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Areas to cover

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Charts: Stem&leaf / Scatter graphs / Correlation / Lines of best fit Histograms Averages: Modal class Estimated grouped mean, Cumulative frequency median. – PowerPoint PPT presentation

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Title: Areas to cover


1
Areas to cover
  • Charts
  • Stemleaf / Scatter graphs / Correlation / Lines
    of best fit
  • Histograms
  • Averages
  • Modal class
  • Estimated grouped mean,
  • Cumulative frequency median.
  • Moving averages
  • Measures of spread.
  • IQR, Box Plots
  • Probability
  • Simple theoretical / Experimental / Historical
    methods
  • Combined/Independent events Tree Diagrams
  • Dependent(conditional) Events
  • Sampling
  • Random / Systemic
  • Stratified

G
F
E
D
C
B
A
A
2
Stem and Leaf
  • Remember the key.
  • Draw stem.
  • Add leaves disordered.
  • Draw second stem for ordered leaves.
  • Use ordered stem and leaf to find median and
    range.

D
3
Example - drawing a stem and leaf
Question- 29 students were set a simple
task. Their completion times to the nearest
second were 47 61 53 43 46 46 68 48 72
57 48 54 41 63 49 42 58 65 45 44 43
51 45 38 46 44 52 43 47 (a) set these
data into a stem and leaf diagram (b) find the
median and range
D
4
Example - drawing a stem and leaf
Answer (a)- (3) re draw diagram with leafs in
numerical order
Completion times
3 4 5 6 7
8 7 3 6 6 8 8 1 9 2 5 4 3 5 6 4 3
7 3 7 4 8 1 2 1 8 3 5 2
Dont forget the key
D
Completion times
47 means 47
3 4 5 6 7
8 1 2 3 3 3 4 4 5 5 6 6 6 7 7 8 8
9 1 2 3 4 7 8 1 3 5 8 2
Median 15th value. 47 (remember the 40).
Range Biggest Smallest 72 38 34
5
Scatter Graphs
A good correlation means the points form a line
called a LINE OF BEST FIT.
10,000
As the line slopes down, this is called a
negative correlation. If the line slopped up then
it would be a positive correlation
8,000
6,000
Value of Car
4,000
D
2,000
A scatter graph is a graph with lots of points,
rather than a line or curve.
0
New
1 Yr
2 Yrs
3 Yrs
4 Yrs
5 Yrs
6Yrs
7 Yrs
Age of Car
Estimated value of a 6 yr old car? Use the line
of best fit to find the value 2200
6
Scatter Graphs - Correlation
This is called a zero correlation as there is
little or no correlation
2.5
100
This is called a positive correlation
2
80
1.5
60
Test Score
Height (metres)
1
40
D
0.5
20
0
Born
2 Yrs
4 Yrs
6 Yrs
8 Yrs
10 Yrs
12Yrs
0kg
15kg
30kg
45kg
60kg
Age of Person
Weight
7
Frequency Polygons
A frequency distribution can be shown using a
frequency polygon.
A Frequency Polygon can be drawn onto an existing
histogram.
1. Mark the mid - points of each bar at the top
with a point.
2. Draw in straight lines connecting points.
Extend lines if necessary ½ a class interval
beyond first and last bars
Test Scores
20
15
D
Frequency
10
5
0
25-30
20-24
5-9
10-14
15-19
Marks
8
The same 55 students sat two separate maths
tests. The scores for each are shown by the
frequency polygons below. Comment on the
differences.
It is often useful to show frequency polygons,
side by side, in order to compare distributions.
Test Scores
20
15
Frequency
10
C
5
0
25-30
20-24
5-9
10-14
15-19
Marks
9
Bar Charts vs Histograms
A
10
Histograms - Construction
6
A
11
Histograms - Construction
A
12
Histograms - Construction
AREA
13
Histogram - Example
60
45
10
10
A
14
Modal Group
  • Mode with grouped data this is called the modal
    group or class.

