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Mathematics Teaching Self-evaluation Tools

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Here you can see a summary of the areas in which you are confident and those in which you are less confident; there are some ideas and suggestions which may help you in your professional learning.

Showing all next steps for the selected topic.Click on a question to show more information.

the mode, modal class interval, the mean and the estimated mean of grouped data?(show/hide all)

What this might look like in the classroom

Question
Find the modal class and the mean of the following grouped frequency distribution. Explain why your answer is only an approximation

Group 20−30 30−35 35−40 40−60
Frequency 5 11 7 2

Answer
The modal class is the 30−35 group as this is where the = 13^{th} value will lie.

The midpoints of the given groups are 25, 32.5, 37.5 and 50 respectively

Group

Frequency

Midpoint

Frequency x midpoint

20-30

5

25

125

30-35

11

32.5

357.5

35-40

7

37.5

262.5

40-60

2

50

100

∑f = 25

∑fx = 845

Mean =

The answer is only an approximation because the midpoints are being used to represent the values

Taking this mathematics further

Although it is usual, to find the modal class rather than the mode of a grouped distribution, the following formula gives a usable approximation.

Mode = x h

where

L = lower limit of the modal class
F = frequency of the modal class
F1 = frequency of the class immediately before the modal class
F2 = frequency of the class immediately after the modal class
h = range of the modal class (upper limit - lower limit)

The task might be developed by looking at other measures for the grouped frequency distribution including standard deviation giving a measure of the spread of the data

Making connections

For grouped data, you cannot find the mode so you find the modal class instead. The modal class is the class interval with the greatest frequency.

Statistics is useful for a number of disciplines including biology, economics, geography, psychology, and sociology. See here for further details.

Question
The following information is collected about the time taken (in minutes) to complete a test :

10

6

9.5

11.5

9

8

10.5

14

11

8

10

8.5

8

13

10

9

12.5

9

11

1

11

6

4

7

16.5

10.5

6.5

14

9.5

11.5

8

12

9.5

3.5

8.5

9

12

5

7.5

10

12

11

8

11

10.5

13.5

9

13

10

8

Complete the table below to show this information

Time t (minutes)

Tally

Frequency

0 ≤ t < 5

5 ≤ t < 8

8 ≤ t < 10

10 ≤ t < 12

12 ≤ t < 15

15 ≤ t < 20

Answer

Time t (minutes)

Frequency

Frequency

0 ≤ t < 5

///

3

5 ≤ t < 8

//// /

6

8 ≤ t < 10

//////////// /

16

10 ≤ t < 12

////////////

15

12 ≤ t < 15

//// ////

9

15 ≤ t < 20

/

1

Tallies are useful ways to collate the information but care must be taken to ensure that the tally is placed in the correct class especially where the class intervals include inequalities

Taking this mathematics further

The task might be developed by undertaking some analyses of the grouped frequency table including finding the mean and standard deviation of the data.

Making connections

To construct a grouped frequency distribution, the data should be sorted into groups called classes.

It is usual to collect the data into 5 to 10 classes and the frequency of the data belonging to each class is recorded.

There are 5 rules for the classes in a grouped frequency distribution.

the difference between a frequency diagram, a histogram and a frequency polygon?(show/hide all)

What this might look like in the classroom

Question 1 The weights of cereal in a box are recorded as follows:

Weight g (grams)

Frequency

245 ≤ g < 248

3

248 ≤ g < 249

4

249 ≤ g < 250

10

250 ≤ g < 252

15

252 ≤ g < 255

6

255 ≤ g < 260

2

Show this information as a

histogram

frequency polygon

Answer 1

Since the class intervals are unequal, it is necessary to calculate the frequency density

Weight g (grams)

Frequency

Frequency density

245 ≤ g < 248

3

3 ÷ 3 = 1.0

248 ≤ g < 249

4

4 ÷ 1 = 4.0

249 ≤ g < 250

10

10 ÷ 1 = 10.0

250 ≤ g < 252

15

15 ÷ 2 = 7.5

252 ≤ g < 255

6

6 ÷ 3 = 2.0

255 ≤ g < 260

2

2 ÷ 5 = 0.4

The histogram will look like this

The frequency polygon is found by joining the midpoints of the top of each bar with a series of straight lines to form a polygon.

