Lesson video

Good day, everyone, and welcome to another numerical summaries lesson with me, Mr. Grason, thank you so much for joining me in this lesson, where we will be looking at a certain type of mean, the weighted mean.

The definitions of weighting factor and the weighted mean will be introduced throughout the lesson.

But first, let's have a look at the mean across groups of different sizes.

In this scenario, we have two groups of runners, each with 4 people in them.

The mean distance that each group ran is 8 kilometres and 10 kilometres respectively.

Andeep says that, "The mean distance that everybody ran is 9 kilometres, this is because the mean of 8 and 10 is 9." For this specific example, I agree with Andeep.

However, in this different scenario, the two groups are of different size.

Group C has 3 people, whilst Group D has 6 people, will this affect how we calculate the mean? Andeep does not seem to think so, he thinks that the mean distance ran across both these groups is still 10 kilometres since the mean of 7 and 13 is 10.

Sam, on the other hand, disagrees.

Sam says, "Simply taking the mean of these two numbers just is not fair.

Group D has more people in it than Group C, twice as many, in fact, and so it is only fair that the mean represents Group D's distances a little bit more than Group C's distances." Andeep, however, thinks that "Because 7 kilometres and 13 kilometres are just both numbers, the mean of the two groups is simply the sum of 7 and 13, then divided by 2." So surely Sam is overcomplicating things, right? Well, actually Sam is correct.

7 and 13 are not just numbers, they are the means of two different sets of data.

Both 7 and 13 represent far more information than just the distance that a single person ran.

We do not know how far each person ran, however, we don't need to know this, as the mean shows the sum of their distances split evenly between all the people in the group.

For Group C, the total distance that they ran can be shared evenly between the 3 people as 7 kilometres each.

Similarly, for Group D, the total distance that they ran can be shared evenly between the 6 people as 13 kilometres each.

Therefore, Group C ran a total of 21 kilometres.

We can distribute the 21 kilometres between the three people evenly so that the mean distance ran per person is 7 kilometres.

But equally, the 3 people might have run some other combination of distances that also sums 21, such as 2, 8, and 11.

This is similar for Group D.

All 6 people ran a total of 78 kilometres, meaning that each person ran a mean distance of 13 kilometres.

However, the exact distance that each person ran may vary.

Andeep, seeing the mean data separated out into 3 sevens and 6 thirteens, seems to understand things a little bit more clearly now.

To find the mean of everybody's distances, we first have to find the total of every person's distance.

That is 3 lots of 7, plus 6 lots of 13, so 99 kilometres in total is the total distance ran by everybody.

Andeep asks, "To find the mean, do we still divide by 2 because there are two groups?" Sam thinks not.

Sam believes we should be dividing by 9 instead, since there are 9 people in total across both groups.

This is the same as saying that there are 9 data points in the dataset of Group C and Group D combined.

Therefore, the mean is 99 divided by 9, which is 11.

The mean distance run by the members of this combined group, every person in Groups C and D, is an average of 11 kilometres per person.

We can see that this number 11 is closer to 13, the mean of Group D, than it is to 7, the mean of Group C.

This is because there are more people in Group D, and so the mean should represent the values in Group D more than the values in Group C.

Onto a check.

For these two other groups, pause here to consider, which of these statements is correct.

All 6 members of Group E ran a total of 24 kilometres, this is because the mean of 6 kilometres represents taking the total of 24 kilometres and distributing that number evenly across all 4 people.

However, this does not imply that each person ran exactly 6 kilometres.

We do not know how far each person ran exactly.

Now we know the total distance run by people in Group E.

Pause here to find the total distance run by everyone, in both groups, both E and F.

Group F ran a total of 66 kilometres, therefore the total distance run by everybody is 90 kilometres.

Hence, what was the mean distance run by all the people in the combined group, E and F? Pause now to figure this out, by first of all finding out how many people in total there are in Groups E and F.

In total, across both groups, there are 10 people.

So the total distance of 90 divided by the 10 people, gives you a mean distance run per person at 9 kilometres.

Brilliant, onto the practise for this task.

For Question 1, pause here to consider different totals and means for these two groups of teddy bears.

And for Question 2, we have three groups or teams of people.

Each team has a different amount of members and the winning team is the team with the highest means score across the members in their team.

Pause here to answer, why the mean and not the total score might be more suitable in this context, and which team could Team B combined with in order for the new combined team to have the biggest increase in mean score, when compared to Team B's current mean score of 1,200? Great stuff, onto the answers.

For Question 1, the total mass of the bears in Group A is 140 grammes times by 4, which is 560 grammes.

