# Estimating the Mean Worksheets, Questions and Revision

GCSE 4 - 5KS3AQAEdexcelOCRWJECFoundationAQA 2022Edexcel 2022OCR 2022WJEC 2022

## Estimating The Mean

When we’re dealing with information displayed in a grouped frequency table, it’s impossible for us to calculate the actual mean because we don’t know each of the individual values. So, instead, we try our best to estimate the mean.

Level 4-5 GCSE    ## How to Estimate the Mean

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

Step 1. Add a new column to the table writing down the midpoint (middle value) of each group.

Step 2. Multiply each midpoint by the frequency of that group and add the results in a new column..

Step 3. Add the values in the midpoint $\textcolor{red}\times$ frequency column.

Step 4. Divide that value by the total frequency to get the estimate of the mean.

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

Level 4-5 GCSE    ## Example 1: Estimating the Mean Cindy asked a collection of people 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.

[4 marks]

Step 1: First we need to calculate the midpoint of each class.

To do this we would add together the upper and lower value of the class and divide by $2$. e.g.

$\dfrac{16+20}{2} = \dfrac{36}{2} = 18$

Step 2: Now we multiply the frequency by the midpoint for each row. It’s helpful to write the result in a new column. Step 3: We now sum the frequency column, to get the total number of people asked, and get:

$12 + 18 + 19 + 6 + 2 = \textcolor{red}{57}$

and sum the $f \times m$ column, and get

$24 + 108 + 190 + 84 + 36 = \textcolor{red}{442}$

Step 4: Now we have all of these values, we can estimate the mean by dividing the sum of $f \times m$ by the total number of people Cindy asked:

$\text{estimated mean} = \dfrac{\textcolor{red}{442}}{\textcolor{red}{57}}$

Putting this (carefully) into a calculator, we get our answer:

$\text{estimated mean }=\textcolor{red}{7.75\text{ hours}}$

Level 4-5GCSE     ## Example 2: Estimating the Mean

The grouped frequency table shows data on the weights of $117$ cats. Calculate an estimate for the mean weight of the cats.

[4 marks]

As before, we need to note down the midpoints of each group.

We do this by adding the maximum and minimum in each class and dividing the result by $2$.

Doing this for the last group we find,

$\dfrac{4.5 + 6}{2}= \dfrac{10.5}{2} = 5.25$

We then multiply each midpoint by it’s frequency and add them all together and divide by the total frequency. The total number of cats is $\textcolor{red}{117}$, and the sum of the $f\times m$ column is

$55 + 45.5 + 146.25 + 123.25 + 68.25 = \textcolor{red}{438.25}$

Therefore, we calculate the estimated mean to be,

$\dfrac{\textcolor{red}{438.25}}{\textcolor{red}{117}} = \textcolor{red}{3.7 \text{ kg}}$ ($1$dp)

Level 4-5GCSE    ## Example Questions

The first thing we need to do is add an additional midpoint column to our table.  Since we have a range of journey times, we have to assume that all journeys are exactly in the middle of the range.  This is probably not the case in reality, which is why this is an estimated mean rather than the actual mean.

The first midpoint, the point which is half-way between $0$ minutes and $10$ minutes, is $5$ minutes.  If you are unsure how to calculate a midpoint, add up the two values and divide by $2$:

$\dfrac{0 + 10}{2}=5$

Once you have calculated all the midpoints, you should have a table that looks like the table below: We then need to multiply the frequency of each group by its midpoint.  This will give us the estimated total journey time for the number of journeys completed within each range of journey time.

$2\times5\text{ minutes}= 10\text{ minutes}$

$45\times15\text{ minutes}= 675\text{ minutes}$

$25\times25\text{ minutes}= 625\text{ minutes}$

$3\times35\text{ minutes}= 105\text{ minutes}$

We then need to add up all these new values.   This will give us the estimated total journey time for all of the journeys completed.

