© christine crisp “teach a level maths” statistics 1

© Christine Crisp

““Teach A Level Teach A Level Maths”Maths”

Statistics 1Statistics 1

Introduction to S1

You met some statistical diagrams when you did GCSE.

The next three presentations and this one remind you of them and point out some details that you may not have met before.

We will start with stem and leaf diagrams ( including back-to-back ).

Stem and leaf diagrams are sometimes called stem plots.

Introduction to S1

Weekly hours of 30 men

5

4

4

3

3

2

2

5 1

4 5 6

4 1 1 2

3 5 5 5 5 5 5 6 6

3 0 0 1 1 2 2 2 2 3 3 3 4 4 4

2 8

2 1

e.g. The table below gives the number of hours worked in a particular week by a sample of 30 men 35 41 33 31 30 45 35 36 51 32

32 30 28 35 34 33 35 36 32 42

33 31 41 21 34 35 34 46 32 35

The stem shows the tens . . .

I’ll use intervals of 5 hours to draw the diagram i.e. 20-25, 26-30 etc.

and the leaves the units

e.g. 46 is 4 tens and 6 units

Introduction to S1


e.g. The table below gives the number of hours worked in a particular week by a sample of 30 men

5 1

4 5 6

4 1 1 2

3 5 5 5 5 5 5 6 6

3 0 0 1 1 2 2 2 2 3 3 3 4 4 4

2 8

2 1


35 41 33 31 30 45 35 36 51 32

32 30 28 35 34 33 35 36 32 42

33 31 41 21 34 35 34 46 32 35


Weekly hours of 30 menThe stem shows the

tens . . . and the leaves the units

Introduction to S1



5 1

4 5 6

4 1 1 2

3 5 5 5 5 5 5 6 6

3 0 0 1 1 2 2 2 2 3 3 3 4 4 4

2 8

2 1


35 41 33 31 30 45 35 36 51 32

32 30 28 35 34 33 35 36 32 42

33 31 41 21 34 35 34 46 32 35



N.B. 35 goes here . . . not in the line below.

The stem shows the tens . . . and the leaves the units

Introduction to S1


5 1

4 5 6

4 1 1 2

3 5 5 5 5 5 5 6 6

3 0 0 1 1 2 2 2 2 3 3 3 4 4 4

2 8

2 1


35 41 33 31 30 45 35 36 51 32

32 30 28 35 34 33 35 36 32 42

33 31 41 21 34 35 34 46 32 35

e.g. 46 is 4 tens and 6 unitsWe must show a key.

Key: 3 5 means 35 hours

Weekly hours of 30 menThe stem shows the

tens . . . and the leaves the units

Introduction to S1

5 1

4 5 6

4 1 1 2

3 5 5 5 5 5 5 6 6

3 0 0 1 1 2 2 2 2 3 3 3 4 4 4

2 8

2 1


If you tip your head to the right and look at the diagram you can see it is just a bar chart with more detail.Points to

notice:• The leaves are in numerical order• The diagram uses raw ( not grouped )

data

Key: 3 5 means 35 hours

Introduction to S1

The data below is a back to back stem and leaf diagram giving the weight in grams of eggs collected from ostriches and emus. This method can be used to compare two sets of data.

Ostrich Emu8 3 1 0 27 2 4 87 6 2 1 28 2 4 6 7 9

4 1 29 0 3 51 0 30 7

Key 27|2 = 272

Introduction to S1

0–4 0 25–9 0 5 5 8

10–14 1 215–19 1 6 7 7 920–24 2 0 2 2 425–29 2 5 5 5 7 7 9 9 30–34 3 35–39 3 6

A grouped data stem and leaf diagramData2, 5, 5, 8, 12, 16, 17, 17, 19, 20, 22, 22, 24, 25, 25, 25, 27, 27, 29, 29, 36

Draw a stem and leaf diagram using groupings 0–4, 5–9, 10–14 etc

Key 1/2 = 12 3/6 = 36

Introduction to S1

Histogram : A bar chart with continuous data. The bars are drawn up to the class boundaries. NO GAPS between bars. The class boundary occurs halfway between the boundaries of two successive groups. (Except in age questions)Groups 0-9 , 10-19 , 20-29 etc. the class boundaries between each group occur at 9.5 , 19.5

So any quantity >9.5 is in group 2 and any quantity <9.5 is in group 1 . The bars are drawn at 9.5 and 19.5 etc.

It is very important that the area under each bar is proportional to the frequency.

Introduction to S1

Suppose the data are grouped so that those below 20 and above 69 are combined.

e.g. The projected population of the U.K. for 2005 ( by age )

Source: USA IDB

090+

280 – 89

470 – 79

660 – 69

850 – 59

940 – 49

930 – 39

720 – 29

810 – 19

70 – 9

(millions)

( years )

FreqAGE

Histograms

Introduction to S1


Source: USA IDB

090+

280 – 89

470 – 79

660 – 69

850 – 59

940 – 49

930 – 39

720 – 29

810 – 19

70 – 9

(millions)

( years )

FreqAGESuppose the data are grouped so that those below 20 and above 69 are combined.