For Part B, the total mass of bears in group B is 2,058 grammes.

For Part C, therefore, the total mass of all the bears is 2,618 grammes.

And for Part D, the mean mass of the 11 bears is 238 grammes.

This value is closer to the mean mass of Group B than Group A because Group B had more bears, and for part E, a larger proportion of Group B's bears are big bears, when compared to Group A.

The mean score is more fair because if the total score was used instead, the team with the most members will have a greater chance of getting the highest score, even if each individual player wasn't even that good.

And finally, 2 B, Team B should join Team C, not Team A.

Pause here to check your calculations and compare them with the ones on screen.

Next up, is something very similar to what we've just seen, although the context is a little bit different.

Let's find out about the weighted mean.

Andeep acknowledges that, "Finding the mean isn't always as straightforward as taking some data, adding it all together, and then dividing by the number of data points collected." Sam agrees, "Sometimes, some data points are more important or impactful, and so need to be counted for more when calculating the mean." Sam uses the example of their maths scores.

If Sam wants to find the mean of all of their scores to see how well they're doing in maths, surely their more recent scores should count for more.

This is because those scores more closely reflect their current ability.

This is an example of a weighted mean.

The weighted mean is a type of mean where some data points are considered more or less important than other data points in the same dataset.

In order to weight a data point, we could count that data point several times.

This means those data points will play a bigger part in the calculation of the mean.

Here's a demonstration of the weighted mean in a way where Sam's more recent test scores will count for more In the calculation of the mean.

Sam could count the year 7 scores once, count the year 8 scores twice, like so, count the year 9 scores three times each, like so, and count the year 10 scores four times each, like so.

If Sam were to weight the scores this way, their year 10 scores would be 4 times as important as their year 7 scores, which makes sense with what Sam is trying to achieve.

Andeep correctly identifies that, "The total of all of these scores is 942." However, Sam did six tests and Andeep says, "942 divided by 6 is 157.

Which is far too high to be the mean!" Since the mean should be a value between 45, the lowest score, and 74, the highest score.

Andeep is correct to question this incorrect answer.

The total weighted mean of 942 is correct, but dividing by 6 is an incorrect step to finding a weighted mean.

Pause now, to think about or discuss possible reasons why this is an incorrect step, and what correct step could be done to find the correct weighted mean? It is true that Sam did 6 tests, however, in the calculation defined, the weighted mean, we counted some of those test scores multiple times in order to weight those scores.

Therefore, when calculating the weighted mean, we divide this weighted total by this modified number of tests in our calculation.

There are 15 tests in our weighted calculation, and so the weighted mean is 942 divided by the 15 weighted tests equals 62.

8.

Sam is satisfied that this seems like a sensible and representative weighted mean.

Firstly, 62.

8 is between the lowest and highest scores, but it is also closer to some of the more recent scores of 60, 65, and 74, than it is to some of Sam's younger scores of 45 and 52, with that year 8 score of 61 being a little bit of an anomaly.

For this check, Aisha wants her year 8 score to count for twice as much as her year 7 scores, and her year 9 score to count for four times as much as her year 7 score.

Pause here to calculate a weighted total for Aisha's weighted scores.

Because some scores were weighted as more important than the others, we consider these to be her weighted scores.

Therefore, her weighted total score is the sum of all eight of these scores, giving 414 in total, and hence, pause here to use all of this information to calculate Aisha's weighted mean.

Okay, the weighted total of 414, divided by the number of weighted scores, which is 8 weighted scores, therefore, 414 divided by 8 is 51.

75, as her weighted mean test score.

We can show how much importance we put on a data point by multiplying that data point by a weighting factor.

Multiplying a data point by a weighting factor is the same as duplicating a data point multiple times in order to make it more significant when calculating a weighted mean.

For example, with Sam's test scores, their year 10 test scores were four times as important as their year 7 test scores, and so the weighting factor for year 10 was 4, and a year 7 weighting factor was 1.

After considering the weighting factors, it is sensible to create a weighted value column, which is the product of the data point and its weighting factor.

The weighted values for Sam's test scores are 45 times 1, 61 times 2, and so on.

The total weighted value of Sam's tests is then 942.

Amazing, Andeep seems to get the idea now.

He is using the earlier representation that Sam showed to identify that we need to take the total weighted score of 942 and divide it by the total weighting factor as that's the number of times that each test score has been counted.

The total weighting factor for this example is 15, and therefore, the weighted mean of Sam's test scores is 942 divided by 15 equals 62.

8.

A value that we agree with from Sam's earlier representation.

Furthermore, Andeep correctly identifies that this table and its calculations look very similar to finding the mean from a frequency table.

