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Published byKimberly Halfpenny Modified over 2 years ago

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Cumulative Frequency How to draw a cumulative frequency graph

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Data xFrequency 10 ≤ x < ≤ x < ≤ x < ≤ x < ≤ x < ≤ x ≤ 702 Draw a cumulative frequency diagram for this data

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DataFrequencyCumulative Frequency 10 – – = 6 30 – = – = – = – = 24 Create a third CUMULATIVE FREQUENCY column like this

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You don’t have to show working DataFrequencyCumulative Frequency 10 – – = 6 30 – = – = – = – = 24

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You don’t have to show working DataFrequencyCumulative Frequency 10 – – – – – – 70224

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DataFrequenc y Cumulative Frequency 10 – – – – – – Plot these numbers Plot the second number in the data column against the number in the cumulative frequency column

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Now, join up the points

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

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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

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Frequency Teach GCSE Maths Grouped Data Rainfall (mm) and the Mean < 0 x < x < x < x < x < x < 40 < < < < <

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e.g.1This table gives the time taken for 30 components to fail. Time to failure (hours), t Number of components f 0 t < t < t < < < < 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”

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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 < t < t < < < < Decide with your partner which number you would use to represent the 1 st class ( 0 t < 20 ). <

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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 < t < t < 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) = ( ) = ( ) = < < <

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Time to failure (hours), t Number of components f Mid-point 0 t < t < t < Time to failure (hours), t Number of components f Mid-point t × f 0 t < t < t < 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 = This column now gives t. t < < < = 38 hours Check: 38 is between 0 and 60.

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1.The table shows the lengths of 25 pieces of wood l < l < l < l < l < 30 Frequency f Length (cm) l < < < < < Exercise (a)Calculate an estimate of the mean length. (b) Which is the modal class?

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Length (cm) l Frequency f 10 l < l < l < l < l < 90 3 Total25 Mid-value l × f Solution: Length (cm) l Frequency f 10 l < l < l < l < l < 90 3 < < < < < (a) mean length = total length ÷ number of pieces = sum of l × f sum of f = = 46·4 cm Check: 46·4 is between 10 and 90. (b) the modal class is 40 l < 50 < 1160

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Changing the Subject of a formula

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Substituting

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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 = x 3 + y = 8 so y = 2 x = 3 and y = 2

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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 x 1 = 0 x = 2 and y =

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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 = 18 x = 5 and y = x 3 - x + 2 x 3 = 11 so x = 5

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

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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 ?

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