$10\text{ minutes}+675\text{ minutes}+625\text{ minutes}+105\text{ minutes} = 1415\text{ minutes}$

This means that the total estimated journey time was $1415$ minutes.  By adding up the frequency column, we know that $75$ journeys were completed $(2+45+25+3=75\text{ journeys})$.  Since there were $75$ journeys with an estimated total time of $1415$ minutes, then the estimated mean can be calculated as follows:

$\dfrac{1415 \text{ minutes}}{75\text{ journeys}} = 19 \text{ minutes to the nearest minute}$

The first thing we need to do is add an additional midpoint column to our table.  Since we have a range of long jump distances, we are assuming that all jumps are exactly in the middle of the range.  This is probably not the case in reality, which is why this is an estimated mean rather than the actual mean.

The first midpoint, the point which is half-way between $0$ cm and $50$ cm, is $25$ cm.  If you are unsure how to calculate a midpoint, add up the two values and divide by $2$:

$\dfrac{0 + 50}{2}=25$

Once you have calculated all the midpoints, you should have a table that looks like the table below: We then need to multiply the frequency of each group by its midpoint.  This will give us the estimated total jump length for the number of jumps completed within each range of jump length.

$2\times25\text{cm}= 50\text{cm}$

$18\times75\text{cm}= 1,350\text{cm}$

$56\times125\text{cm}= 7,000\text{cm}$

$32\times175\text{cm}= 5,600\text{cm}$

$8\times225\text{cm}= 1,800\text{cm}$

We then need to add up all these new values.   This will give us the estimated total jump length for all of the jumps completed.

$50\text{cm}+1,350\text{cm} +7,000\text{cm} +5,600\text{cm} +1,800\text{cm} =15,800\text{cm}$

This means that the total estimated jump length was $15,800$ cm.  By adding up the frequency column, we know that $118$ jumps were completed $(4+18+56+32+8=118\text{ jumps})$.  Since there were $118$ jumps with an estimated total jump length of $15,800$ cm, then the estimated mean can be calculated as follows:

$\dfrac{15,800 \text{cm}}{118\text{ jumps}} = 134 \text{cm to the nearest cm}$

The first thing we need to do is add an additional midpoint column to our table.  Since we have a range of journey times, we are assuming that all journeys are exactly in the middle of the range.  This is probably not the case in reality, which is why this is an estimated mean rather than the actual mean.

The first midpoint, the point which is half-way between $0$  minutes and $10$ minutes, is $5$ minutes.  If you are unsure how to calculate a midpoint, add up the two values and divide by $2$.  For example, working out the midpoint in the $32$$45$ minute range is not that easy:

$\dfrac{32 + 45}{2}=38.5$

Once you have calculated all the midpoints, you should have a table that looks like the table below: We then need to multiply the frequency of each group by its midpoint.  This will give us the estimated total journey time for the number of journeys completed within each range of journey time.

$12\times5\text{ minutes}= 60\text{ minutes}$

$18\times12\text{ minutes}= 216\text{ minutes}$

$34\times17\text{ minutes}= 578\text{ minutes}$

$33\times26\text{ minutes}= 858\text{ minutes}$

$19\times38.5\text{ minutes}= 731.5\text{ minutes}$

We then need to add up all these new values.   This will give us the estimated total journey time for all of the journeys completed.

$60\text{ minutes}+216\text{ minutes}+578\text{ minutes}+858\text{ minutes}+731.5\text{ minutes} = 2443.5\text{ minutes}$

This means that the total estimated journey time was $2443.5$ minutes.  By adding up the frequency column, we know that $116$ journeys were completed $(12+18+34+33+19=116\text{ journeys}).$  Since there were $116$ journeys with an estimated total time of $2443.5$ minutes, then the estimated mean can be calculated as follows:

$\dfrac{2443.5 \text{ minutes}}{116\text{ journeys}} = 21 \text{ minutes to the nearest minute}$

a)  The first thing we need to do is add an additional midpoint column to our table.  Since we have a range of fish size lengths, we are assuming that all fish caught fall exactly in the middle of the range.  This is probably not the case in reality, which is why this is an estimated mean rather than the actual mean.

The first midpoint, the point which is half-way between $0$ cm and $20$ cm, is $10$ cm.  If you are unsure how to calculate a midpoint, add up the two values and divide by $2$.  For example, working out the midpoint in the $90$$300$ cm is not that easy:

$\dfrac{90 + 300}{2}=195$

Once you have calculated all the midpoints, you should have a table that looks like the table below: We then need to multiply the frequency of each group by its midpoint.  This will give us the estimated total fish length for the number of fish caught within each range of fish length.