Sam agrees the calculations are basically the same.

The only difference is that we have weightings for each data point, rather than the frequencies of each data point.

Let's have a look at this new example together, in order to consider the weighting factors, let's add two more columns to help us with our calculations.

One, where we'll put the weighting factor for each event, and the other for the product of the position and the weighting factor.

Pause here to have a look at the scenario and check out the table we'll be using to calculate the weighted mean.

Sometimes, the weighting factors will be given to you directly, but sometimes, you'll have to interpret them from the information given.

This says that hurdles are counted as more important than the sprint, therefore, the hurdles will have a greater weighting factor than the sprint.

Since no exact weightings are given, we can simply say that the sprint has a factor of 1, and therefore, the hurdles have a factor of 1 times 1.

5, or just 1.

5.

We could have picked any value for the weighting factor of sprint, and the weighted mean at the very end of all of our calculations, will have been the same.

However, choosing one helps keep the rest of our calculations relatively simple.

Next up, we have that long distance is 2.

5 times as important as the sprint, therefore, 1 for the sprint, times by 2.

5, equals 2.

5 as the factor weighting factor for long distance.

Okay, after finalising all of the weighting factors, we then multiply each position by its weighting factor, and place the result in that final right-hand column.

An important step is then, to find the weighted total in that rightmost column, and the total weightings of all of the weighting factors in the weighting factors column.

The weighted mean is therefore, 36 divided by 5, which equals 7.

2.

This person's weighted finishing position is the seven point second place.

Everyone in the competition can then compare their weighted finishing positions, and whoever has the lowest weighted finishing position is the overall winner.

For this check, by reading how andeep weights the opinions of different people who he interviews, consider the possible weighting factors of A to C.

Pause now, to go through this scenario and identify information that can be translated into weightings.

Let the weighting factor for tourists be 1, therefore, the weighting factor for the people who work there are 3, and for the locals 5, and pause here to consider what are the values of D to F? Each value in that rightmost column is a product of the ratings and its weighting factor, and pause here to consider two totals, and use those two totals to find the weighted mean rating.

The total weighted rating is 31, and the total weighting factor is 9, 31 divided by 9 is about 3.

4.

Great stuff, onto the practise.

Pause here to complete both tables, and use the relevant values to calculate the weighted means of both data sets.

And pause here for Question 3, where you first have to identify the weighting factors before calculating the weighted mean.

Great work, here are the answers.

For Question 1, Izzy did just about pass the course as her weighted mean was just over 70, at 70.

462.

And for question 2, the family should not buy the house, as the weighted score was just a little bit under 7.

5.

This is quite surprising to me since the room with the highest weighting factor, the kitchen, actually scored a 9 out of 10.

And for Question 3, pause here to compare all of your calculations to the ones on screen.

So we've seen weighted means with specific weighting factors, but how are these weighting factors chosen? And how can they be presented? Let's have a look.

Weighting factors can be considered in many different ways, including as percentages, fractions, or ratios.

For example, Izzy's mum invests money in a business, different outcomes to the money that she invests, have different probabilities of occurring.

There is a 20% chance of a "big hit," giving Izzy's mum a 1,000-pound return on that investment.

There is a 50% chance of a "small hit," giving Izzy's mum a smaller, but still pretty good, 400 pound return on that investment.

But actually, the remaining chance is a "miss" where Izzy's mum actually loses her money.

She loses 500 pounds on her investment.

We can find the weighted mean expected return on Izzy's mum's investment, but how, how can we calculate the weighted means from this example, which includes percentages? Let's have a look through this example together.

Okay, let's use this familiar table to help us calculate the weighted mean.

There are three types of hits that Izzy's mum could make, these three values.

A "big hit" of 1,000 pounds, a "small hit" of 400 pounds, and a "miss" of negative 500 pounds.

Pause here to think about or discuss why this final value is a negative.

If Izzy's mum loses money on the investment, then she's earning a negative amount.

Next up, the weighting factors are these three probabilities.

We've got 20% for the "big hit," I'm not going to factor in the percentage at the moment, so I'm only gonna put in a weighting factor of 20.

Next up, we've got 50% for a "small hit." To find the remaining percentage, we first of all have to consider the total weight factor of 100, and then find out what is missing from 100% in that remaining missing weighting factor.

So we've got a total weight factor of 100, 20 plus 50 plus 30 is 100, and so the missing chance is 30%, and so the missing weighting factor is 30.

We then calculate the product of the value and the weight factor in the fourth column.

We've got 1,000 times by 20 is 20,000, 400 times by 50 is also 20,000, but negative 500 times by 30 is negative 15,000.