Without the histogram the frequency polygon will look like this

Question 2

The histogram shows the marks of students in a college

How many students are there in the college?

Find the median mark

Find the interquartile range

Answer 2

Using the key, there are 200 students in the college.

For the median mark there must be 100 students above the line and 100 students below the line. The median is 40 marks.

For the quartiles, these should be plotted to divide the data into quarters.

The lower quartile is 40 and the upper quartile is 78 (actually 78.3333..)

The interquartile range is 78 – 40 = 38

Taking this mathematics further

The task might be developed by looking at the median and interquartile ranges of more than one data set, so being able to compare two different sets of data.

Making connections

A histogram is used to represent grouped continuous data. A histogram is similar to a bar chart except that it is the area of the bars which gives the frequency rather than the height of the bars.

In the case of a histogram with equal class intervals, then the heights of the bars is indicative of the frequency although the vertical axis will often be identified as frequency density rather than frequency.

A frequency polygon is similar to a histogram. A frequency polygon can be drawn from a histogram by joining the mid points of the top of each bar with a series of straight lines to form a polygon.

Since the area under the graph represents the frequency then the median divides this area into two. Similarly, the quartiles divide the data into four equal parts

Question
The cumulative frequency diagram shows the distribution of marks for 100 students in a mathematics examination

(a) Estimate
(i) the median
(ii) the interquartile range

(b) The pass mark for the examination is 62 marks. How many students pass the examination

(c) The lowest mark recorded was 13. The highest mark recorded was 96

Use this information to draw a box plot of the distribution of marks.

Answer
Using the graph

Please create these diagrams so that the blue lines ‘change direction’ at exactly the point that they touch the cf curve

(a) (i) the median = 54
(ii) the interquartile range = 68 − 39 = 29

(b) The cumulative frequency for 62 marks = 66
So 100−66 = 34 students pass

(c)

Taking this mathematics further

Cumulative frequency graphs can be used to construct box plots to compare the shapes of distributions

Making connections

The cumulative frequency graph is created by calculating the ‘accumulated’ totals for the frequency. The cumulative frequency should be plotted at the upper boundary of the class interval (never in the centre which is a common error).

The cumulative frequency table for the graph in this question looks like this:

Interval

Frequency

Cumulative frequency

0-10

0

0

11-20

5

0 + 5 = 5

21-30

8

5 + 8 = 13

31-40

14

13 + 14 = 27

41-50

16

27 + 16 = 43

51-60

20

43 + 20 = 63

61-70

16

63 + 16 = 79

71-80

12

79 + 12 = 91

81-90

8

91 + 8 = 99

91-100

1

99 + 1 = 100

The shape of the cumulative frequency graph is always similar so you can use this fact to check you have not made any errors.

Cumulative frequency graphs are used to calculate the median and interquartile of a distribution although these can sometimes be found using the original data

Cumulative frequencies are most usually used for continuous and/or or grouped data. Increasingly, discrete data is being shown using cumulative frequency polygons where the points are joined with straight lines.

a scatter graph, a line of best fit and correlation?(show/hide all)

What this might look like in the classroom

Question
Draw a scatter graph for the following table which shows the heights and shoe sizes of 10 students in a class.

Shoe size

3

6

8

5

3

2

5

1

4

7

Height

138

151

154

135

122

109

144

110

142

151

Use the graph to draw a line of best fit and write down what this says about the relationship between shoe size and height

Answer
Plotting this information on a scatter graph gives

Scatter diagram of height and shoe size

The line of best fit should be drawn to give a best fit for the data as follows

The line of best fit suggests a positive correlation between shoe size and height (for this data). That means that taller people have larger shoe sizes − in general.

Taking this mathematics further

All the cases mentioned here involve linear correlation. It is possible to have non−linear correlation, but such cases are not usually considered at KS4.

Find out about more sophisticated measures of correlation including Spearman rank correlation and Pearson’s Product Moment correlation.

Making connections

Scatter graphs can be used to show whether there is any relationship or correlation between the two variables.

Where an increase in one variable is associated with an increase in the other variable, then the correlation is positive or direct.

Where an increase in one variable is associated with a decrease in the other variable, then the correlation is negative or inverse.

When the points on a scatter diagram show some correlation, a line of best fit can be drawn through, or as close to, as many of the points as possible, to approximate the relationship.

