Maths · 10 Statistics

10.5 Averages on a calculator

Before you start

What the mean, median, lower quartile and upper quartile are, and how to find them by hand for a short list (10.4).
The difference between discrete and continuous data (10.3).
How to read a frequency table and a grouped frequency table, including class labels such as $10 < t \leq 20$ .
Rounding to 3 significant figures.

Syllabus objectives covered (2)

Use a graphic display calculator to calculate: mean, median and quartiles for discrete data
Use a graphic display calculator to calculate: mean for grouped data.

Part 1 of 5

What the calculator does, and what is still your job

In 10.4 you found the mean, median and quartiles by hand. In this subtopic the graphic display calculator does the sorting and the arithmetic. Your work is to give it the right data and to read the right result.
Paper 2 is the non-calculator paper, and it does not examine this subtopic. So every question on 10.5 is on a calculator paper, and the calculator method is the one to learn.
All of it goes through one tool: the statistics mode. You type the data into a list, then you ask for one-variable statistics (the menu option is usually labelled 1-Var or 1VAR). The calculator then shows a screen of results.
These are the lines you need. $\overline{x}$ is the mean. $n$ is the number of values. $Σ x$ is the total of the values. Q1 is the lower quartile, Med is the median and Q3 is the upper quartile. minX and maxX are the smallest and the largest values.
The screen shows other lines too, such as $Σ x^{2}$ and $σ x$ . This subtopic does not use them, so do not copy them into your answer.
The calculator cannot tell if you typed a value wrongly. So check the $n$ line every time. If $n$ is not the number of values you expected, every other line on the screen is wrong too.
Before you type new data, clear the old data from the lists. Values left from an earlier question are counted with the new ones, and $n$ comes out too large.

Part 2 of 5

A list of values: mean, median and quartiles

Type each value into List 1. The order does not matter, because the calculator sorts the values before it finds the median and the quartiles.
Set the frequency to 1. This tells the calculator that each value in the list occurs once.
Run one-variable statistics on List 1. Take an example: the 7 values $5, 2, 9, 4, 6, 8, 3$ . The screen gives $n = 7$ , $Σ x = 37$ , $\overline{x} = 5.2857 \dots$ , Q1 $= 3$ , Med $= 5$ and Q3 $= 8$ .
Check that the screen makes sense. In order the values are $2, 3, 4, 5, 6, 8, 9$ . The median is the middle value, the 4th, which is $5$ . The lower quartile is the median of the values below it, $2, 3, 4$ , which is $3$ . The upper quartile is the median of $6, 8, 9$ , which is $8$ .
The mean is not exact, so give it to 3 significant figures: $5.29$ . The median and the quartiles are whole numbers here, so write them as they are.
The same screen gives you the range and the interquartile range, which you met in 10.4. Use the calculator values in the subtractions.
$range = maxX - minX = 9 - 2 = 7, IQR = Q3 - Q1 = 8 - 3 = 5$

Part 3 of 5

A frequency table: give the calculator the frequencies

Discrete data can come as a frequency table, where each value is written once with the number of times it occurs. Take this table of the number of goals scored in 30 matches.
$\begin{matrix} Goals & 0 & 1 & 2 & 3 \\ Frequency & 6 & 10 & 9 & 5 \end{matrix}$
Put the values $0, 1, 2, 3$ in List 1 and the frequencies $6, 10, 9, 5$ in List 2. Then set the frequency to List 2. Now the calculator counts the value $0$ six times, the value $1$ ten times, and so on.
Run one-variable statistics. The screen gives $n = 30$ , $Σ x = 43$ , $\overline{x} = 1.4333 \dots$ , Q1 $= 1$ , Med $= 1$ and Q3 $= 2$ . So the mean is $1.43$ goals to 3 significant figures, and the median is $1$ goal.
Notice what the two count lines mean now. With a frequency list, $n$ is the total frequency, $Σ f = 30$ . And $Σ x$ is the total of all 30 values, $Σ f x = 43$ . These are the numbers you write down as working.
The trap is a frequency setting left at 1. Then the calculator sees only four values, $0, 1, 2, 3$ . It shows $n = 4$ and $\overline{x} = 1.5$ . That mean looks reasonable, but it is wrong. The $n$ line tells you: you expected $30$ , not $4$ .
A second trap is to type the frequencies into List 1 as if they were data. The values are the goals. The frequencies only say how many times each value occurs.

