The numbers of fouls made in eight hockey matches and in eight football matches played at the weekend are shown in the following back-to-back stem-and-leaf diagram. Draw two box-and-whisker diagrams using the same scales. Write a sentence to compare the numbers of fouls committed in the two sports.

Given back to back stem and leaf plot: A back-to-back stem-and-leaf plot is a method for comparing two data distributions by attaching two sets of 'leaves' to the same 'stem' in a stem-and-leaf plot. The leaf goes in the centre, with numbers running from least (at the top) to highest (at the bottom). Numbers in the right leaf run lowest to highest, with the lowest number placed closest to the stem. The data on the left side of the stem runs in the opposite direction (like a mirror) with the numbers running from lowest to highest starting at the stem. Let number of data value be N . Here, the number of data values of football and hockey are 8 ⇒ N = 8 For football matches: If N is even then the formula of median = Average of N 2 th and N 2 + 1 th data values, where N is number of data values. ⇒ Median is average of 8 2 th and 8 2 + 1 th data values = Average of 4 th and 5 th data values. So, median = 19 + 21 2 = 20 Now, lower quartile Q 1 is the median of the lower 50 % of the data. Here, there are 8 data values. If N is even, then the formula to find lower quartile is Q 1 = Average of N 4 th and N 4 + 1 th data values, where N is number of data values. ⇒ Lower quartile is average of 8 4 th and 8 4 + 1 th data values = Average of 2 nd and 3 rd data values. So, Q 1 = 18 + 19 2 = 18 . 5 Now, upper quartile Q 3 is the median of the upper 50 % of the data. Here, there are 8 data values. If N is even, then the formula to find upper quartile is Q 3 = Average of 3 N 4 th and 3 N 4 + 1 th data values, where N is number of data values. ⇒ Upper quartile is average of 3 × 8 4 th and 3 × 8 4 + 1 th data values = Average of 6 th and 7 th data values. So, Q 3 = 22 + 23 2 = 22 . 5 Now, interquartile range is IQR = Q 3 - Q 1 = 22 . 5 - 18 . 5 = 4 . We know that, the range is the difference between the maximum and minimum values of the data set. Here, the maximum value is 23 and the minimum value is 10 . So, Range = 23 - 10 = 13 . For Hockey players: If N is even then the formula of median = Average of N 2 th and N 2 + 1 th data values, where N is number of data values. ⇒ Median is average of 8 2 th and 8 2 + 1 th data values = Average of 4 th and 5 th data values. So, median = 16 + 18 2 = 17 Now, lower quartile Q 1 is the median of the lower 50 % of the data. Here, there are 8 data values. If N is even, then the formula to find lower quartile is Q 1 = Average of N 4 th and N 4 + 1 th data values, where N is number of data values. ⇒ Lower quartile is average of 8 4 th and 8 4 + 1 th data values = Average of 2 nd and 3 rd data values. So, Q 1 = 12 + 14 2 = 13 Now, upper quartile Q 3 is the median of the upper 50 % of the data. Here, there are 8 data values. If N is even, then the formula to find upper quartile is Q 3 = Average of 3 N 4 th and 3 N 4 + 1 th data values, where N is number of data values. ⇒ Upper quartile is average of 3 × 8 4 th and 3 × 8 4 + 1 th data values = Average of 6 th and 7 th data values. So, Q 3 = 19 + 21 2 = 20 Now, interquartile range is IQR = Q 3 - Q 1 = 20 - 13 = 7 . We know that, the range is the difference between the maximum and minimum values of the data set. Here, the maximum value is 11 and the minimum value is 21 . So, Range = 21 - 11 = 10 . In a box and whisker plot: The ends of the box are the upper and lower quartiles, so the box spans the interquartile range. The median is marked by a vertical line inside the box The whiskers are the two lines outside the box that extend to the highest and lowest observations. So, at the end of the right whisker is the maximum value and at the end of the left whisker is the minimum value. The extended horizontal lines are called whiskers. Make a box by drawing horizontal lines connecting the quartiles. Connect the top or the first quartile to the top of the third quartile, going through the median. Connect the bottom of the first quartile to the bottom of the third quartile, making sure to go through the median. General diagram of Box and whisker plot: Thus, the box-and whisker plot for the given data: Thus, there were fewer fouls on average in hockey (medians are 17 and 20 ) and the numbers of fouls in hockey are more varied than in football.

