Calculating Averages from a Grouped Frequency Table

Example 1

Solution

a)

b)

c) We use interpolation:

Example 2

Solution

a)

b) Again, we use interpolation:

c)

Calculating Quartiles

When we have extreme values or outliers in a set of data, the range and mean are affected. It is

useful to ‘cut out’ these values by looking at the middle 50% of data and working out the

interquartile range (IQR).

For example:

10 58 59 60 61 68 69 69 70 71 72 73 74 76 77 77 78

Median Q2 = Upper Quartile Q3 =

Lower Quartile Q1 =

We then work out the Interquartile Range (IQR):

Q3 - Q1 =

When we are calculating quartiles from large sets of data (i.e., from frequency tables), we just use:

Lower Quartile Q1 =

Median Q2 =

Upper Quartile Q3 =

Example 1

Lower Quartile Q1 =

Median Q2 =

Upper Quartile Q3 =

IQR =

When we are looking at grouped frequency tables use interpolation, again, to calculate an

estimate of the interquartile range.

Example 2

Solution

Lower Quartile Q1 =

Interpolation:

Upper Quartile Q3 =

Interpolation:

IQR = Q3 - Q1

=

Identifying Outliers

An outlier is an extreme value that lies outside the overall pattern of the data.

The following rule is used to identify outliers:

Greater than Q3 + 1.5 x IQR

Less than Q1 - 1.5 x IQR

Example

Solution

a)

b)

Box Plots (Box and Whisker Diagrams)

Without a box plot – How to Tell If The Data is Skewed

Example

Solution

a)

b)

c)

d)

Example

Example 2

Histograms

Example 1

Solution

a)

Time (mins)

Frequency (F)

Class width (CW)

FD (F/CW)

25 - 55

30 - 39

35 - 68

40 - 32

50 – 80 6

b)

Example 2

Solution

a)

b)

c)

d)

Scatter Graphs and Correlation

Scatter graphs show any relation between two variables.

As one variable increases, the other increases.

Eg,

As one variable increases, the other decreases.

Eg,

No relationship between variables.

Eg,

Correlation can be given as a value between -1 and 1.

This is called the product moment correlation coefficient (r)

(-1 is perfect negative, 0 is no correlation and 1 is perfect positive)

This value, r, is found by using the following formulae:

Where For variables x and y

Example 1

A gardener wishes to know if plants which produce only a few potatoes produce larger ones.

He selects five plants at random, sieves out the small ones, counts those remaining

and weighs the largest one.

Number of potatoes, x 5 5 7 8 10

Weight of the largest, y (grams) 240 232 227 222 215

a) Calculate the correlation coefficient, r, between x and y.

b) Comment on the result.

Solution

a) n = 5

x y x2 y2 xy

5 240

5 232

7 227

8 222

10 215

b)

Calculating Mean, SD, etc, on Calculator (Casio)

You will probably be given a list or table of values, from which you will be asked to calculate

mean, SD.

For a list of values

Eg. Work out the mean and standard deviation for these values

120 125 134 156 159 173 184 190

Solution

On calculator:

Firstly, to clear your calculator: SHIFT

9

3

=

AC

Next, enter the data: MODE

2:STAT

1:1-VAR

SHIFT

SETUP (then scroll down)

3:STAT

Frequency? 2:OFF

MODE

2:STAT

1:1-VAR

You will see something like this:

X

1

2

3

Now enter data: key in each number, followed by = after each one.

ie, 120 = 125 = …

Store data: Press AC button

Access statistical values (mean, SD, etc.): SHIFT

1 (says STAT above it)

5:VAR

Now…

Pressing 1:n = will give you the number of values you entered (8)

Pressing 2:�̅� = will give you the mean (155.125)

Pressing 4:𝑥𝜎𝑛-1 = will give you the sample standard deviation (26.654)

(Ignore 3: 𝑥𝜎𝑛…this is for a population)

Press the following after one value is read, to access the next one:

SHIFT

1 (says STAT above it)

5:VAR

For a Table of Values

Work out the mean and standard deviation for the following information:

x f

18 3

19 5

20 9

21 2

Solution

Apply same method as above only at the frequency? 1:ON 2:OFF stage

press 1:ON

Your table should look like this:

X FREQ

1

2

3

Enter X column first as before, then scroll across to FREQ column and enter same way. (FREQ

column automatically entered as 1 until you edit it) Remember to press your AC button after to

store the data.