Part 4 of 5

Grouped data: the mean from mid-points

In a grouped frequency table you do not know the individual values. You know only how many values are in each class.
So you let the mid-point of each class stand for every value in that class. The mid-point is half-way between the two ends of the class.
$mid-point = \frac{lower end + upper end}{2}$
For the class $10 < t \leq 20$ the mid-point is $\frac{10 + 20}{2} = 15$ . The symbols $<$ and $\leq$ do not change the mid-point: for $10 \leq t < 20$ it is also $15$ .
Classes can have different widths. Use the same rule for each class. For $40 < t \leq 60$ the mid-point is $\frac{40 + 60}{2} = 50$ .
Grouped discrete data has classes such as 10–19. The mid-point is $\frac{10 + 19}{2} = 14.5$ . It does not matter that a count such as $14.5$ cannot occur. The mid-point is only a value that stands for the class.
On the calculator, type the mid-points in List 1 and the frequencies in List 2. Set the frequency to List 2 and run one-variable statistics. Read the $\overline{x}$ line.
Because each value was replaced by a mid-point, the result is an estimate of the mean, not the exact mean. That is why the question asks you to calculate 'an estimate of the mean'.
Read only the $\overline{x}$ line. The screen also shows Med, Q1 and Q3, but the calculator found them from the mid-points, not from the real data. They are not the median and quartiles of the data. For grouped data, this objective asks for the mean only.
Do a quick check. The estimate must lie between the smallest mid-point and the largest mid-point. If it does not, a value was typed wrongly.

Part 5 of 5

What to write down on a calculator question

A method mark is given for a valid method. A correct answer implies the method. But if you typed one value wrongly and wrote only the answer, there is no method on the page that an examiner can give a mark for.
So write one short line of working from the screen. For a grouped mean, write the mid-points, then $Σ f x$ and $Σ f$ from the screen, then the division.
$\overline{x} = \frac{Σ f x}{Σ f} = \frac{1310}{50} = 26.2$
For a list, write the total and the count, for example $184 \div 11 = 16.7$ .
For the median and the quartiles there is nothing to calculate. Write each value with its name: 'median $= 17$ ', 'lower quartile $= 13$ '. A value with no name could be read as a different quantity.
Give an answer that is not exact to 3 significant figures. An answer that is exact to 4 or 5 significant figures is not rounded: a mean of $2.175$ stays $2.175$ .
If a later part uses the mean, use the full value from the calculator, not your rounded value.
If the question adds new data, you do not have to start again. Add the new values to the lists and run the statistics again. Or add the new $Σ f x$ to the old $Σ f x$ , add the new frequency to the old $Σ f$ , and divide.

Worked examples

Example 1. Eleven students did a test. Their scores were $14, 9, 22, 17, 11, 19, 25, 13, 16, 20, 18$ . Use your graphic display calculator to find the mean, the median and the lower and upper quartiles.

Show the working

Clear List 1, then type the 11 scores into it. The order does not matter.
Set the frequency to 1, because each score occurs once. Run one-variable statistics on List 1.
Check $n$ first. The screen shows $n = 11$ , which is the number of students, so the data went in correctly.
Read the mean. The screen shows $Σ x = 184$ and $\overline{x} = 16.7272 \dots$ . This is not exact, so give 3 significant figures.
$\overline{x} = \frac{184}{11} = 16.7$
Read the median and the quartiles: Med $= 17$ , Q1 $= 13$ , Q3 $= 20$ .
Check them against the ordered list $9, 11, 13, 14, 16, 17, 18, 19, 20, 22, 25$ . The 6th value is $17$ . The median of $9, 11, 13, 14, 16$ is $13$ , and the median of $18, 19, 20, 22, 25$ is $20$ . The screen agrees.

Answer: Mean $= 16.7$ (3 s.f.), median $= 17$ , lower quartile $= 13$ , upper quartile $= 20$ .

