the mean of three from each π‘₯-value. That gives negative three, negative two, zero, one, three.

In the next row of the table, we square each of these values, giving nine, four, zero, one, and nine. Finally, we multiply each of these values by the frequency for that π‘₯-value. So we’re multiplying the values in the fifth row of our table by the values in the second row, giving 45, eight, zero, seven, and 36.

Now, returning to the formula for the standard deviation, the numerator of the fraction underneath the square root is the βˆ‘ of each π‘₯ 𝑖 value minus the mean squared multiplied by the frequency. Summing the values in the final row of the table gives 96. So this will be the numerator of the fraction. And we’ve already worked out that the total frequency is 25.

So we have that the standard deviation 𝜎 sub 𝑋 is equal to the square root of 96 over 25. In exact form, that simplifies to four root six over five. But we’ve been asked to give our answer to three decimal places. So evaluating this as a decimal gives 1.9595 continuing. The value in the fourth decimal place is a five, so we round up.

We found then that the standard deviation of the number of goals scored in the first half of the football season, to three decimal places, is 1.960.

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