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Either paper 2 marks
Here are Aisha's marks in five tests.
\(10,\ 19,\ 13,\ 6,\ 17\)
Find the mean mark.
Show answer
\(13\)Add the five marks together:
\(10 + 19 + 13 + 6 + 17 = 65\)
There are 5 marks, so divide the total by 5:
\(65 \div 5 = 13\)
- M1the five marks added: \(65\)
- A1\(65 \div 5 = 13\)
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Either paper 6 marksThe table shows the number of cakes sold by four children at a school fair.
Child Number of cakes Owen 13 Priya 3 Callum 18 Zara 19 (a)Calculate the median number of cakes.
(b)Calculate the range.
(c)Draw a bar chart on the grid to show the information in the table.
Each square on the vertical axis represents 1 cake.
You cannot draw on this page. Draw (c) on paper, then show the answer and check your drawing against the model.
Show answer
(a) \(15.5\)
(b) \(16\)
(c) Four bars: Owen to 13, Priya to 3, Callum to 18 and Zara to 19, on a vertical axis numbered 1 to a square(c)Put the numbers in order:
\(3,\ 13,\ 18,\ 19\)
There are four numbers, so the median is halfway between the second and the third:
\((13 + 18) \div 2 = 31 \div 2 = 15.5\)
The range is the largest number take away the smallest:
\(19 - 3 = 16\)
One square is 1 cake, so each bar is as many squares high as its number.
The heights in squares, in the order of the table, are \(13,\ 3,\ 18,\ 19\).
Number the vertical axis in steps of 1, then draw each bar up to its number.
- (a) M1numbers in order, middle two \(13\) and \(18\)
- (a) A1halfway between them: \(15.5\)
- (c) B1the vertical axis numbered 1 to a square
- (c) B1at least two bars drawn to the right height
- (c) B1all four bars drawn to the right height: 13, 3, 18 and 19 cakes
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Either paper 4 marks
A game is played 40 times.
The vertical line chart shows the winning scores.
(a)Write down the mode.
(b)The game is played again.
Use the chart to estimate the probability that the winning score is \(54\).
(c)Use the chart to estimate the probability that the winning score is \(51\) or more.
Show answer
(a) \(50\)
(b) \(\frac{3}{20}\)
(c) \(\frac{4}{5}\)The tallest line is at the score \(50\), so the mode is \(50\).
The line at \(54\) has height \(6\), so \(54\) was the winning score in \(6\) of the \(40\) games.
\(P = \frac{ 6 }{ 40 } = \frac{3}{20}\)
\(51\) or more means the lines at 51, 52, 53, 54, 55, 56. Their heights are \(4,\ 6,\ 5,\ 6,\ 6,\ 5\), so the winning score was \(51\) or more in \(32\) games.
The total number of games is \(40\), which is also all seven line heights added up.
\(P = \frac{ 32 }{ 40 } = \frac{4}{5}\)
- (c) M1adds the heights of the lines at \(51\) and above, \(4,\ 6,\ 5,\ 6,\ 6,\ 5\), to get \(32\)
- (c) A1probability \(\frac{4}{5}\)
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Either paper 3 marks
The list shows the ages, in years, of the seven members of a chess club.
\(13,\ 20,\ 45,\ 13,\ 27,\ 30,\ 38\)
(a)State the median.
(b)State the mode.
(c)Find the range.
Show answer
(a) \(27\)
(b) \(13\)
(c) \(32\)Put the values in order first:
\(13,\ 13,\ 20,\ 27,\ 30,\ 38,\ 45\)
There are 7 values, so the median is the 4th one along:
\(27\)
The mode is the value that appears most often. 13 appears twice and every other age appears once, so the mode is 13.
The range is the largest value take away the smallest:
\(45 - 13 = 32\)
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Either paper 3 marks
Here are the marks scored by eight pupils in a test.
\(10,\ 37,\ 68,\ 59,\ 42,\ 51,\ 21,\ 30\)
(a)Find the median.
(b)A ninth value of \(80\) is added to the list.
Find the median of all nine values.
Show answer
(a) \(39.5\)
(b) \(42\)Put the values in order:
\(10,\ 21,\ 30,\ 37,\ 42,\ 51,\ 59,\ 68\)
There are eight values, so there is no single middle one. The median is halfway between the fourth and the fifth:
\((37 + 42) \div 2 = 79 \div 2 = 39.5\)
\(80\) is larger than all eight, so it goes on the end and there are now nine values in order.
With nine values the median is the fifth one, which is \(42\).
- (a) M1values in order, middle two \(37\) and \(42\)
- (a) A1halfway between them: \(39.5\)
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Either paper 3 marks
Five boxes were weighed, in kilograms. Here are four of the five values.
\(24,\ 15,\ 29,\ 7\)
The fifth value, which is the largest of the five, has been rubbed out.
The range of all five values is \(33\) kg.
(a)Work out the value that has been rubbed out.
(b)Work out the mean of the five values.
Show answer
(a) \(40\ \text{kg}\)
(b) \(23\ \text{kg}\)The range is the largest take away the smallest, and the smallest of the values shown is \(7\).