Example 2. The table shows the number of books that each of 40 students read in one month. $\begin{matrix} Number of books & 0 & 1 & 2 & 3 & 4 & 5 \\ Frequency & 4 & 9 & 12 & 8 & 5 & 2 \end{matrix}$ Calculate the mean, and find the median and the quartiles.

Show the working

Type the numbers of books $0, 1, 2, 3, 4, 5$ into List 1 and the frequencies $4, 9, 12, 8, 5, 2$ into List 2.
Set the frequency to List 2, so that each number of books is counted as many times as the table says. Run one-variable statistics.
Check $n$ . The screen shows $n = 40$ , which is the total frequency. If it showed $n = 6$ , the frequency setting would still be 1.
Read the mean. With frequencies, $Σ x$ is $Σ f x$ .
$\overline{x} = \frac{Σ f x}{Σ f} = \frac{87}{40} = 2.175$
The value $2.175$ is exact and has 4 significant figures, so do not round it.
Read the median and the quartiles: Med $= 2$ , Q1 $= 1$ , Q3 $= 3$ .
Compare what a frequency setting of 1 gives: $n = 6$ and $\overline{x} = 2.5$ , the mean of $0, 1, 2, 3, 4, 5$ . That is the mean of the column headings, not the mean number of books.

Answer: Mean $= 2.175$ books, median $= 2$ books, lower quartile $= 1$ , upper quartile $= 3$ .

Example 3. The table shows the times, $t$ minutes, that 50 people waited at a clinic. $\begin{matrix} Time (t min) & 0 < t \leq 10 & 10 < t \leq 20 & 20 < t \leq 30 & 30 < t \leq 40 & 40 < t \leq 60 \\ Frequency & 5 & 11 & 16 & 12 & 6 \end{matrix}$ Calculate an estimate of the mean waiting time.

Show the working

You do not know the exact times, so use the mid-point of each class: $5, 15, 25, 35, 50$ .
The last class is wider than the others. Its mid-point is $\frac{40 + 60}{2} = 50$ , from the same rule as the other classes.
Type the mid-points into List 1 and the frequencies $5, 11, 16, 12, 6$ into List 2. Set the frequency to List 2 and run one-variable statistics.
Check $n = 50$ , the total frequency. The screen shows $Σ x = 1310$ , which is $Σ f x$ .
Write the working from the screen, then the answer.
$\overline{x} = \frac{Σ f x}{Σ f} = \frac{1310}{50} = 26.2$
Check: $26.2$ is between the smallest mid-point, $5$ , and the largest, $50$ . The answer is an estimate because each time was replaced by the mid-point of its class.

Answer: An estimate of the mean waiting time is $26.2$ minutes.

Example 4. Four more people are added to the clinic data in the previous example. Each of them waited for a time in the class $60 < t \leq 80$ . Revise the estimate of the mean waiting time for all 54 people.

Show the working

The new class has mid-point $\frac{60 + 80}{2} = 70$ .
Method 1: add a row to the lists, $70$ in List 1 and $4$ in List 2. Run the statistics again. Check that $n = 54$ .
Method 2: change the totals. The four new people add $4 \times 70 = 280$ to $Σ f x$ , and they add $4$ to $Σ f$ .
$\overline{x} = \frac{1310 + 280}{50 + 4} = \frac{1590}{54} = 29.444 \dots$
Both methods give the same value. It is not exact, so give 3 significant figures.
Check: the new people waited longer than everyone else, so the mean must go up. It goes up from $26.2$ to $29.4$ .

Answer: The revised estimate of the mean is $29.4$ minutes (3 s.f.).

Where people go wrong

Mistake: Leaving the frequency setting at 1 when the data are in a frequency table.

Instead: Put the frequencies in List 2 and set the frequency to List 2. Then check that $n$ equals the total frequency.

With the setting at 1, the calculator sees each value only once. It then gives the mean of the column headings, which can look reasonable and is wrong.

Mistake: Typing the lower end of each class, or the class width, into List 1 for grouped data.

Instead: Type the mid-point of each class: (lower end + upper end) ÷ 2.

