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Either paper 4 marks
The box plot shows information about the times taken by a group of students to solve a puzzle.
(a)State the median time.
(b)Find the range of the times.
(c)Find the interquartile range of the times.
Show answer
(a) \(27\ \text{minutes}\)
(b) \(32\ \text{minutes}\)
(c) \(11\ \text{minutes}\)The median is the line drawn inside the box.
Reading straight down from it to the scale gives \(27\) minutes.
The range is the largest value take away the smallest, and those are the two ends of the whiskers: \(11\) and \(43\).
\(43 - 11 = 32\) minutes
The two ends of the box are the quartiles. Reading them off the scale, the lower quartile is \(20\) and the upper quartile is \(31\).
Interquartile range \(= 31 - 20 = 11\) minutes
The interquartile range uses only the middle half of the data, so one unusually large value cannot stretch it the way it stretches the range.
- (c) M1upper quartile \(31\) and lower quartile \(20\) read from the ends of the box
- (c) A1interquartile range \(11\) minutes
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Either paper 4 marks
The stem-and-leaf diagram shows the number of minutes eleven trains were delayed.
Stem Leaf 2 0 3 6 9 3 1 2 4 7 4 0 3 9 Key: 2 | 0 means 20 minutes
(a)Calculate the median.
(b)Calculate the interquartile range.
Show answer
(a) \(32\ \text{minutes}\)
(b) \(14\ \text{minutes}\)A stem-and-leaf diagram is already in order, which is what makes it worth drawing: reading the leaves left to right, row by row, gives
\(20,\ 23,\ 26,\ 29,\ 31,\ 32,\ 34,\ 37,\ 40,\ 43,\ 49\)
There are eleven trains, so the median is the \(6\)th value: \(32\) minutes.
The median is the \(6\)th value, leaving five values below it and five above it.
The lower quartile is the middle of the bottom five, the \(3\)rd value: \(26\) minutes.
The upper quartile is the middle of the top five, the \(9\)th value: \(40\) minutes.
\(40 - 26 = 14\) minutes
- (a) M1values read in order \(20,\ 23,\ 26,\ 29,\ 31,\ 32,\ 34,\ 37,\ 40,\ 43,\ 49\), 6th taken
- (a) A1median \(32\) minutes
- (b) M1lower quartile \(26\) and upper quartile \(40\)
- (b) A1interquartile range \(14\) minutes
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Either paper 3 marks
Here are the marks of 11 students in a test, written in order.
\(1,\ 8,\ 11,\ 17,\ 22,\ 28,\ 31,\ 36,\ 44,\ 48,\ 53\)
(a)State the lower quartile.
(b)State the upper quartile.
(c)Work out the interquartile range.
Show answer
(a) \(11\)
(b) \(44\)
(c) \(33\)There are 11 values, so the median is the 6th, \(28\).
The lower quartile is the middle of the five values below the median, which is the 3rd value along:
\(11\)
The upper quartile is the middle of the five values above the median, which is the 9th value along:
\(44\)
The interquartile range is the upper quartile take away the lower quartile:
\(44 - 11 = 33\)
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Either paper 5 marks
Here are the numbers of goals scored by eleven hockey players last season.
\(13,\ 27,\ 34,\ 20,\ 37,\ 42,\ 23,\ 29,\ 5,\ 10,\ 17\)
(a)Calculate the median.
(b)Calculate the interquartile range.
(c)A box plot is to be drawn for this data.
Write down the five values it needs.
Show answer
(a) \(23\ \text{goals}\)
(b) \(21\ \text{goals}\)
(c) \(5\), \(13\), \(23\), \(34\), \(42\)In order the eleven values are
\(5,\ 10,\ 13,\ 17,\ 20,\ 23,\ 27,\ 29,\ 34,\ 37,\ 42\)
With eleven values the median is the 6th one, because there are five below it and five above it.
Median \(= 23\) goals
The median splits the ordered list into a lower half of five values and an upper half of five values.
The lower quartile is the middle of the lower half, the 3rd value: \(13\).
The upper quartile is the middle of the upper half, the 9th value: \(34\).
Interquartile range \(= 34 - 13 = 21\) goals
A box plot needs the smallest value, the lower quartile, the median, the upper quartile and the largest value, in that order:
\(5\), \(13\), \(23\), \(34\), \(42\)
The whiskers reach out to \(5\) and \(42\), the box runs from \(13\) to \(34\), and the median is drawn inside it at \(23\).
- (b) M1lower quartile \(13\) and upper quartile \(34\)
- (b) A1interquartile range \(21\)
- (c) B1median \(23\) and quartiles \(13\) and \(34\)
- (c) B1smallest \(5\) and largest \(42\) found from the ordered data
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Either paper 3 marks
A vet writes the masses of the animals treated last month in order, smallest first. There are \(31\) masses in the list.
(a)Write down the position of the median in the list.
(b)Write down the position of the lower quartile.
(c)Write down the position of the upper quartile.
Show answer
(a) the 16th value
(b) the 8th value
(c) the 24th valueThe median is at position \(\frac{ 32 }{2}\), because \(31 + 1 = 32\).
That is the 16th value.
The lower quartile is a quarter of the way along: \(\frac{ 32 }{4} = 8\).
That is the 8th value.
The upper quartile is three quarters of the way along: \(\frac{ 3 \times 32 }{4} = 24\).
That is the 24th value.
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Either paper 2 marks
The box plot shows information about the marks scored by a class in a science test.
One of the students had a mark of \(30\) marks.
Is that mark in the top 25% of the marks? Explain your answer.
Show answer
NoThe top 25% of the data is everything above the upper quartile, which is \(31\) marks.