The line of best fit (or regression line) can be used to predict other values from the given data. Care should always be taken in predicting values outside the available range.

The following scatter graph demonstrates strong positive correlation

The following scatter graph demonstrates negative correlation

The following scatter graph demonstrates no correlation

explain how to calculate a moving average?(show/hide all)

What this might look like in the classroom

Question 1
A hotel recoded the number of bookings it took each quarter over a two year period.
The figure for the first quarter of 2008 is missing.

2007

2008

Quarter

1^{st}

2^{nd}

3^{rd}

4^{th}

1^{st}

2^{nd}

3^{rd}

4^{th}

Number of bookings

26

53

41

16

63

49

22

Why is a four point moving average most appropriate?

Show that the value of the first four point moving average is 34

The value of the second four point moving average is 35. Calculate the number of bookings in the first quarter of 2008

Answer 1

(a) A four point moving average is appropriate for this data as the data is collected quarterly

(b) First four point moving average = 26 + 53 + 41 + 16 = 34
4

(c) Second four point moving average = 53 + 41 + 16 + x = 35
4

So

110 + x = 140
x = 30

Question 2

The table shows the costs of gas bills between December 2007 and March 2009.

Date

Dec07

Mar08

Jun08

Sep08

Dec08

Mar09

Cost

133.50

127.50

109.70

116.30

147.50

137.50

(a) Calculate the value of the three four-point moving averages

(b) Show this information on a graph and explain what the trend line tells you about the price of the gas bills

Answer 2

First four point moving average

= 133.50 + 127.50 + 109.70 + 116.30
4

= 467
4

= (£)116.75

Second four point moving average

= 127.50 + 109.70 + 116.30 + 147.50
4

= 501
4

= (£)125.25

Third four point moving average

= 109.70 + 116.30 + 147.50 + 137.50
4

= 511
4

= (£)127.75

Showing this information on a graph:

Remember that the four point moving averages are plotted at the centre of the group of four points (i.e. between the second and third point.)

Taking this mathematics further

Find out about more sophisticated moving averages, including rolling averages that can be calculated for a range of different circumstances

Making connections

A moving average is used to ‘smooth’ out the fluctuations in a time series. For example, a four point moving average is found by averaging successive groups of four data readings

Moving averages are particularly useful in stock markets to spot trends as demonstrated here

The relative frequency of the dice showing the numbers 1 to 4 are given in the table below

The probability of throwing a five is twice that of throwing a four.

(a) Use this fact to complete the table

Jenny is going to throw the dice 500 times.

(b) About how many times would she expect the dice to show a three?

John has an unbiased six-sided dice.

(c) Who is most likely to throw a four? Give a reason for your answer

Answer

(a)

Number on dice

1

2

3

4

5

6

Relative frequency

0.14

0.18

0.13

0.10

0.20

0.25

The sum of the relative frequencies must add up to one as the events are mutually exclusive.

(b) In 500 throws, Jenny would expect 500 x 0.13 threes; i.e. 65 threes

(c) The probability that Jenny throws a four is 0.10. The probability that John throws a four is

1

6

= 0.16666…So John is more likely to throw a four.

Taking this mathematics further

Research the history of probability – a comparatively recent development in mathematics that began in earnest with the correspondence between the mathematicians Blaise Pascal and Pierre de Fermat in the 1600’s. While its early origins can be shown to be linked to gambling, it is invaluable in the modern world in terms of its links, among many others, to medicine, insurance and quality control.

Further work on probability including the development of probability theory. See here.

Making connections

Events are mutually exclusivewhen they cannot happen at the same time.

The sum of the probabilities of all the mutually exclusive events is 1.

So if the probability of something happening is p, then the probability of it not happening is equal to

1 – p.

If event A and event B are mutually exclusive, then P(A or B) = P(A) + P(B)

Question
Peter, Paul and Mary are taking their entrance examinations

The probability that Peter will pass is 0.7
The probability that Paul will pass is 0.4
The probability that Mary will pass is 0.9

What is the probability that all three of them will pass?

What is the probability that Peter and Mary will pass and Paul will fail?

What is the probability that all three of them will fail?

