## What you need to know

When we’re dealing with information displayed in a grouped frequency table – Revise grouped frequency tables if you’re not sure what that is – it’s impossible for us to calculate the mean because we don’t know each of the individual values.

So, instead, we try our best to estimate the mean.

Example: Cindy asked a collection of people who like playing video games how many hours they spent playing video games in the last week. Her results are shown in the table below. Estimate the mean number of hours spent playing video games from Cindy’s sample.

So, we know that 12 of the people Cindy asked had spent between 0 and 4 hours playing games in the last week, but we don’t know how many hours each of those people *actually* played for.

So, what do we do? We assume that every person in the group spent the “midpoint” number of hours playing video games. Let’s break that down – firstly, let’s define midpoint:

• Midpoint: the point that lies halfway between the lower bound and upper bound of a group.

For example, for the first group we have

Lower bound = 0, upper bound = 4

Halfway between 0 and 4 is 2. So, the midpoint of the first class is 2. This means we’re going to assume everybody in the first class, 0<t\leq 4, spent exactly 2 hours playing video games.

The idea is that for every person who spent more than 2 hours there will be someone who spent less than two hours and it’ll roughly even out.

Before continuing, let’s add another column to our table which is just for the midpoints. If you aren’t sure what the midpoint of a group is then this calculation will do the job:

\left(\text{lower bound}+\text{upper bound}\right) \div 2Now we have all the midpoints we can calculate our estimate! Since we’re assuming that all 12 people in the first group played for 2 hours, then we have 12 lots of 2 to add up.

Then, in the next group we are assuming that all 18 people played for 6 hours. This means, similarly, that we have 18 lots of 6 to add up.

This will be the same for every group. In other words, we need to do

\text{frequency}\times\text{midpoint}For every group, add up the results, and divide by the number of people. This looks like

\text{estimated mean}=\dfrac{(12\times 2)+(18\times 6)+(19\times 10)+(6\times 14)+(2\times 18)}{12+18+19+6+2}Putting this (carefully) into a calculator, we get our answer:

\text{estimated mean }=7.75\text{ hours}METHOD

The method for estimating the mean can be outlined as follows.

1. Add a new column to the table containing the midpoints of each group.

2. Multiply each midpoint by the frequency of that group and add up the results.

3. Divide that value by the totals of all the frequencies (i.e. the number of people/things involved in the scenario) to get the estimate of the mean.

Remember, it is just an estimate. The actual mean could be noticeably different.

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### Example Questions

Firstly, we must add on a column of midpoints to our table. The first midpoint, halfway between 0 and 10, is 5. We can check this by doing

(0+10)\div 2=5

Continuing this for the remainder of the numbers, we get the table below.

To find the estimate of the mean, we must multiply the frequency of each group by the midpoint, add up all those values, and divide by the total frequency. This looks like

\begin{aligned}\text{estimate of the mean}&=\dfrac{(2\times 5)+(45\times 15)+(25\times 25)+(3\times 35)}{2+45+25+3} \\ &=\dfrac{1,415}{75} \\ &= 18.9\text{ minutes}\end{aligned}

Therefore, our estimate for the mean journey time is 18.9 minutes.

Firstly, we must add on a column of midpoints to our table. The first midpoint, halfway between 0 and 50, is 25. We can check this by doing

(0+50)\div 2=25

Continuing this for the remainder of the numbers, we get the table below.

To find the estimate of the mean, we must multiply the frequency of each group by the midpoint, add up all those values, and divide by the total frequency. This looks like

\begin{aligned}\text{estimate of the mean}&=\dfrac{(4\times 25)+(18\times 75)+(56\times 125)+(32\times 175)+(8\times 225)}{4+18+56+32+8} \\ &=\dfrac{15,850}{118} \\ &= 134\text{ cm}\end{aligned}

Therefore, our estimate for the mean long jump distance is 134 cm.

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