Answers you should get:

n = 19

mean = 19.526

SD = 0.905

*** If you are given grouped data, enter the mid points into column X ***

Calculating Mean, SD, etc, on Calculator (Casio fx-991EX)

Taken from the link:

https://www.youtube.com/watch?v=Y1gsxlSbojs

Finding values from data from a table or a list will be exactly the same method, only, at the

FREQUENCY? 1:ON 2:OFF stage, press 1 for tabled data and 2 for list.

To input data into calculator, press: Shift

Setup

(Scroll down) 3:Statistics

Frequency? 1:on for table

2:off for list

MENU

6:Statistics (or scroll to the bar chart symbol)

1:Variable

Now enter data followed by = after each data entry

Press AC now to store this data!

Your calculator display will now say Statistics

1:Variable

To access values like mean, SD etc, press: OPTN

3:Data

OPTN

3:1-Variable

All values should now be available!

A B

A B 1

2 5 4

3

6

Probability

Notation used:

𝑃(𝐴 ∪ 𝐵) means probability of event A OR B happening (sometimes called the union of A and B)

𝑃(𝐴 ∩ 𝐵) means probability of BOTH events A AND B happening

Venn Diagrams to Illustrate Probability

The ‘OR’ Rule

Mutually Exclusive Events A and B

𝑷(𝑨 𝒐𝒓 𝑩) = 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩)

𝑷(𝑨 𝒐𝒓 𝑩) = 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩)

= 3

6+

1

6=

4

6=

2

3

Non – Mutually Exclusive Events A or B

A and B can happen at the same time. So there is an overlap for both.

A B

A and B cannot happen at the same time. So there is no overlap.

Example 1

What is the probability of getting a 5 or an even number on a dice?

A Even no. on a dice

B 5 on a dice

Understanding parts of the venn diagram…

𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝑨 ∩ 𝑩)

𝑷(𝑨 𝒐𝒓 𝑩) = 𝑷(𝑨 ∪ 𝑩)

This is called the ‘union’ of A and B

𝑷(𝑵𝒐𝒕 𝑨) = 𝑷(𝑨′)

𝑷(𝑶𝒏𝒍𝒚 𝑨) = 𝑷(𝑨) − 𝑷(𝑨 ∩ 𝑩)

The AND rule for independent even

Example 2

What is the probability of rolling a dice and getting an even number or a number greater than 4?

If A Even no. on a dice

B Number greater than 4

The venn diagram:

𝑃(𝐴) =3

6

𝑃(𝐵) =2

6

𝑃(𝐴 𝑜𝑟 𝐵) =

𝑃(𝐴 ∪ 𝐵) =

𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) =

3

6 +

2

6 −

1

6 =

4

6 (𝑜𝑟

2

3)

So, we have the addition rule…

𝑷(𝑨 𝒐𝒓 𝑩) =

𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩)

It is useful to know we can rearrange this to find the intersection:

𝑷(𝑨 𝒂𝒏𝒅 𝑩) =

𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∪ 𝑩)

Example 3

When you add the 2 circles, you have

added the 6 twice, so you take it away

at the end

Two events are independent if they can happen at the same time.

𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨) × 𝑷(𝑩)

Eg, the probability of getting a 3 on a dice and heads on a coin

𝑃(3 ∩ 𝐻) = 𝑃(3) × 𝑃(𝐻) = 1

6×

1

2=

1

12

We can see this from a two way table or sample space diagram:

There is one possible outcome out of 12.

1 2 3 4 5 6

Head H,1 H,2 H,3 H,4 H,5 H,6

Tail T,1 T,2 T,3 T,4 T,5 T,6

Past Paper Question

On a venn diagram

Tree Diagrams – Probability of Events from 2 Trials

When 2 (or 3) events are happening, it is useful to draw a tree diagram.

Eg. A box contains 8 red and 6 blue marbles. 2 marbles are selected at random, one after the other,

without replacement.