Rishi and Daisy take the same seven tests in Mathematics, and both students' marks improve on successive tests. Their percentage marks are as follows. Rishi ' s marks 15 24 28 33 39 42 50 Daisy ' s marks 51 65 69 72 78 83 86 Explain why it would not be useful to use the range or the interquartile range alone as measures for comparing the marks of the two students.

Given data: Rishi ' s marks 15 24 28 33 39 42 50 Daisy ' s marks 51 65 69 72 78 83 86 Rish's marks: Arrange the data in ascending order 15 , 24 , 28 , 33 , 39 , 42 , 50 We know that, formula to find interquartile range is IQR = Q 3 - Q 1 , where Q 1 is the first quartile of the data set and Q 3 is third quartile of the data set. Now, upper quartile Q 3 is the median of the upper 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find upper quartile is Q 3 = 3 N + 1 4 th data value, where N is number of data values. ⇒ upper quartile Q 3 = 3 7 + 1 4 th = 6 th data value. The 6 th data value in the ascending order of the data is 42 . We know that, lower quartile Q 1 is the median of the lower 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find lower quartile is Q 1 = N + 1 4 th data value, where N is number of data values. ⇒ Lower quartile, Q 1 = 7 + 1 4 th = 2 nd data value. Hence, the lower quartile of the given data is Q 1 = 24 . Now, interquartile range is IQR = Q 3 - Q 1 = 42 - 24 = 18 . Daisy's marks: Arrange the data in ascending order 51 , 65 , 69 , 72 , 78 , 83 , 86 We know that, formula to find interquartile range is IQR = Q 3 - Q 1 , where Q 1 is the first quartile of the data set and Q 3 is third quartile of the data set. Now, upper quartile Q 3 is the median of the upper 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find upper quartile is Q 3 = 3 N + 1 4 th data value, where N is number of data values. ⇒ upper quartile Q 3 = 3 7 + 1 4 th = 6 th data value. The 6 th data value in the ascending order of the data is 83 . We know that, lower quartile Q 1 is the median of the lower 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find lower quartile is Q 1 = N + 1 4 th data value, where N is number of data values. ⇒ Lower quartile, Q 1 = 7 + 1 4 th = 2 nd data value. Hence, the lower quartile of the given data is Q 1 = 65 . Now, interquartile range is IQR = Q 3 - Q 1 = 83 - 65 = 18 . We know that, the range is the difference between the maximum and minimum values of the data set. Rish's marks: Here, the maximum value is 50 and the minimum value is 15 . So, Range = 50 - 15 = 35 . Daisy's marks: Here, the maximum value is 86 and the minimum value is 51 . So, Range = 86 - 51 = 35 . Thus, Ranges and IQR's are the same 35 and 18 but the two students' marks are quite different.

Rishi and Daisy take the same seven tests in Mathematics, and both students' marks improve on successive tests. Their percentage marks are as follows. Rishi ' s marks 15 24 28 33 39 42 50 Daisy ' s marks 51 65 69 72 78 83 86 Name two measures that could be used together to give a meaningful comparison of the two students' marks.