670+660 - 69850 - 59940 - 49930 - 39720 - 29

150 - 19

AGE(years)

Freq(million

s)

To draw the diagram we must have an upper class

value

Introduction to S1


Source: USA IDB

Suppose the data are grouped so that those below 20 and above 69 are combined.

I chose a sensible figure

670 - 109660 - 69850 - 59940 - 49930 - 39720 - 29

150 - 19

Freq(million

s)

AGE(years)

Source: USA IDB

090+

280 – 89

470 – 79

660 – 69

850 – 59

940 – 49

930 – 39

720 – 29

810 – 19

70 – 9

(millions)

( years )

FreqAGE

Introduction to S1e.g. The projected population of the U.K. for 2005 ( by

age )

670 - 109660 - 69850 - 59940 - 49930 - 39720 - 29

150 - 19

Freq(million

s)

AGE(years)

-2

2

4

6

8

10

12

14

16

18

10 20 30 40 50 60 70 80 90 1000X->

|̂Y

If we use the data below to draw an age/frequency graph then it is very misleading as the 1st and last bar dominate

So frequencies are represented by

areas

Bar1 1 should represent just over twice as many people as bar 2 but it appears to be about 4 times as many

Introduction to S1

A histogram shows frequencies as areas. To draw the histogram, we need to find the width and height of each column.The width is the class width: upper class boundary (u.c.b.) minus lower class boundary (l.c.b.).

670 - 109660 - 69850 - 59940 - 49930 - 39720 - 29

150 - 19

Freq(million

s)

AGE(years)

Classwidth

20

Since these are ages, the 1st class, for example,has u.c.b.= 20 and the l.c.b.= 0, so the width is 20.

Introduction to S1

A histogram shows frequencies as areas. e.g. The projected population of the U.K. for 2005 ( by

age )

height= frequencywidth

The width is the class width: upper class boundary (u.c.b.) minus lower class boundary (l.c.b.).

Area of a rectangle = width height

To draw the histogram, we need to find the width and height of each column.

So, frequency= width height

670 - 109660 - 69850 - 59940 - 49930 - 39720 - 29

20150 - 19

Classwidth

Freq(million

s)

AGE(years)

401010101010

Introduction to S1

40670 - 10910660 - 6910850 - 5910940 - 4910930 - 3910720 - 2920150 - 19

Classwidth

Freq(million

s)

AGE(years)

Freqdensity


age )

The height is called the frequency

density


e.g. For the 1st class,freq. density =


height= frequencywidth

Introduction to S1

40670 - 10910660 - 6910850 - 5910940 - 4910930 - 3910720 - 2920150 - 19

Freqdensity

Classwidth

Freq(million

s)

AGE(years)


age )


density


e.g. For the 1st class,freq. density =


75020

15

height= frequencywidth 0 ·75

Introduction to S1

40670 - 10910660 - 6910850 - 5910940 - 4910930 - 3910720 - 2920150 - 19

Freqdensity

Classwidth

Freq(million

s)

AGE(years)


age )


We can now draw the histogram.



density

height= frequencywidth 0 ·75

0 ·150 ·60 ·80 ·90 ·90 ·7

Introduction to S1

AGE(years)

Freq(million

s)

Classwidth

Freqdensity

0 - 19 15 20 0 ·7520 - 29 7 10 0 ·730 - 39 9 10 0 ·940 - 49 9 10 0 ·950 - 59 8 10 0 ·860 - 69 6 10 0 ·6

70 - 109 6 40 0 ·15

The projected population of the U.K. for 2005 ( by age )

Notice that the frequencies for the last 2 classes are the same. On the histogram the areas showing these classes are the same.If we had plotted frequency on the y-axis, the diagram would be very misleading. ( It would suggest there are 6 million in each age group 70 – 79, 80 – 89, 90 – 99 and 100 – 109. )

Introduction to S1

SUMMARY

Frequency is shown by area.

The y-axis is used for frequency density.

Histograms are used to display grouped frequency data.

Class width is given by

u.c.b. – l.c.b.where, u.c.b. is upper class boundary

andl.c.b. is lower class boundary

frequency density = width class

frequency

waterclear

float ducksfluffy

Introduction to S1Exercise95 components are tested until they fail. The table gives the times taken ( hours ) until failure.

Time to failure (hours)

0-19 20-29 30-39 40-44 45-49 50-59 60-89

Number of components

5 8 16 22 18 16 10

Find 3 things wrong with the histogram which represents the data in the table.

Introduction to S1Answer:

Time to failure (hours)

0-19 20-29 30-39 40-44 45-49 50-59 60-89

Number of components

5 8 16 22 18 16 10

• Frequency has been plotted instead of frequency density.

• There is no title.

• There are no units on the x-axis.

