Number problems - frequency
The table shows the time it takes some people to complete a questionnaire.
| Time, \(x\) , in mins | Frequency |
| \(0 \leq x \textless 4\) | 4 |
| \(4 \leq x \textless 8\) | ? |
| \(8 \leq x \textless 12\) | 19 |
| \(12 \leq x \textless 16\) | 10 |
| \(16 \leq x \textless 20\) | 10 |
| Time, \(x\) , in mins | \(0 \leq x \textless 4\) |
|---|---|
| Frequency | 4 |
| Time, \(x\) , in mins | \(4 \leq x \textless 8\) |
|---|---|
| Frequency | ? |
| Time, \(x\) , in mins | \(8 \leq x \textless 12\) |
|---|---|
| Frequency | 19 |
| Time, \(x\) , in mins | \(12 \leq x \textless 16\) |
|---|---|
| Frequency | 10 |
| Time, \(x\) , in mins | \(16 \leq x \textless 20\) |
|---|---|
| Frequency | 10 |
An estimate of the meanThe mean is calculated by adding all of the data and dividing by the number of items of data. time taken is 11 minutes.
How many people took between 4 and 8 minutes?
1. What do I have to do?
Read the question through twice.
Highlight or underline the important pieces of information in the question.
The table shows the time it takes some people to complete a task.
2. What information do I need?
The highlighted words are the most important ones.
The most important part of this question is that an estimate of the mean is 11.
What is the question asking?
The question asks for the number of people who took between 4 and 8 minutes.
This is the value that should be the question mark in the frequency column.
The answer should be a whole number as it is not possible to have part of a person.
3. What information donât I need?
Everything in this question is relevant to working out the answer.
4. What maths can I do?
An estimate of the mean is 11 minutes.
This provides the way into the question as the midpoint method of finding an estimate of the mean can be used.
Step A
Add two more columns onto the table and swap the â?â for a letter.
Find the midpoint of each of the groups listed in the first column.
Multiply the midpoint by the frequency.
| Time, \(x\) , in mins | Frequency | Midpoint | Midpoint \(\times\) frequency |
| \(0 \leq x \textless 4\) | 4 | 2 | 8 |
| \(4 \leq x \textless 8\) | \(y\) | 6 | \(6y\) |
| \(8 \leq x \textless 12\) | 19 | 10 | 190 |
| \(12 \leq x \textless 16\) | 10 | 14 | 140 |
| \(16 \leq x \textless 20\) | 10 | 18 | 180 |
| Time, \(x\) , in mins | \(0 \leq x \textless 4\) |
|---|---|
| Frequency | 4 |
| Midpoint | 2 |
| Midpoint \(\times\) frequency | 8 |
| Time, \(x\) , in mins | \(4 \leq x \textless 8\) |
|---|---|
| Frequency | \(y\) |
| Midpoint | 6 |
| Midpoint \(\times\) frequency | \(6y\) |
| Time, \(x\) , in mins | \(8 \leq x \textless 12\) |
|---|---|
| Frequency | 19 |
| Midpoint | 10 |
| Midpoint \(\times\) frequency | 190 |
| Time, \(x\) , in mins | \(12 \leq x \textless 16\) |
|---|---|
| Frequency | 10 |
| Midpoint | 14 |
| Midpoint \(\times\) frequency | 140 |
| Time, \(x\) , in mins | \(16 \leq x \textless 20\) |
|---|---|
| Frequency | 10 |
| Midpoint | 18 |
| Midpoint \(\times\) frequency | 180 |
Step B
Add up the frequency column.
\(4 + y + 19 + 10 + 10 = 43 + y\)
Add up the midpoint \(\times\) frequency column.
\(8 + 6y + 190 + 140 + 180 = 518 + 6y\)
Step C
An estimate for the mean is found by dividing the midpoint \(\times\) frequency by the frequency.
The question says the estimate for the mean is 11 so set up an equation to show this:
\(\frac{518 + 6y}{43 + y} = 11\)
Step D
Solve the equation to find out the value of \(y\).
\(\frac{518 + 6y}{43 + y} = 11\)
- Multiply both sides by \(43 + y\): \(518 + 6y = 11(43 + y)\)
- Expand the bracket: \(518 + 6y = 473 + 11y\)
- Subtract \(6y\) from both sides: \(518 = 473 + 5y\)
- Subtract 473 from both sides: \(45 = 5y\)
- Divide both sides by 5: \(9 = y\)
Therefore the value of \(y\) is 9, which means 9 people took between 4 and 8 minutes to complete the task.
5. Is my solution correct?
It is important to check any calculations at the end.
Go through and check them, even if a calculator was used.
Check whether an estimate of the mean is 11.
| Time, \(x\) , in mins | Frequency | Midpoint | Midpoint \(\times\) frequency |
| \(0 \leq x \textless 4\) | 4 | 2 | 8 |
| \(4 \leq x \textless 8\) | \(y\) | 6 | \(6y\) |
| \(8 \leq x \textless 12\) | 19 | 10 | 190 |
| \(12 \leq x \textless 16\) | 10 | 14 | 140 |
| \(16 \leq x \textless 20\) | 10 | 18 | 180 |
| 52 | 572 |
| Time, \(x\) , in mins | \(0 \leq x \textless 4\) |
|---|---|
| Frequency | 4 |
| Midpoint | 2 |
| Midpoint \(\times\) frequency | 8 |
| Time, \(x\) , in mins | \(4 \leq x \textless 8\) |
|---|---|
| Frequency | \(y\) |
| Midpoint | 6 |
| Midpoint \(\times\) frequency | \(6y\) |
| Time, \(x\) , in mins | \(8 \leq x \textless 12\) |
|---|---|
| Frequency | 19 |
| Midpoint | 10 |
| Midpoint \(\times\) frequency | 190 |
| Time, \(x\) , in mins | \(12 \leq x \textless 16\) |
|---|---|
| Frequency | 10 |
| Midpoint | 14 |
| Midpoint \(\times\) frequency | 140 |
| Time, \(x\) , in mins | \(16 \leq x \textless 20\) |
|---|---|
| Frequency | 10 |
| Midpoint | 18 |
| Midpoint \(\times\) frequency | 180 |
| Time, \(x\) , in mins | |
|---|---|
| Frequency | 52 |
| Midpoint | |
| Midpoint \(\times\) frequency | 572 |
Estimate of the mean is \(\frac{572}{52} = 11\).
6. Have I completed everything?
The answer is supposed to be a whole number, which it is.
It seems reasonable and fits in with the data.
Nothing else was asked for.