intermediate 2 – additional question bank unit 2 : graphs, charts & tables you have chosen to...

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

UNIT 2 : Graphs, Charts & Tables

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1 2 3 4

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5 6

Stem& Leaf

Dot Plot

CumFreqTable

Dot toboxplot

Stem toboxplot

Piechart

(a) Use this information to find the (i) median

(ii) lower & upper quartiles

(iii) the semi-interquartile range

(b)What is the probability that someone chosen at random earns less than £180?

The following stem & leaf diagram shows the distribution of wages for employees in a small factory …..

16 2 3 6 9

17 1 1 1 8 8 9

18 2 3 3 5 6 7 7

19 1 2 8

20 1 5 5 6 n = 25

21 8 17 4 = £174

GRAPHS, CHARTS, TABLES : Question 1

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16 2 3 6 9

17 1 1 1 8 8 9

18 2 3 3 5 6 7 7

19 1 2 8

20 1 5 5 6 n = 25

21 8 17 4 = £174


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Graphs etc Menu

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Use median position = (n+1) / 2 to find median

Q1 is midpoint from start to

median

Q3 is midpoint from median to

end

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16 2 3 6 9

17 1 1 1 8 8 9

18 2 3 3 5 6 7 7

19 1 2 8

20 1 5 5 6 n = 25

21 8 17 4 = £174


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median = £183

Q1 = £171

Q3 = £195= £12

= 2/5


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Question 1

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16 2 3 6 917 1 1 1 8 8 918 2 3 3 5 6 7 719 1 2 820 1 5 5 6 n = 2521 8 17 4 = £174

(i) Median(ii) lower & upper quartiles(iii) the semi-interquartile range

1. Use median = (n+1) / 2 to find median

(a)(i) Since n = 25 then the median is

13th value ie median = £183

(ii) Both 6th & 7th values are £171 so Q1 = £171

2. There are 12 values before median so Q1 position = 13 - (12 + 1) / 2

3. There are 12 values after median so Q3 position = 13 + (12 + 1) / 2

19th is £192 & 20th is £198 so Q3 = £195


(NOT 3!!!)

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Question 1

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16 2 3 6 917 1 1 1 8 8 918 2 3 3 5 6 7 719 1 2 820 1 5 5 6 n = 2521 8 17 4 = £174

(i) Median(ii) lower & upper quartiles(iii) the semi-interquartile range

4. Use SIQR = ½ (Q3 – Q1 ) / 2

(iii) SIQR = ½(Q3 – Q1)

= (£195 - £171) 2

= £12

Question 1 5. Use P = no of favourable / no of data

16 2 3 6 917 1 1 1 8 8 918 2 3 3 5 6 7 719 1 2 820 1 5 5 6 n = 2521 8 17 4 = £174

(b)What is the probability that

someone chosen at random

earns less than £180?

No of favourable ( under £180) = 10

No of data = n = 25

(b) Prob(under £180) = 10/25 = 2/5 .

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3. There are 12 values after median so Q3 = 13 + (12 + 1) / 2

19th is £192 & 20th is £198 so Q3 = £195

Median: the middle number in the ordered list. 25 numbers in the list.

1 – 12 13 14 - 25

12 numbers on either side of the median median is the 13th number in order.

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19th is £192 & 20th is £198 so Q3 = £195

To find the upper and lower quartiles deal with the numbers on either side of the median separately.

Q1 12 numbers before median.6 numbers either side of Q1

is midway between the 6th and 7th number.

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19th is £192 & 20th is £198 so Q3 = £195

To find the upper and lower quartiles deal with the numbers on either side of the median separately.

Q3 12 numbers after median.6 numbers either side of Q3

is midway between the 19th and 20th number.

Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows

28 29 29 30 31 30 28 30 29 28

29 30 30 28 28 29 29 29 29 28

a) Illustrate this using a dot plot.

b) What type of distribution does this show?

c) If a bag is chosen at random what is the probability it will be heavier than the modal weight?

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28 29 29 30 31 30 28 30 29 28

29 30 30 28 28 29 29 29 29 28




Establish lowest & highest values and

draw line with scale.

Plot a dot for each piece of data and label

diagram.

For probability use:

P = no of favourable / no of data

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28 29 29 30 31 30 28 30 29 28

29 30 30 28 28 29 29 29 29 28




Tightly clustered

3/10

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Question 2 1. Establish lowest & highest values

and draw line with scale.28 29 29 30 31 30 28 30 29 2831 30 30 28 2829 29 29 29 28

Illustrate this using a dot plot.

26 28 30 32

(a) Lowest = 28 & highest = 31.

Weights in g

2. Plot a dot for each piece of data and label diagram.

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Question 2 3. Make sure you know the possible

descriptions of data.28 29 29 30 31 30 28 30 29 2831 30 30 28 2829 29 29 29 28

26 28 30 32

Weights in g

What type of distribution does

this show?