Given data: Rishi ' s marks 15 24 28 33 39 42 50 Daisy ' s marks 51 65 69 72 78 83 86 Rish's marks: Arrange the data in ascending order 15 , 24 , 28 , 33 , 39 , 42 , 50 Let n represents the number of data values. Now, Median = n + 1 2 th data value = 7 + 1 2 th data value = 4 th data value Thus, median = 33 . Now, Mean = sum of the data values Total number of data values = 15 + 24 + 28 + 33 + 39 + 42 + 50 7 = 33 We know that, formula to find interquartile range is IQR = Q 3 - Q 1 , where Q 1 is the first quartile of the data set and Q 3 is third quartile of the data set. Now, upper quartile Q 3 is the median of the upper 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find upper quartile is Q 3 = 3 N + 1 4 th data value, where N is number of data values. ⇒ upper quartile Q 3 = 3 7 + 1 4 th = 6 th data value. The 6 th data value in the ascending order of the data is 42 . We know that, lower quartile Q 1 is the median of the lower 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find lower quartile is Q 1 = N + 1 4 th data value, where N is number of data values. ⇒ Lower quartile, Q 1 = 7 + 1 4 th = 2 nd data value. Hence, the lower quartile of the given data is Q 1 = 24 . Now, interquartile range is IQR = Q 3 - Q 1 = 42 - 24 = 18 . Daisy's marks: Arrange the data in ascending order 51 , 65 , 69 , 72 , 78 , 83 , 86 Let n represents the number of data values. Now, Median = n + 1 2 th data value = 7 + 1 2 th data value = 4 th data value Thus, median = 72 . Now, Mean = sum of the data values Total number of data values = 51 + 65 + 69 + 72 + 78 + 83 + 86 7 = 72 We know that, formula to find interquartile range is IQR = Q 3 - Q 1 , where Q 1 is the first quartile of the data set and Q 3 is third quartile of the data set. Now, upper quartile Q 3 is the median of the upper 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find upper quartile is Q 3 = 3 N + 1 4 th data value, where N is number of data values. ⇒ upper quartile Q 3 = 3 7 + 1 4 th = 6 th data value. The 6 th data value in the ascending order of the data is 83 . We know that, lower quartile Q 1 is the median of the lower 50 % of the data. Here, there are 7 data values. If N is odd, then formula to find lower quartile is Q 1 = N + 1 4 th data value, where N is number of data values. ⇒ Lower quartile, Q 1 = 7 + 1 4 th = 2 nd data value. Hence, the lower quartile of the given data is Q 1 = 65 . Now, interquartile range is IQR = Q 3 - Q 1 = 83 - 65 = 18 . We know that, the range is the difference between the maximum and minimum values of the data set. Rish's marks: Here, the maximum value is 50 and the minimum value is 15 . So, Range = 50 - 15 = 35 . Daisy's marks: Here, the maximum value is 86 and the minimum value is 51 . So, Range = 86 - 51 = 35 . Thus, Ranges and IQR's are the same 35 and 18 but the two students' marks are quite different. Thus, One of median or mean, and one of the range or IQR could be used together to give a meaningful comparison of the two students' marks.

The following table shows the maximum speeds, s km / h , of some vintage cars. Maximum speed s km / h No. vintage cars f s < 35 0 35 ⩽ s < 40 20 40 ⩽ s < 45 65 45 ⩽ s < 50 110 50 ⩽ s < 55 27 55 ⩽ s < 70 13 70 ⩽ s < 75 5 Construct a cumulative frequency polygon and a box-and-whisker diagram to illustrate the data.

Given data: Maximum speed s km / h No. vintage cars f s < 35 0 35 ⩽ s < 40 20 40 ⩽ s < 45 65 45 ⩽ s < 50 110 50 ⩽ s < 55 27 55 ⩽ s < 70 13 70 ⩽ s < 75 5 If the frequency of first class interval is added to the frequency of second class and this sum is added to third class and so on then frequencies so obtained are known as Cumulative Frequency. A table which displays the manner in which cumulative frequencies are distributed over various classes is called a cumulative frequency distribution or cumulative frequency table. Cumulative frequency table for this information: Maximum speed s km / h No. vintage cars f Cummulative freuqncy s < 35 0 0 35 ⩽ s < 40 20 20 40 ⩽ s < 45 65 85 45 ⩽ s < 50 110 195 50 ⩽ s < 55 27 222 55 ⩽ s < 70 13 235 70 ⩽ s < 75 5 240 Steps to draw less than type cumulative frequency curve:A curve that represents the cumulative frequency distribution of grouped data on a graph is called a Cumulative Frequency Curve or an Ogive (i) Draw and mark the horizontal and vertical axes. (ii) Take the cumulative frequencies along the y -axis (vertical axis) and the upper-class limits on the x -axis (horizontal axis). (iii) Against each upper-class limit, plot the corresponding cumulative frequencies. (iv) Connect the points with a continuous curve From the cumulative frequency curve: To find the median, we draw a horizontal line across from the frequency axis to the curve at the n 2 th value and then down to the x -axis to determine the median data point. Here n represents total number of data values. Here, n = 240 So, n 2 th value is 120 th value. Draw a line across from 120 on the vertical axis( y -axis) until it hits the curve and then draw a line vertically down until it touches the horizontal axis( x -axis). This is the median Q 2 . Thus, the median of the given data is 46 . 5 . To determine the lower quartile Q 1 , we draw a horizontal line across from the frequency axis to the curve at the n 4 th value and then down to the x -axis. i.e., 240 4 th = 60 th value. Draw a line across from 60 on the vertical axis until it hits the curve and then draw a line vertically down until it touches the horizontal axis. This is the lower quartile. Thus, lower quartile of the given data is Q 1 = 43 . To determine the upper quartile Q 3 , we draw a horizontal line across from the frequency axis to the curve at the 3 n 4 th value and then down to the x -axis. i.e., 3 × 240 4 th = 180 th value Draw a line across from 180 value on the vertical axis until it hits the curve and then draw a line vertically down until it touches the horizontal axis. This is the upper quartile. Thus, the upper quartile of the given data is Q 3 = 50 . 1 . The lower limit of the minimum interval is 35 and upper limit of the maximum interval is 75 . Here, Minimum First quartile Median Third quartile Maximum 35 43 46 . 5 50 . 1 75 Here, make a box by drawing horizontal lines connecting the quartiles. Connect the top or the first quartile to the top of the third quartile, going through the median. Connect the bottom of the first quartile to the bottom of the third quartile, making sure to go through the median. General diagram of Box and whisker plot: Thus, Less than type cumulative frequency curve and box plot for the given data is