$16\times10\text{cm}= 160\text{cm}$

$27\times25\text{cm}= 675\text{cm}$

$9\times40\text{cm}= 360\text{cm}$

$13\times60\text{cm}= 7,800\text{cm}$

$8\times80\text{cm}= 640\text{cm}$

$6\times195\text{cm}= 1,170\text{cm}$

We then need to add up all these new values.   This will give us the estimated total length for all of the fish caught.

$160\text{cm}+675\text{cm}+360\text{cm}+7,800\text{cm}+640\text{cm}+1,170\text{cm}=10,805\text{cm}$

This means that the estimated total length of all the fish caught was $10,805$cm.  By adding up the frequency column, we know that $79$ fish were caught $(16+27+9+13+8+6=79\text{ fish caught}.)$  Since there were $79$ fish caught with an estimated combined length of $10,805$cm, then the estimated mean can be calculated as follows:

$\dfrac{10,805 \text{cm}}{79\text{ fish}} = 137\text{cm to the nearest cm}$

b)  If we do not take into consideration the $6$ fish that were in the $90 – 300$ cm category, then the total number of fish caught is reduced from $79$ to $73$.

The estimated total length of all the $79$ fish caught was $10,805$cm.  We need to subtract the $6$ fish in the $90 – 300$cm category:

$10,805\text{cm} - (6 \times 196\text{cm}) = 9,629 \text{cm}$

Therefore, excluding the $6$ biggest fish, the estimated total length of the other $73$ fish was $9,629$cm.  The estimated mean can therefore be calculated as follows:

$\dfrac{9,629 \text{cm}}{73\text{ fish}} = 132\text{cm to the nearest cm}$

a)  The first thing we need to do is add an additional midpoint column to our table.  Since we have a range of number of bullseyes hit, we are assuming that the number of bullseyes hit by each participant falls exactly in the middle of the range.  This is probably not the case in reality, which is why this is an estimated mean rather than the actual mean.

The first midpoint, the point which is half-way between $0$ cm and $10$ cm, is $5$ cm.  If you are unsure how to calculate a midpoint, add up the two values and divide by $2$.  In this table, the midpoints are all quite easy to calculate .

Once you have calculated all the midpoints, you should have a table that looks like the table below: We then need to multiply the frequency of each group by its midpoint.  This will give us the estimated total number of bullseyes hit by participants who fall into this category.

$3\times5\text{ hits}= 15\text{ hits}$

$25\times15\text{ hits}= 375\text{ hits}$

$28\times25\text{ hits}= 700\text{ hits}$

$19\times35\text{ hits}= 665\text{ hits}$

$8\times45\text{ hits}= 360\text{ hits}$

$2\times55\text{ hits}= 110\text{ hits}$

We then need to add up all these new values.   This will give us the estimated total number of times the bullseye was hit.

$15\text{ hits}+375\text{ hits}+700\text{ hits}+665\text{ hits}+360\text{ hits}+110\text{ hits}=2,225\text{ hits}$

This means that the estimated total number of bullseyes hit was $2,225$.  By adding up the frequency column, we know that there was a total of $85$ participants $(3+25+28+19+8+2=85\text{ participants}.)$  Since there were $85$ participants with an estimated total of $2,225$ bullseyes hit, then the estimated mean can be calculated as follows:

$\dfrac{2,225 \text{ bullseyes hit}}{79\text{ participants}} = 26\text{ hits to the nearest whole number}$

b)  By organising the data in batches of $20$ hits, rather than batches of $10$ hits, each row in Flolella’s table will combine $2$ of Suzanna’s rows, so the table will be half the size.  It will look as follows: In order to calculate the estimated mean, we will need to work out new midpoints, as follows: Using Flolella’s table, the estimated mean can be calculated as follows:

$\dfrac{(28\times10) + (47\times30) + (10\times 50)}{85}=26\text{ hits}$

When the answer is rounded to the nearest whole number, the estimated mean is the same.

## Worksheet and Example Questions

### (NEW) Estimating the Mean Exam Style Questions - MME

Level 4-5 GCSENewOfficial MME

Level 4-5 GCSE

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