2. `

3. 14 42 115 217 294 329 350

4. Cumulative Frequency How to draw a cumulative frequency graph

5. Data xFrequency 10 ≤ x < 202 20 ≤ x < 304 30 ≤ x < 405 40 ≤ x < 507 50 ≤ x < 604 60 ≤ x ≤ 702 Draw a cumulative frequency diagram for this data

6. DataFrequencyCumulative Frequency 10 – 2022 20 – 3042 + 4 = 6 30 – 4052 + 4 + 5 = 11 40 – 5072 + 4 + 5 +7 = 18 50 – 6042 + 4 + 5 +7 + 4 = 22 60 – 7022 + 4 + 5 +7 + 4 + 2 = 24 Create a third CUMULATIVE FREQUENCY column like this

7. You don’t have to show working DataFrequencyCumulative Frequency 10 – 2022 20 – 3042 + 4 = 6 30 – 4052 + 4 + 5 = 11 40 – 5072 + 4 + 5 +7 = 18 50 – 6042 + 4 + 5 +7 + 4 = 22 60 – 7022 + 4 + 5 +7 + 4 + 2 = 24

8. You don’t have to show working DataFrequencyCumulative Frequency 10 – 2022 20 – 3046 30 – 40511 40 – 50718 50 – 60422 60 – 70224

9. DataFrequenc y Cumulative Frequency 10 – 2022 20 – 3046 30 – 40511 40 – 50718 50 – 60422 60 – 70224 Plot these numbers Plot the second number in the data column against the number in the cumulative frequency column

10. Now, join up the points

11. You can use this to find the middle half ¾ (18) ½ (12) ¼ (6 ) Lower Q (30) Median (42) Upper Q (53)

12. The Lower Quartile is 30 Median is 42 The Upper Quartile is 53 This means that the middle half is between 30 and 53. Called the inter-quartile range. 53 – 30 = 23

14. Frequency Teach GCSE Maths Grouped Data Rainfall (mm) 2 7 18 17 5 1 and the Mean < 0 x < 20 20 x < 30 30 x < 35 40 x < 50 50 x < 70 35 x < 40 < < < < <

15. e.g.1This table gives the time taken for 30 components to fail. Time to failure (hours), t Number of components f 0 t < 20 5 20 t < 40 8 40 t < 60 17 < < < means t can also equal 0. Decide with your partner if t can equal 20 in the 1 st class. BUT, the extra line... 0 t < 20 < Tip: Tilt your head to the right and you can see the extra line making an equals sign. Ans: No. Measurements of t = 20 are in the 2 nd class. Since the quantity is time, a t has been used instead of x. The t written between 0 and 20 means that the time is between 0 and 20 hours ! The numbers 20, 40 and 60, at the top of the classes, are called the “upper class boundaries”

16. Tell your partner why, using the table, we cannot find the exact value of the mean. Suppose we want to find the mean time that a component lasts. To calculate an estimate of the mean, we need to choose one number in each class that represents the class. Ans: We don’t know the exact value of each time. For example, in the 1 st class there are 5 failures. They could all have been in the 1 st hour, or be equally spaced, or be 13·5, 16·2, 17, 18·7, 19·9... or any times between 0 and 20. Ans: t = 10. It is the mid-point of the class, the average of 0 and 20. To represent a class, we use the mid-point of the class. Time to failure (hours), t Number of components f 0 t < 20 5 20 t < 40 8 40 t < 60 17 < < < Decide with your partner which number you would use to represent the 1 st class ( 0 t < 20 ). <

17. We will need an extra column for the mid-points ( which can also be called t ). Time to failure (hours), t Number of components f 0 t < 20 5 20 t < 40 8 40 t < 60 17 In this question, the mid-points are easy to spot but we need to remember that a mid-point is the average of the numbers at each end of the class ( the boundary values ). (0 + 20) = 10 1 2 (20 + 40) = 30 1 2 (40 + 60) = 50 1 2 < < <

18. Time to failure (hours), t Number of components f Mid-point 0 t < 20 5 20 t < 40 8 40 t < 60 17 Time to failure (hours), t Number of components f Mid-point t × f 0 t < 20 5 20 t < 40 8 40 t < 60 17 Totals Now we can calculate an estimate of the mean time. mean time = total time ÷ number of components = sum of t × f sum of f = 50 240 850 30 1140 This column now gives t. t 10 30 50 < < < 1140 30 = 38 hours Check: 38 is between 0 and 60.

19. 1.The table shows the lengths of 25 pieces of wood. 3 60 l < 90 6 50 l < 60 9 40 l < 50 4 30 l < 40 3 10 l < 30 Frequency f Length (cm) l < < < < < Exercise (a)Calculate an estimate of the mean length. (b) Which is the modal class?

20. Length (cm) l Frequency f 10 l < 30 3 30 l < 40 4 40 l < 50 9 50 l < 60 6 60 l < 90 3 Total25 Mid-value 20 35 45 55 75 60 140 405 330 225 l × f Solution: Length (cm) l Frequency f 10 l < 30 3 30 l < 40 4 40 l < 50 9 50 l < 60 6 60 l < 90 3 < < < < < (a) mean length = total length ÷ number of pieces = sum of l × f sum of f = 1160 25 = 46·4 cm Check: 46·4 is between 10 and 90. (b) the modal class is 40 l < 50 < 1160

21. Changing the Subject of a formula

22. Substituting 35 -3 -31 149 -24

23. Same Sign Subtract Solve 2x + y = 8 and 5x + y = 17 3x + 0 = 9 x = 3 Substitute x = 3 in Check in (not used directly to find y) 5 x 3 + 2 = 17 1 2 21 1 2 - 2 x 3 + y = 8 so y = 2 x = 3 and y = 2

24. Different Signs Add Solve 3x + 2y = 8 x - 2y = 0 4x + 0 = 8 so x = 2 Substitute x = 2 in to find y 3 x 2 + 2y = 8 so 2y = 2 so y = 1 Check in 2 - 2 x 1 = 0 x = 2 and y = 1 1 2 12 1 2 +

25. Different amounts of x and y Solve x + 2y = 11 and 3x + y = 18 Need either same number of x’s or y’s so gives 3x + 6y = 33 (SSS) 0 + 5y = 15 so y = 3 Sub y = 3 in Check in 3 x 5 + 3 = 18 x = 5 and y = 3 1 2 31 32 1 2 x 3 - x + 2 x 3 = 11 so x = 5

26. Sometimes... We need to multiply both equations Solve 5x + 2y = 15 and 3x - 3y = 51 1 2 We could do x 3 then 1 2 2 x 34 We would then have two new equations & which can be added to cancel out y as before

27. Word problems 2.A fruit machine contains 200 coins. These are either 20p or 50p. The total value of the coins is £65.20 How many of each coin are in the machine? What is the value in pence of x 20p’s ?