\(30\) marks is below \(31\) marks, so the answer is No.
- M1the upper quartile, \(31\) marks
- A1No, comparing \(30\) marks with \(31\) marks
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Either paper 5 marks
Here are the scores of 11 players on Monday.
\(4,\ 11,\ 53,\ 6,\ 21,\ 46,\ 42,\ 37,\ 27,\ 32,\ 16\)
(a)Work out the median of the Monday points.
(b)Work out the interquartile range of the Monday points.
(c)On Tuesday, the median score was \(32\) points and the interquartile range was \(24\) points.
Compare the two sets of points.
Show answer
(a) \(27\)
(b) \(31\)
(c) Tuesday was higher on average — its median of 32 is greater than Monday's median of 27. Tuesday was also more consistent, because its interquartile range of 24 is smaller than Monday's 31.In order, the points are
\(4,\ 6,\ 11,\ 16,\ 21,\ 27,\ 32,\ 37,\ 42,\ 46,\ 53\)
There are 11 values, so the median is the 6th one along:
\(27\)
The lower quartile is the middle of the five values below the median, the 3rd value: \(11\)
The upper quartile is the middle of the five values above the median, the 9th value: \(42\)
\(\text{IQR} = 42 - 11 = 31\)
Compare the medians first, because that is the comparison of averages:
\(32 > 27\), so Tuesday was higher on average.
Then compare the interquartile ranges, because that is the comparison of spread:
\(24 < 31\), so the middle half of the Tuesday points is more tightly bunched — Tuesday was more consistent.
- (b) M1lower quartile \(11\) and upper quartile \(42\)
- (b) A1interquartile range \(31\)
- (c) B1Tuesday higher on average, median 32 against 27
- (c) B1more consistent by the interquartile range, 24 against 31
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Either paper 4 marks
The times, in minutes, that 11 patients waited to see a doctor are
\(31,\ 25,\ 19,\ 21,\ 17,\ 22,\ 26,\ 12,\ 29,\ 80,\ 14\)
(a)Find the range of the waiting times.
(b)Find the interquartile range of the waiting times.
(c)Which of the range and the interquartile range better describes how spread out the waiting times were? Give a reason for your answer.
Show answer
(a) \(68\ \text{minutes}\)
(b) \(12\ \text{minutes}\)
(c) The interquartile range, because it uses only the middle half of the data, so the single long wait of 80 minutes does not affect it.\(80 - 12 = 68\) minutes
In order, the times are
\(12,\ 14,\ 17,\ 19,\ 21,\ 22,\ 25,\ 26,\ 29,\ 31,\ 80\)
There are 11 values, so the median is the 6th, \(22\), the lower quartile is the 3rd, \(17\), and the upper quartile is the 9th, \(29\).
\(\text{IQR} = 29 - 17 = 12\) minutes
Ten of the eleven patients waited between 12 and 31 minutes. One waited 80 minutes, and that single value stretches the range out to 68 minutes, which describes nobody's experience.
The interquartile range ignores the lowest quarter and the highest quarter of the data, so the long wait does not affect it. At 12 minutes it describes the spread of the times much better.
- (b) M1lower quartile \(17\) and upper quartile \(29\)
- (b) A1interquartile range \(12\) minutes
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Either paper 4 marks
The box plots show information about the marks scored by the boys and by the girls in a test.
(a)Find the interquartile range of the marks for Boys.
(b)Compare the marks for Boys with the marks for Girls.
Make one comparison about the average and one about the spread.
Show answer
(a) \(10\ \text{marks}\)
(b) The median for Girls is higher, \(20\) marks against \(17\) marks, so on average the marks for Girls were greater. The interquartile range for Boys is smaller, \(10\) marks against \(14\) marks, so the marks for Boys were more consistent.The ends of the Boys box are the quartiles: \(14\) and \(24\).
Interquartile range \(= 24 - 14 = 10\) marks
One comparison about the average and one about the spread, each backed up by a number off the diagram.
Average: the median for Girls is \(20\) marks and the median for Boys is \(17\) marks, so the marks for Girls were greater on average.
Spread: the interquartile range for Boys is \(10\) marks and for Girls it is \(14\) marks, so the marks for Boys were more consistent.
Comparing the two ranges would also be a comparison of spread, but the interquartile range is the safer one: it ignores the two whisker ends, where a single unusual value sits.
- (a) M1quartiles \(14\) and \(24\) read for Boys
- (a) A1interquartile range \(10\) marks
- (b) B1median for Girls higher, \(20\) marks against \(17\) marks
- (b) B1interquartile range for Boys smaller, \(10\) marks against \(14\) marks, so more consistent
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Either paper 3 marks
The table shows information about the heights of the plants in a greenhouse. No diagram has been drawn.
Smallest height \(6\) cm Lower quartile \(10\) cm Median \(15\) cm Range \(22\) cm Interquartile range \(13\) cm (a)Calculate the largest height.
(b)Calculate the upper quartile.
(c)A box plot is to be drawn from this information.
Write down, in order, the five values it needs.
Show answer
(a) \(28\ \text{cm}\)
(b) \(23\ \text{cm}\)
(c) \(6\), \(10\), \(15\), \(23\), \(28\)The range is the largest value take away the smallest, so the largest is the smallest plus the range.
\(6 + 22 = 28\) cm
The interquartile range is the upper quartile take away the lower quartile, so the upper quartile is the lower quartile plus the interquartile range.
\(10 + 13 = 23\) cm
Three of the five are given and two came from parts (a) and (b):
\(6\), \(10\), \(15\), \(23\), \(28\)
The range and the interquartile range are *not* two of the five — they are the distances between them.