YEAR 10
REVISION BLASTER
GCSE MATHS REVISION
• UNIT 1 – all about DATA
• Probability
• Averages
• Cumulative frequency
• Histograms
• Percentages
Probability
KEY POINTS
•The probability scale goes from 0 to 1.
•Probability of a certain event = 1
•Probability of an impossible event= 0
•P(event)= No. of ways that event can happen
Total no. of ALL possible outcomes
22 Sweets in a bag, 7 are red, 6 are blue, 9 are green.
What is the probability of randomly picking a blue sweet?
•P(event)= No. of ways that event can happen
Total no. of ALL possible outcomes
P (BLUE) =
622
=
11
3
Calculate the Probability of NOT picking a blue sweet.
P(event Not happening)=1 – P(event happening)
P (NOT BLUE) = 1 – P(picking blue)
P (NOT BLUE) = 1 –
P (NOT BLUE) =
11
3
11
8
Calculating the Probability of combined events.STARTER:
P(garlic bread)=0.2; P(dough balls)= 0.8
MAIN:
P(pasta)=0.2; P(Salad)=0.2 P(Pizza) = 0.6
•Calculate the probability of having; garlic bread followed by pizza.
•Calculate the probability of having ANY starter followed by salad.
0.2
P(garlic bread)=0.2; P(dough balls)= 0.8P(pasta)=0.2; P(Salad)=0.2 P(Pizza) = 0.6
STARTER
Garlic Bread
Dough
Balls
0.2
0.8
Pizza
Pasta
Salad
0.2
0.2
0.6
Pasta
Salad
0.2
0.2
0.6Pizza
MAINCalculate the probability of having; garlic bread followed by pizza.
0.20.6
0.2 x 0.6 = 0.12
AND RULE (Going ALONG the branches)
We MULTIPLY the Probabilities.
Answer= 0.12
•Calculate the probability of having ANY starter followed by salad.
0.2
0.2 x 0.2 = 0.04
0.80.2 0.8 x 0.2 =
0.16
There are TWO possibilities here; garlic bread and then salad, OR dough balls and the salad.
OR RULE= ADD PROBABILITIES
SO we add together the possibilities:
0.04 + 0.16 = 0.20
Probability of having ANY starter followed by salad is 0.20
Relative Frequency: An estimate for theoretical probabilityRelative Frequency = Frequency of the event
Total Number of trials
Calculate the relative frequency of the toast landing jam side up.
No. of drops
No. of jam side up
Relative Frequency
12 6 12/6 =2
Relative Frequency = Frequency of the event
Total Number of trials
The relative frequency of seeing a red car go past your window is 0.6.
You watch 100 cars go past your window. How many would you expect to be red?
0.6
=Frequency of the event
100
0.6 x 100 = Frequency
60 = Frequency
Probability Answers• 1. (a) ½• (b) 300• 2. (a) (i) 11/12 or 0.92• (ii) 8/12 or 2/3• (b) (i) 15/22 (allow 14 or 15 on top 22 or 21 on
bottom)• (ii) 5/11 or any equivalent fraction
• 3. (a) 0.4 on first branch & all other branches correct
• (b) (i) 0.16 or 4/25 or 16 %• (ii) 0.84 or 21/25 or 84 %• (a)•• (b) 15• (c) More than expected with a suitable qualification
(allow expect 10)
20
3,
20
3,
20
3,
20
5,
20
3,
20
3
Averages
MediaMedia
nnM
od
Mod
ee
MEAMEA
NN
3, 5, 4, 3, 5, 4, 4, 6, 73, 3, 4, 4, 4, 5, 5, 6, 7
3, 3, 4, 4, 4, 5, 5, 6, 7
3+3+4+4+4+5+5+6+7=41 41
9= 4.55…
0 × 20 = 0
1 × 17 = 17
2 × 15 = 30
3 × 10 = 30
4 × 9 = 36
5 × 3 = 15
6 × 2 = 12
140
Mean = sum of ALL the values sum of frequencies
Calculate the mean.
26
3
9
10
15
17
20
Frequency
4
5
3
2
1
0
Numbers of sports played
TOTAL 76
Frequency x Numbers of sports played
= 140 = 1.84 76
2105263
1 × 7.5 = 7.5
8 × 12.5 = 100
12 × 17.5 = 210
10 × 22.5 = 225
3 × 27.5 = 82.5
1 × 32.5
1 × 37.5
= 32.5
= 37.5
695
7.5
12.5
17.5
22.5
27.5
32.5
37.5
Mean = sum of ALL frequency x midpoint sum of frequencies
= 695 = 19.3 36
055555…
135 ≤ d < 40
1
3
10
12
8
1
Frequency
30 ≤ d < 35
25 ≤ d < 30
20 ≤ d < 25
15 ≤ d < 20
10 ≤ d < 15
5 ≤ d < 10
Javelin distances in
metres
36TOTAL
Midpoint Frequency × midpoint
Calculate an estimate for the mean.