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Find the range and the interquartile range of the dataset represented in the following box plot.

Important Questions on Measures of Variation

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 2

What type of skewness would you expect this set of data to have?

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 3

The following stem-and-leaf diagram shows the marks out of $50$ obtained by $15$ students in a Science test.

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Find the range and interquartile range of the marks.

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 3

The following stem-and-leaf diagram shows the marks out of $50$ obtained by $15$ students in a Science test.

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Five-point summary for the given data set:

Minimum	Lower quartile	Median	Upper quartile	Maximum
$9$	$28$	$37$	$46$	$50$

The box-and whisker plot for the given data:

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For this set of data, express $Q_{3}$ in terms of $Q_{1}$ and $Q_{2} .$

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 4

The numbers of fouls made in eight hockey matches and in eight football matches played at the weekend are shown in the following back-to-back stem-and-leaf diagram.

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Is it true to say that the numbers of fouls in the two sports are equally varied? Explain your answer.

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 4

The numbers of fouls made in eight hockey matches and in eight football matches played at the weekend are shown in the following back-to-back stem-and-leaf diagram.

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Draw two box-and-whisker diagrams using the same scales. Write a sentence to compare the numbers of fouls committed in the two sports.

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 5

Rishi and Daisy take the same seven tests in Mathematics, and both students' marks improve on successive tests. Their percentage marks are as follows.

$\begin{matrix} Rishi' s marks \\ 15 24 28 33 39 42 50 \end{matrix}$ $\begin{matrix} Daisy' s marks \\ 51 65 69 72 78 83 86 \end{matrix}$

Explain why it would not be useful to use the range or the interquartile range alone as measures for comparing the marks of the two students.

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 5

Rishi and Daisy take the same seven tests in Mathematics, and both students' marks improve on successive tests. Their percentage marks are as follows.

$\begin{matrix} Rishi' s marks \\ 15 24 28 33 39 42 50 \end{matrix}$ $\begin{matrix} Daisy' s marks \\ 51 65 69 72 78 83 86 \end{matrix}$

Name two measures that could be used together to give a meaningful comparison of the two students' marks.

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Cambridge International AS & A Level Mathematics : Probability & Statistics 1 Course Book>Chapter 3 - Measures of Variation>EXERCISE 3A>Q 6

The following table shows the maximum speeds, $s km / h,$ of some vintage cars.

Maximum speed $(s km / h)$	No. vintage cars $(f)$
$s < 35$	$0$
$35 ⩽ s < 40$	$20$
$40 ⩽ s < 45$	$65$
$45 ⩽ s < 50$	$110$
$50 ⩽ s < 55$	$27$
$55 ⩽ s < 70$	$13$
$70 ⩽ s < 75$	$5$

Construct a cumulative frequency polygon and a box-and-whisker diagram to illustrate the data.

Find the range and the interquartile range of the dataset represented in the following box plot.

Important Questions on Measures of Variation

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