Introduction to S1

Time taken for 95 components to fail

Incorrect diagram

Correct diagram

Introduction to S1

Length of millipede Class boundaries

Frequency Class width Freq. Density

0 – 9 0 – 9.5 6 9.5 0.6310 – 19 9.5 – 19.5 18 10 1.820 – 39 19.5 – 39.5 14 20 0.7

Note Bars drawn at 9.5, 19.5 and 39.Freqdensity

length

Histogram showing length of millipede

Introduction to S1

Source: USA IDB

60090+

60280 – 89

58470 – 79

54660 – 69

48850 – 59

40940 – 49

31930 – 39

22720 – 29

15810 – 19

770 – 9

(millions)

(millions)

( years )

Cu.FFreqAGE

ANS: The data are given to the nearest million. The projected figure was 113,000.

Why does this appear as 0?

In drawing the diagram I shall miss out this group.

e.g. The projected population of the U.K. for 2005, by age:

Cumulative Frequency Graphs

Introduction to S1

Source: USA IDB

60280 – 89

58470 – 79

54660 – 69

48850 – 59

40940 – 49

31930 – 39

22720 – 29

15810 – 19

770 – 9

(millions)

(millions)

( years )

Cu.FFreqAGE

• Points are plotted at upper class boundaries (u.c.bs.)

Points to notice:

e.g. the u.c.b. for 0 9 would normally be 9·5

• There is no gap between 9 and 10 as the data are continuous.


Introduction to S1

Source: USA IDB

60280 – 89

58470 – 79

54660 – 69

48850 – 59

40940 – 49

31930 – 39

22720 – 29

15810 – 19

770 – 9

(millions)

(millions)

( years )

Cu.FFreqAGE Points to notice:

e.g. the u.c.b. for 0 9 would normally be 9·5 Age data have different

u.c.bs. Can you say why this is?

ANS: If I ask children their ages, they reply 9 even if they are nearly 10, so, the 0-9 group contains children right up to age 10 NOT just nine and a half.




Introduction to S1


Points to notice:

The u.c.bs. for this data set are 10, 20, 30, . . .Source: USA IDB

60280 – 89

58470 – 79

54660 – 69

48850 – 59

40940 – 49

31930 – 39

22720 – 29

15810 – 19

770 – 9

(millions)

(millions)

( years )

Cu.FFreqAGE

e.g. the u.c.b. for 0 9 would normally be 9·5



Introduction to S1


Age (yrs)

The median age is estimated as the age corresponding to a cumulative frequency of 30 million.The median age is 39

years( Half the population of the U.K. will be over 39 in 2005. )


Source: USA IDB

9060280 – 89

8058470 – 79

7054660 – 69

6048850 – 59

5040940 – 49

4031930 – 39

3022720 – 29

2015810 – 19

10770 – 9

( yrs )

(m)(m)( yrs )

u.c.b.

Cu.f

fAGE

Introduction to S1


Age (yrs)

The quartiles are found similarly:

lower quartile: 20 yearsupper quartile: 56 years



Source: USA IDB

9060280 – 89

8058470 – 79

7054660 – 69

6048850 – 59

5040940 – 49

4031930 – 39

3022720 – 29

2015810 – 19

10770 – 9

( yrs )

(m)(m)( yrs )

u.c.b.

Cu.f

fAGE

The interquartile range is 36 years

LQ = (n+1)th item of data

UQ = (n+1)th item of data

Introduction to S1


Age (yrs)

If the retirement age were to be 65 for everyone, how many people would be retired?

ANS: ( 60 – 51 ) million= 9 million


51

Source: USA IDB

9060280 – 89

8058470 – 79

7054660 – 69

6048850 – 59

5040940 – 49

4031930 – 39

3022720 – 29

2015810 – 19

10770 – 9

( yrs )

(m)(m)( yrs )

u.c.b.

Cu.f

fAGE

Introduction to S1ExerciseThe table and diagram show the number of flowers in a sample of 43 antirrhinum plants.

431160-179

421140-159

411120-139

405100-119

35780-99

281260-79

161040-59

6620-39

Cu.ffx

Source: O.N.Bishop

Number of flowers on antirrhinum plants

Estimate the median number of plants and the percentage of plants that have more than 90 flowers.

Introduction to S1


431160-179

421140-159

411120-139

405100-119

35780-99

281260-79

161040-59

6620-39

Cu.ffx

The u.c.bs. ( where we plot the points ) are at 39·5, 59·5 etc.

Solution:

Number with more than 90 flowers =

There are 43 observations, so the median is given by the 21·5th one.Median = 70

32

Introduction to S1


431160-179

421140-159

411120-139

405100-119

35780-99

281260-79

161040-59

6620-39

Cu.ffx

Solution:

32

There are 43 observations, so the median is given by the 21·5th one.

Number with more than 90 flowers =

Median = 70

Percentage with more than 90 flowers

43 – 32 = 11

43

11 26%

Introduction to S1

© christine crisp “teach a level maths” statistics 1

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