Note, the products are also allowed to be negative.

The sum total of everything in that fourth column is 25,000 because 20,000 plus 20,000 is 40,000, then take away that 15,000, it gives you 25,000 remaining.

The weighted mean total return is 25,000, the weighted total, divided by 100, the total weighting factor, because we are dealing with percentages, which are out of 100s.

The result of this is then 250.

The weighted mean amount that Izzy's mum is expected to receive from her investment is 250 pounds.

Note, that she will never earn exactly that amount for one investment, but if she were to do this investment many, many times, she will be likely to get around 250 pounds return for each investment that she does.

Okay, for this check, pause here to consider what the weight factors A, B, and C are.

The weighting factors are 15, 40, and 45.

Notice that the sum of all three weighting factors is 100 because our weighting factors were given as percentages.

Okay, for this check, there's a lot for you to do, by first of all, finding the three values in the rightmost column of that table, find two sums and use those two sums to calculate the weighted mean amount that Lucas's dad is expected to receive from his investment.

First up, the three values in that rightmost column are 18,000, 12,000, and negative 31,500.

These three values are products of the return on investment and the weighting factor for each day datum.

And next up, the two totals we have to look at are the totals of these two columns.

Therefore, the weighted mean is going to be negative 15,000 divided by 100 equals negative 15.

Oh, no, Lucas's dad is actually expected to make a loss because the final weighted mean is a negative number.

As we've seen, weighting factors can be based off of factual or statistical values.

However, weighting factors can also be based on opinion and intent.

Different people may choose different weighting factors, even when trying to find the weighted mean of the same thing.

Here's an example for you.

A food critic and a customer both review the same meal at a restaurant in terms of price, taste, and presentation.

The food critic weighs these criteria in the ratio, 3:5: and 2.

Whilst the customer weighs the same criteria in the ratio, 4:4:1.

Notice how the ratio parts do not need to have the same total.

Let's consider both these reviews separately.

Okay, for this demonstration and check, I will demonstrate calculating the weighted mean from scratch for the food critic.

After each set of steps, attempt the same instructions for the customer's weighted mean.

So here are the food critics ratings out of 10.

Let's construct columns to show the rating, the weighting factor, and the product of these two values, like so.

We can then put in the weighting factors, the parts of the ratio exactly as they are written in the ratio, 3 for price, 5 for taste, and 2 for presentation.

Pause now, to construct your weighted table and try this for yourself.

Here's what you should have done.

Next up, let's find the product of each rating and its associated weighting factor.

Pause now, to try this for yourself.

The three products that you should have got are 12, 32, and 4.

And finally, find the sum of both the product column and the weighting factor column, in order to find the weighted mean.

For the food critic, their weighted total score was 61, divided by the total weighting of 10, giving a weighted mean score of 6.

1 out of 10.

Pause now to do this, for the customer ratings.

The two sums are 48 and 9.

We then do 48 divided by 9, which is approximately 5.

3.

The customer's weighted mean rating is 5.

3 out of 10.

Notice how both weighted means are different, even though the sum of their ratings were 15 each.

This is due to the different importance that each person put on their individual criteria.

Fantastic effort, onto the final practise questions.

Pause here, for Questions 1 and 2 where you have to first modify or construct a table with appropriate columns, in order to calculate the weighted mean for these two scenarios.

And finally, Question 3.

Time for you to calculate the weighted mean from your own review.

As a set of percentages or ratios, consider what you would prioritise in terms of cuteness, shape, and realism when rating the drawing of an animal.

Then rate both the bear and lion on these criteria out of 10 per criterion, and then calculate your weighted mean rating to see which drawing that you think is better.

Pause now, for this question.

Onto the answers.

For Question 1, pause here to check your answers and compare them to the ones on screen.

And know that Sophia had 3.

55 higher weighted marks when compared to Andeep.

For Question 2, pause here to check your calculations that got you to a weighted mean expected return of 227 pounds and 27 pence.

And well done if you are able to rate the quality of the two drawings.

Here's an example of my weighted mean rating for the lion.

I got 6.

68 out of 10.

How much did my rating differ from yours? Amazing effort, everyone in this lesson, where we have considered that group size matters when trying to find the combined mean of two groups when given the mean of each group.

We've looked at the weighted mean, which gives more or less importance to certain data points, by assigning weighting factors to each data point.

Weighting factors can be given as numbers, percentages, ratios, or fractions, and can be based off of fixed numerical, or statistical values, or from a person's opinion.

Once again, thank you all so much for joining me, Mr. Grason for this lesson.

Take care, and until our next maths lesson together, goodbye!.