Averages Answers
1. 99.7 (allow 100)
2. (a) 131.6 (allow 131 to 132)(b) 110 £ t < 130
3. (a) 26.4 (allow 26)(b) Frequency polygon from (10, 42) to (70, 3) joined with approximately straight lines(c) Comment - eg. Higher mean on Saturday, or
Larger range on Saturday, orMore money spent on Saturday
AD BREAKYear 10 Unit 1 Exam. Tuesday 9th November-
TOMORROW!!
Don’t Forget: CALCULATORS
PEN
PENCIL
RULER
REVISE
CUMULATIVE
FREQUENCY
Cumulative Frequency
The cumulative frequency is the
RUNNING TOTAL OF FREQUENCIES
Used for finding the MEDIAN and UPPER AND LOWER QUARTILES
Marks in a Yr 10 Maths TestMarks Frequency Cumulative
Frequency
11-20
21-30 11
31-40 19
41-50 36
42
61-70 31
71-80 13
81-90 6
110 people scored 60 marks OR less
2 21332
68110141154160
51-60
Plot the top value in each group against cumulative frequency
Median shows that ½ of people score less
than 53 marks and ½ score more
Lower Quartile shows 25% of people scored 44 marks or
lessUpper Quartile shows 75% of people scored 65 marks or
less
Interquartile range =65 -44
=21
Cumulative Frequency Answers
1. (a) Any value from 55 to 57 inclusive(b) 6 ± 0.4
2. (a) 12, 27, 56, 72, 83, 90(b) Correct graph ‘increasing’ (S shape)(c) (i) Median line from 45(.5)
e.g. 26-28(ii) IQR lines from Q1 (22.5) and Q3 (67.5)
and subtract their answers (eg 16 – 20) (d) Approx 48
HISTOGRAMS
Finding lengths given areas
• If the width is 10– What must the height be
to make an area of 25?
10
252.5As 10 x 2.5 = 25
Bar Graphs
• Bar graphs are great to use when you have equal class intervals:
Histograms
• Histograms are used when you have different sized class intervals.
• The area of the rectangle represents the frequency.
• The width is the size of the class interval.
• The height is what we call the frequency density.
Number of kick-ups
Frequency Frequency
Density
0-5 10
5-15 25
15-30 45
30-40 15
The table below shows the number of kick-ups completed in a competition at a local primary school.
We have different sized class intervals
Our class interval width is 5, and our frequency (area) is 10.
So, Frequency Density (height) = 10/5
2
2.53
1.5
Number of
kick-ups
Frequency Frequency
Density
0-5 10 25-15 25 2.515-30 45 330-40 15 1.5
Estimate how many students managed ten kick-ups or less.
Frequency =5 x 2.5 = 12.5
Frequency=5 x 2 = 10
So, 10 + 12.5 = 22.5We can estimate that 22.5 students made ten kick-upsor less.
Comparing Histograms and Bar Graphs
Histograms Answers
1. (a) 0.4, 0.4, 0.5, 0.1
(b) 21
2. (a) Correct histogram:
Widths: 15, 5, 5, 10, 15
Heights: 0.6, 4.2, 4.8, 3.1, 1
(b) 42
Percentage Increase
and Decrease
Increase £20 by 15%
Method 1Find 15%10% = £2 5% = £1 so 15% = £3
Now add it on to £20£20 + £3 = £23
Increase by 15% so
1.15 is the multiplier
Method 2Method 2
I need to find 115% I need to find 115% OF original amountOF original amount115% of £20 is 115% of £20 is 1.15 x £20 = £231.15 x £20 = £23
Decrease £60 by 5%
Method 1
Find 5% of £60
10% = £6 5% = £3
Subtract from £60
£60 - £3 = £57
Method 2
A decrease of 5% is same as 95% of original amount
So 0.95 x £60 = £57 95% of original amount means 0.95 is the multiplier
Compound Interest
• £1,000 in bank earns 5% interest per year. How much will you have in 3 years?
After 1 year 1000 x 1.05 = 1050
After 2 years 1050 x 1.05= 1102.50
After 3 years 1102.50 x 1.05 = 1157.63
OR 1000 x 1.05 x 1.05 x1.05=
1000 x 1.05³=£1157.63
1.05 is the multiplier!
Percentages Answers
1.£2140 (allow £140)
2.(a) 1.029
(b) 1 087 401 937 (allow 1 087 000 000)
3.1st year is £20; 2nd year is £20.80
Interest is £40.80 so Amir is wrong
4.(a) 2.04
(b) 6 (windmills)
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