Time (seconds) Frequency
10 lt t 20 1
20 lt t 30 2
30 lt t 40 2
40 lt t 50 9
50 lt t 60 13
60 lt t 70 17
70 lt t 80 3
80 lt t 90 2
90 lt t 100 2
D
Modal group 60 lt t 70
15
Estimated Mean
  • Draw frequency table if necessary.
  • Find mid-point of each group and add these in a
    separate column.
  • Multiply each mid-point by its frequency, and
    add these calculations in another separate
    column.
  • Total the frequency column.
  • Total the mid-point multiplied by frequency
    column.
  • Divide the Mid-point x Frequency Total by the
    Frequency Total. Check that it looks sensible.
    This answer is the Estimated Mean

C
16
Time Taken by 200 Dansteed and Portway students
to run 600 mFind estimated mean for this data.
Time (seconds) Frequency
124 t lt 126 7
126 t lt 128 144
128 t lt 130 49
Mid- point
125
127
129
Mid-point x Frequency
875
18288
6321
C
Total 200 25484
Estimated mean sum of mid-point x freq
total frequencies

25484 127.4 seconds 200
17
Cumulative Frequency Curves
Cumulative frequency table
Example 1. A P.E teacher
records the distance jumped by each of 70 pupils.

Cumulative Frequency
Upper Limit
No of pupils
Distance d (cm)
d ? 190
2
180 ? d ? 190
2
d ? 200
6
190 ? d ? 200
8
17
d ? 210
9
200 ? d ? 210
24
d ? 220
7
210 ? d ? 220
d ? 230
15
220 ? d ? 230
39
B
d ? 240
18
230 ? d ? 240
57
d ? 250
8
240 ? d ? 250
65
70
d ? 260
5
250 ? d ? 260
Cumulative frequency just means running total.
18
Plotting The Curve
IQR 237 212 25 cm
Plot the end point of each interval against
cumulative frequency, then join the points to
make the curve.
B
Find the Lower Quartile.
Get an estimate for the median.
Find the Upper Quartile.
Find the Inter Quartile Range.(IQR UQ - LQ)
19
Remember
  • The method of constructing a cumulative frequency
    graph enables you to find the median, UQ, LQ and
    IQ range
  • The advantage of finding the interquartile range
    is that it eliminates extreme values and bases
    the measure of spread on the middle 50 of the
    data.
  • The cumulative frequency is always the vertical
    (y) axis.
  • To plot the top point of each group against the
    corresponding cumulative frequency

B
20
52
60 24 36
B
21
Box Plot from Cumulative Frequency Curve
B
0
10
20
30
40
50
60
22
Moving Average
  • Moving Averages, when graphed, allow us to see
    any trends in data that are cyclical
  • By calculating the average of 2 or more items in
    the data, any peaks and troughs are smoothed out.