(b) Tightly clustered distribution.Comments

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Question 2

28 29 29 30 31 30 28 30 29 2831 30 30 28 2829 29 29 29 28

26 28 30 32

Weights in g

4. Use P = no of favourable / no of data

No of favourable ( bigger than 29) = 6

No of data = n = 20

(c) Prob(W > mode) = 6/20 = 3/10 .

If a bag is chosen at random

what is the probability it will

be heavier than the modal

weight?

Mode!

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3. Make sure you know the possible descriptions of data.

26 28 30 32

Weights in g

(b) Tightly clustered distribution.

Other types of distribution:

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Comments

Probability =

Number of favourable outcomes Number of possible outcomes

Weights in g


No of favourable ( bigger than 29) = 6

No of data = n = 20

(c) Prob(W > mode) = 6/20 = 3/10 .

Mode!

26

28

30

32

To calculate simple probabilities:

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Charts, Graphs & Tables : Question 3 The results for a class test were

18 14 16 17 14 16 13 11 13 13 16 14 13 18 15

10 14 17 13 15 15 18 14 17 13 16 10 14 13 17

(a) Construct a cumulative frequency table for this data.

(b) What is the median for this data?

(c) What is the probability that a pupil selected at random

scored under 14?

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18 14 16 17 14 16 13 11 13 13 16 14 13 18 15

10 14 17 13 15 15 18 14 17 13 16 10 14 13 17




scored under 14?

Establish lowest & highest values and

draw table. Complete each row 1 step at a

time, calculating

running total as you go.

For probability use:

P = no of favourable / no of data

Use median = (n+1) / 2 to establish in which row

median lies.

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18 14 16 17 14 16 13 11 13 13 16 14 13 18 15

10 14 17 13 15 15 18 14 17 13 16 10 14 13 17




scored under 14?

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Median = 14

1/3

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Question 3 1. Establish lowest & highest values and draw a table.

(a) Lowest = 10 & highest = 1818 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10

14 13 17

(a)Construct a cumulative frequency table for this data.

Mark Frequency Cum Frequency

10 11 12 13 14 15 16 17 18

2 21 30 3

6 163 194 234 273 30

7 10

2. Complete each row 1 step at a time, calculating running total as you go.

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Question 3 3. Use median = (n+1) / 2 to establish in which row median lies.

18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10

14 13 17


10 11 12 13 14 15 16 17 18

2 21 30 3

6 163 194 234 273 30

7 10

(b) What is the median for

this data?

For 30 values median is between

15th & 16th both of which are in row 14.

Median Mark = 14

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Question 3

18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10

14 13 17


10 11 12 13 14 15 16 17 18

2 21 30 3

6 163 194 234 273 30

7 10(c) What is the probability that a pupil selected at random scored under 14?

No of favourable ( under 14) = 10

No of data = n = 30

(c) Prob(mark<14) = 10/30 = 1/3 .


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Comments

For 30 values median is between

15th & 16th both of which are in row 14.

Median = 14

Mark Freq Cum Freq

10 11 12 13 14 15 16 17 18

2 21 30 3

6 16

4 234 273 30

7 10

3 19

Median:

1 – 15 Q2 16 - 30

Find the mark at which the cumulative frequency first reaches between 15th and 16th number.

Median = 14

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Mark Freq Cum Freq

10 11 12 13 14 15 16 17 18

2 21 30 3

6 16

4 234 273 30

7 10

3 19

No of favourable ( under 14) = 10

No of data = n = 30

(c) Prob(mark<14) = 10/30 = 1/3 .

Probability =

Number of favourable outcomes Number of possible outcomes

To calculate simple probabilities:

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Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes.

48 50 52 54 56 58

(a) Find the (i) median (ii) lower quartile (iii) upper quartile

(b) Construct a boxplot using this data.

(c) In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample?

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48 50 52 54 56 58






median


end

remember bigger SIQR means more

variation (spread) in

data.

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48 50 52 54 56 58




Median = 50

So Q1 = 49

So Q3 = 52

the data is distributed more widely than (or not as clustered as)

the above data

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Question 4

48 50 52 54 56 58

(a) Find the (i) median

(ii) lower quartile

(iii) upper quartile

(a) (i) Sample size = 23

so median position is 12.

ie (23+1)2

Median = 50




(ii) Middle of 1st 11 is position 6.

So Q1 = 49

(iii) Middle of 2nd 11 is position 18.

So Q3 = 52

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Question 4

48 50 52 54 56 58

(b) Construct a boxplot using

this data.

4. Draw number line with scale.

Make sure you note highest & lowest

as well as Q1, Q2, Q3.

48 50 52 54 56 58

(b)Lowest = 48, Q1 = 49, Q2 = 50,

Q3 = 52 & Highest = 58.