The method lets one value stand for every value in the class, and the mid-point is the value half-way through it. The lower end makes every value too small.

Mistake: Giving the Med, Q1 and Q3 lines from the screen as the median and quartiles of grouped data.

Instead: For grouped data, read only the $\overline{x}$ line, and call it an estimate of the mean.

The calculator found Med, Q1 and Q3 from the mid-points, not from the real values, so they are not the median and quartiles of the data.

Mistake: Typing new data into lists that still hold values from an earlier question.

Instead: Clear the lists first, then check that $n$ is the number of values you typed.

The calculator uses every value in the list. Old values change every line on the screen.

Mistake: Rounding an exact answer such as $2.175$ to $2.18$ .

Instead: Write an answer that is exact to 4 or 5 significant figures in full.

The 3 significant figure rule is for answers that are not exact. An exact answer of 4 or 5 significant figures is not rounded.

Check yourself

Use your calculator to find the mean, the median and the quartiles of $3, 8, 5, 12, 7, 9, 4$ .
Answer
$n = 7$ , $Σ x = 48$ . Mean $= 48 \div 7 = 6.86$ (3 s.f.), median $= 7$ , lower quartile $= 4$ , upper quartile $= 9$ .
A frequency table has six values, and its frequencies add up to 30. After you run one-variable statistics, the screen shows $n = 6$ . What went wrong, and how do you fix it?
Answer
The frequency setting was left at 1, so each value was counted once. Put the frequencies in List 2, set the frequency to List 2, run it again and check that $n = 30$ .
The table shows the scores of 20 people. $\begin{matrix} Score & 1 & 2 & 3 & 4 & 5 \\ Frequency & 2 & 5 & 8 & 4 & 1 \end{matrix}$ Find the mean and the median.
Answer
With the frequency set to List 2: $n = 20$ , $Σ f x = 57$ . Mean $= 57 \div 20 = 2.85$ (exact, so not rounded). Median $= 3$ .
When the frequency is set to List 2, what do the $Σ x$ line and the $n$ line show?
Answer
$Σ x$ shows $Σ f x$ , the total of all the values. $n$ shows $Σ f$ , the total frequency.
Write down the mid-point of the class $20 < m \leq 35$ and the mid-point of the class 11–20.
Answer
$\frac{20 + 35}{2} = 27.5$ and $\frac{11 + 20}{2} = 15.5$ .
Calculate an estimate of the mean of these data. $\begin{matrix} x & 0 < x \leq 4 & 4 < x \leq 8 & 8 < x \leq 12 \\ Frequency & 3 & 7 & 10 \end{matrix}$
Answer
Mid-points $2, 6, 10$ . $Σ f x = 148$ , $Σ f = 20$ , so the estimate is $148 \div 20 = 7.4$ .
Explain why the mean you calculate from a grouped frequency table is only an estimate.
Answer
The exact values in each class are not known. Each value is replaced by the mid-point of its class, so the result is only close to the true mean.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The masses, in grams, of 15 apples are $148, 162, 135, 189, 155, 140, 166, 158, 149, 197, 152, 143, 160, 138, 129$ .

(a) Calculate the mean mass. 2 marks

Mark scheme

M1 2321 ÷ 15 soi
A1 155 or 154.7 or better (g)

(b) Write down the median mass. 1 mark

Mark scheme

B1 152 (g)

(c) Find the lower quartile and the upper quartile. 2 marks

Mark scheme

B1 lower quartile 140
B1 upper quartile 162

(d) Work out the interquartile range. 1 mark

Mark scheme

B1 22 (g); FT their upper quartile − their lower quartile from (c)

Question 2. The table shows the number of people in each of 60 cars that passed a checkpoint. $\begin{matrix} Number of people & 1 & 2 & 3 & 4 & 5 \\ Number of cars & 23 & 17 & 9 & 7 & 4 \end{matrix}$