Tree diagram looks like this:

7

13

8

14

6

13

8

13

6

14

5

13

1st selection 2nd selection

14 marbles 13 marbles

a) 𝑃(1𝑅 ∩ 1𝐵) = 𝑃(𝑅 ∩ 𝐵) 𝑜𝑟 𝑃(𝐵 ∩ 𝑅)

= (8

14×

6

13) + (

6

14×

8

13)

=48

91 or 0.527 (3sf)

b) 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1𝑅) = 1 − 𝑃(𝑁𝑂 𝑅)

= 1 − 𝑃(𝐵 ∩ 𝐵)

= 1 − (6

14×

5

13)

= 76

91 or 0.835 (3sf)

Past Paper Questions

1)

R

R

G

y

G

R

G

Y

Y

R

G

Y

2)

i) 𝑃(𝑅 ∩ 𝑅 ∩ 𝑅) = 0.25 × 0.25 × 0.25

= 0.0156 (3 s.f.)

SO

SO

SO

L

R

L

SO

L

R

R

SO

L

R

L

SO

SO

L

R

L

SO

L

R

R

SO

L

R

R

SO

SO

L

R

L

SO

L

R

R

SO

L

R

Binomial Distribution

Tree diagrams can be used to work out probabilities of combined outcomes of up to 3 trials, eg

from the question previous regarding directions 3 drivers take. But when we have more than 3

trials, it would be cumbersome to create a tree diagram as there would be too many branches and

also too many combinations or permutations of different outcomes. Therefore, we use a binomial

model to calculate such probabilities.

We use a binomial model to test the probability of an event:

When the outcome is deemed either success or failure. (Two outcomes bi-nomial)

The trials are independent. (One trial does not affect the others)

A finite number of trials are carried out. Eg. 5 or 6 or 10 or 12 etc

The probability, p, of a successful outcome is the same for each trial.

Example

Bulbs are tested in a factory. 10% of bulbs are said to be faulty. If 3 bulbs are selected at random

from a collection of bulbs, what is the probability that 2 of them are faulty?

Assuming that one faulty bulb does not affect the probability of another bulb being faulty, they are

independent.

So, P(Faulty) = P(F) = 0.1 and P(not Faulty) = 𝑃(�̅�) = 0.9

To calculate the probability we could use a tree diagram because there are only 3 trials, i.e. 3 bulbs.

𝑃(2 𝑓𝑎𝑢𝑙𝑡𝑦) = 𝑃(2𝐹 ∩ 1�̅�)

= 𝑃(𝐹, 𝐹, �̅�) 𝑜𝑟 𝑃(𝐹, �̅�, 𝐹) 𝑜𝑟 𝑃(�̅�, 𝐹, 𝐹)

= (0.1 × 0.1 × 0.9) + (0.1 × 0.9 × 0.1) + (0.9 × 0.1 × 0.1)

= 3 × 0.12 × 0.9 (3 combinations of 0.1 x 0.1 x 0.9)

= 0.027

However, if there were 8 bulbs selected at random, we could not use a tree diagram as there are

more than 3 trials, therefore too many combinations. Let’s see how Binomial works…

What is the probability that out of 8 bulbs selected, 2 are faulty?

One combination of 2 faulty and 6 not faulty is:

𝐹, 𝐹, �̅�, �̅�, �̅�, �̅�, �̅�, �̅� The probability of this single combination is:

𝑃(𝐹, 𝐹, �̅�, �̅�, �̅�, �̅�, �̅�, �̅�) = 0.12 × 0.96

But there are so many combinations or arrangements of this outcome, each with a probability

of 0.12 × 0.96.

The number of arrangements is given by 𝐶2.8 Where 8 is the number of bulbs selected (trials) and

2 is the number of faulty bulbs (successes)

On your calculator press 8 then SHIFT then ÷ (nCr) then 2 then =

You should get 28. This means there are 28 outcomes of 2 faulty bulbs out of 8 bulbs selected.

So 𝑃(𝑒𝑥𝑎𝑐𝑡𝑙𝑦 2 𝑓𝑎𝑢𝑙𝑡𝑦) = 𝐶28

× 0.12 × 0.96 = 28 × 0.12 × 0.96 = 0.149 𝑡𝑜 3 𝑠. 𝑓.