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Question 4

48 50 52 54 56 58

5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data.

(c) In a second sample the

semi-interquartile range was

2.5. How does this compare?

(c)For above sample

SIQR = (52 - 49) 2 = 1.5

In a sample where the SIQR is 2.5 the data is distributed more widely than (or not as clustered as) the above data

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ie (23+1)2

Median = 50





So Q1 = 49


So Q3 = 52

23 numbers in the list:

1 - 11 12 13 - 23

Q2

11 numbers on either side of the median

The median:

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Comments



ie (23+1)2

Median = 50





So Q1 = 49


So Q3 = 52

For quartiles:

1 - 5 6 7 - 11 12

Q2Q1

Q3

13 - 17 18 19 - 2312

Q2

Now count through the list until you reach the 6th, 12th,and 18th number in the list.

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Comments

5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data.

(c)For above sample

SIQR = (52 - 49) 2 = 1.5

In a sample where the SIQR is 2.5 the data is distributed more widely than or not as clustered as the above data

The semi-interquartile range is a measure of the range of the “middle” 50%.

S.I.R. = (Q3 - Q1) 12

Remember: when asked to compare data always consider average and spread.

It is a measure of how spread-out and so how “consistent” or “reliable” the data is.

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Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club.

(a) Find the median, lower & upper quartiles for this data.

(b) Use the data to construct a boxplot.

(c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results.

6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3

11 4 = 114kg

60 70 80 90 100 110 120

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6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3

11 4 = 114kg

60 70 80 90 100 110 120


position


median


end

When comparing

two data sets comment on spread and

average

Full solution

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6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3

11 4 = 114kg

60 70 80 90 100 110 120

median = 87

Q1 = 77

Q3 = 99 Full solution

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Question 5 1. Use median = (n+1) / 2 to find

median


between 13th & 14th value ie median = 87

(ii) so Q1 = 77

2. There are 13 values before median so Q1 position is 6th value

3. There are 13 values after median so Q3 position is 20th position

so Q3 = 99

6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3

11 4 = 114kg

(a)Find the median, lower & upper quartiles for this data.

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Question 5

6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3

11 4 = 114kg




(b)Lowest = 60, Q1 = 77, Q2 = 87,

Q3 = 99 & Highest = 123.

60 70 80 90 100 110 120

(b) Use the data to construct

a boxplot.

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Question 5 5. Compare spread and relevant average.

(c) The boxplot below shows

the weight distribution for

these people after several

months.

Compare the two &

comment on the results.

60 70 80 90 100 110 120

(c) Lightest has put on weight –

lowest now 65,

heaviest 3 have lost weight –

highest now 115,

median same but overall

spread of weights has decreased

as Q3-Q1 was 22

but is now only 15.

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(b)Lowest = 60, Q1 = 77, Q2 = 87,

Q3 = 99 & Highest = 123.

60 70 80 90 100 110 120

Box Plot :

Lowest HighestQ1 Q2 Q3

Remember:

To draw a boxplot you needa “five-figure summary”:

five-figure summary

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Charts, Graphs & Tables : Question 6

The pie chart below shows the breakdown of how a sample of

630 people spent their Saturday nights.

clubbing

144°x° theatre

cinemaWatching TV(a) How many people

went clubbing?

(b) If 84 people went to the

theatre then how big is x°?

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clubbing

144°x° theatre


went clubbing?



angle360°

= amount630

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clubbing

144°x° theatre


went clubbing?



= 252

= 48°

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Question 6

clubbing

144°x° theatre

cinemaWatching TV

How many people went clubbing?

1. Set up ratio of angles and sectors and cross multiply.

(a) The angle is 144° so …..

angle360°

= amount630

144°360°

= amount630

360 x amount = 144 x 630

amount = 144 x 630 360

= 252

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Question 6

clubbing

144°x° theatre

cinemaWatching TV


(b) The amount is 84 so …..

angle360°

= amount630

angle360°

= 84630

630 x angle = 360° x 84

angle = 360° x 84 630

= 48°

(b) If 84 people went to the theatre then how big is x°?

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(a) The angle is 144° so …..

angle360°

= amount630

144°360°

= amount630

360 x amount = 144 x 630

amount = 144 x 630 360

= 252

Can also be tackled by using proportion:

Amount = x 630 144360

360

angle at centreamount total sample

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Comments

Can also be tackled by using proportion:

360

angle at centreamount total sample


(b) The amount is 84 so …..

angle360°

= amount630

angle360°

= 84630

630 x angle = 360° x 84

angle = 360° x 84 630

= 48°

84 = x 630 x360

630 x = 84 x 360

x = 84 360 630

x

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End of graphs, charts etc.

intermediate 2 – additional question bank unit 2 : graphs, charts & tables you have chosen to...

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