(a) Calculate the mean number of people in a car. 2 marks

Mark scheme

M1 132 ÷ 60, or Σfx = 132 seen
A1 2.2 cao

(b) Find the median. 1 mark

Mark scheme

B1 2

(c) Find the upper quartile. 1 mark

Mark scheme

B1 3

(d) A student types only the numbers of people, $1, 2, 3, 4, 5$ , into the calculator and forgets to use the numbers of cars. Explain how the line $n$ on the calculator screen shows that this is wrong. 1 mark

Mark scheme

B1 n shows 5 but there are 60 cars, so each number of people has been counted only once oe

Question 3. The table shows the number of text messages that each of 40 people sent in one day. $\begin{matrix} Number of messages & 0 – 9 & 10 – 19 & 20 – 29 & 30 – 39 & 40 – 59 \\ Frequency & 8 & 14 & 11 & 5 & 2 \end{matrix}$

(a) Write down the mid-point of the class 10–19. 1 mark

Mark scheme

B1 14.5

(b) Calculate an estimate of the mean number of messages. 4 marks

Mark scheme

M1 at least four of the mid-points 4.5, 14.5, 24.5, 34.5, 49.5 soi
M1 Σfx using their mid-points and the frequencies, 780 if correct
M1 dep on the previous M1, their Σfx ÷ 40
A1 19.5 cao

(c) Explain why your answer to part (b) is an estimate. 1 mark

Mark scheme

B1 the exact numbers of messages are not known, and the mid-point is used for every value in a class oe

(d) Ten more people are asked. Each of them sent a number of messages in the class 60–79. Revise your estimate of the mean for all 50 people. 2 marks

Mark scheme

M1 (their Σfx + 10 × 69.5) ÷ 50 oe; FT their Σfx from (b)
A1 29.5 cao

How this is examined

An estimate of the mean from a grouped frequency table, using mid-points: A Calculate part after a grouped frequency table, worded 'Calculate an estimate of the mean', worth about 4 marks The mid-points written down (e.g. $5, 15, 25, \dots$ for the classes $0 < t \leq 10$ , $10 < t \leq 20$ , $20 < t \leq 30$ ), then $Σ f x$ and $Σ f$ from the screen, e.g. $1310 \div 50$ , then the value, $26.2$ . A correct value implies the method; a wrong value with no mid-points or division on the page has no method to credit.
The median and quartiles of discrete data, read from the calculator: Short Write down or Find parts on a list of values or a frequency table, which can lead to a part on the interquartile range Each value written with its name, e.g. 'lower quartile $= 13$ , upper quartile $= 20$ ', from a calculator run with the correct $n$ . For a frequency table the run must use the frequencies; an $n$ equal to the number of rows gives wrong values.
The mean of discrete data from a list or a frequency table: A 2-mark Calculate part, given a list of values or a frequency table The total and the count written down, e.g. $Σ f x = 87$ and $Σ f = 40$ so $87 \div 40$ , then the value, $2.175$ . A value that is not exact goes to 3 significant figures; one that is exact to 4 or 5 significant figures is written in full.
A mean revised after more data are added: A 2-mark Revise part after a mean has been found, with new values or a new class added The new total over the new count, e.g. $\frac{1310 + 4 \times 70}{50 + 4}$ , then the value, $29.4$ . Taking the mean of the old mean and the new mid-point is not a valid method.
Why a mean from grouped data is only an estimate: A 1-mark Explain part after the estimate of the mean has been calculated A sentence that says the exact values are not known and that the mid-point is used for every value in a class. 'Because it is grouped' alone does not give the reason.

Words to know

Graphic display calculator: A calculator that can draw graphs and has a statistics mode, where data are typed into lists.
List: A column in the calculator's statistics mode where you type data values or frequencies.
One-variable statistics: The calculator option that finds the mean, median, quartiles and totals of one set of data.
Mean: The total of the values divided by the number of values; $\overline{x}$ on the calculator screen.
Median: The middle value when the data are in order; Med on the calculator screen.
Lower quartile: The median of the lower half of the ordered data; Q1 on the calculator screen.
Upper quartile: The median of the upper half of the ordered data; Q3 on the calculator screen.
Mid-point: The value half-way between the two ends of a class: (lower end + upper end) ÷ 2.
Estimate of the mean: The mean found from a grouped frequency table, using the mid-point of each class in place of the unknown values.

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