B
23
Year 1996 1996 1996 1996 1997 1997 1997 1997 1998 1998 1998 1998
Quarter 1 2 3 4 1 2 3 4 1 2 3 4
Sales 189 244 365 262 190 266 359 250 201 259 401 265
265
265.25
270.75
269.25
4 Period Moving Average
B
24
Quarters 1-4 2-5 3-6 4-7 5-8 6-9 7-10 8-11 9-12
Moving Average 265 265.25 270.75 269.25 266.25 269 267.25 277.75 281.5
Year 1996 1996 1996 1996 1997 1997 1997 1997 1998 1998 1998 1998
Quarter 1 2 3 4 1 2 3 4 1 2 3 4
Sales 189 244 365 262 190 266 359 250 201 259 401 265
500
x
400
x
x
300
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
200
x
x
x
B
100
1
2
3
4
1
2
3
4
1
2
3
4
1998
1996
1997
25
For example 10 coloured beads in a bag 3 Red,
2 Blue, 5 Green. One taken, colour noted,
returned to bag, then a second taken.
2nd
1st
RR
R
B
RB
R
G
RG
INDEPENDENT EVENTS
R
BR
BB
B
B
B
G
BG
R
GR
G
GB
B
G
GG
26
Probabilities
2nd
1st
RR
R
P(RR) 0.3x0.3 0.09
0.3
0.2
B
RB
P(RB) 0.3x0.2 0.06
R
G
0.3
0.5
RG
P(RG) 0.3x0.5 0.15
R
BR
P(BR) 0.2x0.3 0.06
0.3
0.2
0.2
BB
P(BB) 0.2x0.2 0.04
B
B
0.5
G
BG
P(BG) 0.2x0.5 0.10
R
GR
P(GR) 0.5x0.3 0.15
0.3
0.5
B
G
0.2
GB
P(GB) 0.5x0.2 0.10
B
G
GG
P(GG) 0.5x0.5 0.25
0.5
27
Probabilities
Probability of at least one red.
2nd
1st
RR
R
P(RR) 0.3x0.3 0.09
A D D T O G E T H E R
0.3
0.2
B
RB
P(RB) 0.3x0.2 0.06
R
G
0.3
0.5
RG
P(RG) 0.3x0.5 0.15
R
BR
P(BR) 0.2x0.3 0.06
0.3
0.2
0.2
BB
P(BB) 0.2x0.2 0.04
B
B
0.5
G
BG
P(BG) 0.2x0.5 0.10
R
GR
P(GR) 0.5x0.3 0.15
0.3
0.5
B
G
0.2
GB
P(GB) 0.5x0.2 0.10
B
0.51
G
GG
P(GG) 0.5x0.5 0.25
0.5
28
CONDITIONAL PROBABILITY
Occurs when the probability of one event is
altered by another prior event.
If a card is drawn from a pack and is not
returned, then this will alter the probability
of any subsequent card drawn from the pack.
A
29
CONDITIONAL PROBABILITY
Coloured disks in a bag, 7 green and 3 red. One
is taken at random and not replaced, then a
second is taken.
A
30
CONDITIONAL PROBABILITY
1st event
Red
3 10
Green
7 10
A
31
CONDITIONAL PROBABILITY
2nd event
1st event
2 9
Red
Red
3 10
Green
7 9
Green
7 10
A
32
Sample
  • Key features - a sample must be
  • Random / Systemic
  • Stratified representative of the population
    not skewed by gender, age, etc.
  • Random / systemic and stratified help to minimise
    bias.
  • Advantage
  • Quicker
  • More manageable
  • Disadvantage
  • Conclusions can be unreliable due to size of
    sample
  • Conclusions can be unreliable due to the impact
    of outliers

C
33
Worked Example
You are to complete a survey of opinions
regarding the new Pitstop cafeteria from the
students in Saxon Hall. A summary of the number
of students in the Saxon is included below. A
sample of 50 students (5.9) is taken. Calculate
the number of students to be sampled in each
group?
Year Boys Girls
7 70 64
8 106 77
9 84 93
10 94 101
11 72 80
A
34
Solution
Total number of Boys 426, Girls 415, all
students 841. A sample of 50 represents 5.9 of
the total number of Students. Take 5.9 of each
year group, boys and girls, and round off
appropriately
Year Boys Boys Number in Survey Number in Survey Number in Survey Girls Girls Number in Survey Number in Survey
7 70 70 4 4 4 64 64 4 4
8 106 106 6 6 6 77 77 5 5
9 84 84 5 5 5 93 93 5 6
10 94 94 6 6 6 101 101 6 6
11 72 72 4 4 4 80 80 5 5

A
35
Solution
Total number of Boys 426, Girls 415, all
students 841.
Year Boys of Total Number in Survey Girls of Total Number in Survey
7 70 8.33 4 64 7.62 4
8 106 12.62 6 77 9.17 5
9 84 10.00 5 93 11.07 6
10 94 11.19 6 101 12.02 6
11 72 8.57 4 80 9.52 5

A
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