\(7 + 33 = 40\)
Add all five values, including the one from part (a):
\(7 + 15 + 24 + 29 + 40 = 115\)
\(115 \div 5 = 23\)
- (b) M1sum of all five values \(115\)
- (b) A1mean \(23\)
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Either paper 3 marks
Here are the marks scored by six pupils in a test.
\(39,\ 53,\ 34,\ 24,\ 50,\ 16\)
Idris says the mean is \(59\).
(a)Without working out the mean, explain how you can tell that Idris is wrong.
(b)Calculate the correct mean.
Show answer
(a) The mean must lie between the smallest value \(16\) and the largest value \(53\), and \(59\) is bigger than the largest, \(53\).
(b) \(36\)A mean is a sharing-out of the values, so it can never be smaller than the smallest of them or larger than the largest.
Here the values run from \(16\) to \(53\), and Idris's \(59\) is bigger than the largest, \(53\), so it cannot be the mean.
Add the six values:
\(16,\ 24,\ 34,\ 39,\ 50,\ 53\) add to \(216\)
Then share that between the six:
\(216 \div 6 = 36\)
- (b) M1the six values added: \(216\)
- (b) A1\(216 \div 6 = 36\)
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Either paper 3 marks
Six trains have their delays, in minutes, listed in order of size. Three of the delays are missing.
\(4, \ ?, \ 6, \ ?, \ 11, \ ?\)
Complete the list of six delays so that the mode is \(4\), the median is \(7.5\) and the range is \(9\).
Show answer
\(4,\ 4,\ 6,\ 9,\ 11,\ 13\)The mode is \(4\), so \(4\) appears twice. The 1st number is already \(4\) and it is the smallest number in the list, so the missing 2nd number is also \(4\).
The median is the mean of the 3rd and 4th numbers, so they add up to twice the median:
\(2 \times 7.5 = 15\)
The 3rd number is \(6\), so the missing 4th number is \(15 - 6 = 9\).
The range is the largest number take away the smallest, so the missing largest number is the smallest number plus the range:
\(4 + 9 = 13\)
This is the missing 6th number, so the completed list is \(4,\ 4,\ 6,\ 9,\ 11,\ 13\).
- B1missing 2nd number \(4\)
- B1missing 4th number \(15 - 6 = 9\)
- B1missing 6th number \(4 + 9 = 13\)
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Either paper 6 marks
The bar chart shows information about the goals scored in 20 hockey games. Each game has four quarters.
(a)Which quarter was the mode for the away goals? Circle your answer.
1st 2nd 3rd 4th
(b)There were 20 games. Work out the mean number of goals per game, home and away goals together.
(c)In total, how many more home goals were scored than away goals?
(d)Idris says, “More home teams must have won because there were more home goals.”
Is Idris correct? Give a reason for your answer.
Show answer
(a) 3rd
(b) \(4.25\)
(c) \(7\)
(d) No. The totals only show how many goals were scored, not how many games each team won. For example, one home win by 10 goals and three away wins by 1 goal each would give the home teams 7 more goals in total, yet the away teams would have won more games.The mode is the quarter with the most away goals. The tallest away bar is in the 3rd quarter, with \(11\) goals.
Add the heights of all eight bars:
\(10 + 15 + 9 + 12 + 8 + 10 + 11 + 10 = 85\)
Divide by the number of games:
\(85 \div 20 = 4.25\)
Home goals: \(10 + 15 + 9 + 12 = 46\)
Away goals: \(8 + 10 + 11 + 10 = 39\)
\(46 - 39 = 7\)
The home teams did score 7 more goals in total, but goals scored do not say how many games each side won.
One large win can outweigh several narrow defeats: a home win by \(10\) goals and three away wins by \(1\) goal each gives the home teams \(10 - 3 = 7\) more goals, but the away teams have won three games to one.
So Idris is not correct.
- (b) M1total of all eight bars \(85\)
- (b) A1mean \(4.25\)
- (c) M1home total \(46\) and away total \(39\)
- (c) A1difference \(7\)
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Either paper 2 marks
In one month, the number of hours of exercise taken by 10 people are
\(45,\ 43,\ 56,\ 56,\ 48,\ 52,\ 54,\ 9,\ 50,\ 47\)
Which is the appropriate average to use in this situation? Tick a box.
mean \(\square\) median \(\square\) mode \(\square\)
Give one reason for each of the other two averages as to why they are not appropriate.
Show answer
The median. The mean is not appropriate because the extreme value 9 pulls it down to 46. The mode is not appropriate because the only repeated value is the highest value in the list, so the mode is not in the middle of the data.In order, the values are
\(9,\ 43,\ 45,\ 47,\ 48,\ 50,\ 52,\ 54,\ 56,\ 56\)
One value, \(9\), is far away from the other nine, so it is extreme.
The mean is \(460 \div 10 = 46\). The extreme value \(9\) pulls the mean down, so it is not typical of the data. The median is \(49\), the middle of the ordered list, and it is hardly affected by \(9\).
Mode: the only repeated value is the highest value in the list, so the mode is not in the middle of the data.
The appropriate average is the median.
- B1the median chosen, with a reason against the mean (it is distorted by the extreme value, such as 9 pulling it down to 46)
- B1a reason against the mode: the only repeated value is the highest value in the list, so the mode is not in the middle of the data