Answer The events are independent, and the probability of something happening is equal to 1 minus the probability that it does not happen. Therefore:

0.7 × 0.4 × 0.9 = 0.252

0.7 × 0.9 × 0.6 = 0.378

0.3 × 0.6 × 0.1 = 0.01

Taking this mathematics further

While developing the early theory of probability, Blaise Pascal used the (previously known) triangular array of numbers which has since become known as Pascal’s Triangle. It is crammed full of intriguing connections and patterns − but how is it applied within probability, and how can it tell you the probability of winning the lottery?

Further work on probability including the development of probability theory. See here.

Making connections

Events are dependent when the outcome of one event affects the outcome of the other.

Events are independent when the outcome of one event does not affect the outcome of the other.

Learners will need to be clear about the appropriate use of multiplication and addition of probabilities. If event A and event B are mutually exclusive, then P(A or B) = P(A) + P(B).

Multiplication of probabilities needs a bit more care though. It is true that if event A and event B are independent, then P(A and B) = P(A) × P(B). But if a tree diagram is set up correctly for the (rather traditional) case of drawing two counters from a bag without replacement, then it is still appropriate to multiply probabilities ‘along the branches’ even though the events are not independent. Can you explain why? See the ‘conditional probability’ section for more information.

Question 1
A bag contains five blue and three red discs. A disc is drawn from the bag and replaced, then a second disc is drawn from the bag.

Draw a tree diagram to show the various possibilities that can occur.

Use your tree diagram to calculate the probability of the two discs being different colours giving your answer in its lowest terms

Answer 1
(a) The following tree diagram shows the various possibilities that can occur. After the first draw, the disc is replaced.

(b) The probability of the two discs being different colours is the same as the probability of a blue disc followed by a red disc or a red disc followed by a blue disc

=

5

8

x

3

8

+

3

8

x

5

8

=

15

64

+

15

64

=

30

64

=

15

32

Question 2
A bag contains five blue and three red discs. A disc is drawn from the bag and then a second disc is drawn from the bag.

Draw a tree diagram to show the various possibilities that can occur.

Use your tree diagram to calculate the probability of the two discs being different colours giving your answer in its lowest terms

Answer 2

The following tree diagram shows the various possibilities that can occur. After the first draw, the disc is not replaced so this affects the probability of the second disc.

The probability of the two discs being different colours is the same as the probability of a blue disc followed by a red disc or a red disc followed by a blue disc

=

5

8

x

3

8

+

3

8

x

5

8

=

15

56

+

15

56

=

30

56

=

15

28

Taking this mathematics further

This is only the beginnings of conditional probability − at KS5 this idea is taken further, and includes finding a probability law for the case of conditional probability. Find out what the probability laws are.

Making connections

Sampling with replacement means that you replace the item selected after each draw. The two sample values are independent which means that what you get on the first draw does not affect what you get on the second draw.

Sampling without replacement means that you do not replace the item selected after each draw. The two sample values are not independent which means that what you got on the first draw affects what you can get on the second draw.

If the sample is very small relative to the population, then sampling with replacement is virtually the same as sampling without replacement.

Question
Ann and Eve play each other in a tennis match. The match consists of three sets. The winner of the match is the first player to win two sets.

The tree diagram show all the possible outcomes

The probability that Ann wins the first game is 0.5. If Ann wins a game, the probability that she wins the next game is 0.6. If Eve wins a game, the probability that she wins the next game is 0.7.

Complete the tree diagram

What is the probability that Ann wins the tennis match

Answer (a)

(b) The probability that Ann wins the tennis match
= p(AA) or P(AEA) or p(EAA)
= 0.5 x 0.6 + 0.5 x 0.4 x 0.3 + 0.5 x 0.3 x 0.6
= 0.30 + 0.06 + 0.09
= 0.45

Taking this mathematics further

Find out about game trees and decision trees. You might want to start here.

Making connections

A tree diagram may be used in early work on probability as a systematic approach to finding all the outcomes for an experiment. For example, to find all the outcomes for flipping two coins, learners could use a list, a sample space diagram or a tree diagram. If the problem involves flipping three or more coins a list becomes more difficult to keep track of, a sample space diagram doesn’t work (in the conventional sense), but a tree diagram is still useful.

A tree diagram is a useful tool for calculating probabilities. The probabilities are written on the branches of the tree and the probabilities on the branches should add up to 1

Encourage learners to find the probability of each route through the tree diagram: not just the one needed to solve a problem. If they leave any fractions un−simplified it is then a straightforward check to ensure that the sum of the probabilities is 1.

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