If 14 bulbs are selected, the probability that 2 bulbs are faulty and 12 bulbs not faulty would be:

𝑃(𝑒𝑥𝑎𝑐𝑡𝑙𝑦 2 𝑓𝑎𝑢𝑙𝑡𝑦) = 𝐶214

× 0.12 × 0.912 = 91 × 0.12 × 0.912 = 0.257 𝑡𝑜 3 𝑠. 𝑓.

The discrete random variable, X, in this example is the number of bulbs tested.

For any experiment with conditions for a binomial model, the distribution is written:

𝑋~𝐵𝑖𝑛(𝑛, 𝑝) 𝑜𝑟 𝑋~𝐵(𝑛, 𝑝) 𝑤ℎ𝑒𝑟𝑒 𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑜. 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 𝑎𝑛𝑑 𝑝 𝑖𝑠 𝑡ℎ𝑒

𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠

(𝒏 𝑎𝑛𝑑 𝒑 𝑎𝑟𝑒 𝑘𝑛𝑜𝑤𝑛 𝑎𝑠 𝑡ℎ𝑒 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓𝒔 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛)

If 𝑋~𝐵𝑖𝑛(𝑛, 𝑝), the probability of obtaining 𝑟 successes in 𝑛 trials is 𝑃(𝑋 = 𝑟)

where

𝑃(𝑋 = 𝑟) = 𝐶𝑟 × 𝑝𝑟 × 𝑞𝑛−𝑟𝑛 𝑓𝑜𝑟 𝑟 = 0,1,2,3 … . 𝑛

𝒑 is success and 𝒒 is failure.

Example 1

If a random variable, 𝑋~𝐵(8,1

3) so, 8 trials and prob of success is a third

What is:

a) 𝑃(𝑋 = 2) (exactly 2 successes out of 8 trials)

𝑃(𝑋 = 2) = 𝐶2 × (1

3)2 × (

2

3)68

= 28 ×1

9×

64

729

= 0.273 𝑡𝑜 3 𝑠. 𝑓.

b) 𝑃(𝑋 ≤ 1) (0 or 1 successes out of 8 trials)

𝑃(𝑋 ≤ 1) = 𝐶0 × (1

3)

0

× (2

3)

8

+ 8 𝐶1 × (1

3)

1

× (2

3)

78

= 1 × 1 ×256

6561 + 8 ×

1

3×

128

2187

= 0.195 𝑡𝑜 3 𝑠. 𝑓.

Example 2

For a particular type of plant 25% have blue flowers. A garden centre sells these plants in trays of

15 plants of mixed colours. A tray is selected at random. Find the probability that the number of

blue flowers in this tray is

a) Exactly 4 b) At most 3 c) Between 3 and 6 (inclusive)

Always start by writing the distribution function

𝑋~𝐵(15,0.25)

a) 𝑃(𝑋 = 4) = 𝐶4 × 0.254 × 0.751115 = 0.225 𝑡𝑜 3 𝑠. 𝑓.

b) 𝑃(𝑋 ≤ 3) = 𝑃(𝑋 = 0, 1, 2, 3) = 𝐶0 × 0.250 × 0.751515 + 𝐶1 × 0.251 × 0.751415 +

𝐶2 × 0.252 × 0.751315 + 𝐶3 × 0.253 × 0.751215

= 0.0133634 … + 0.0668173 … + 0.155907 … + 0.225199 … = 0.461 𝑡𝑜 3 𝑠. 𝑓.

c) 𝑃(3 ≤ 𝑋 ≤ 6) = 𝑃(𝑋 = 3, 4, 5, 6) = 𝐶3 × 0.253 × 0.751215 + 𝐶4 × 0.254 × 0.751115 +

𝐶5 × 0.255 × 0.751015 + 𝐶6 × 0.256 × 0.75915

= 0.225199 … + 0.225199 … + 0.165145 … + 0.0917477 … = 0.707 𝑡𝑜 3 𝑠. 𝑓.

Exam Question

And we are told this is equal to 3 × 𝑃(𝑋 = 1)

So

105𝑝2

45𝑝=

(1 − 𝑝)14

(1 − 𝑝)13

7𝑝

3= 1 − 𝑝

7𝑝 = 3 − 3𝑝

10𝑝 = 3

𝑝 = 0.3

Mean and Variance of a Binomial Distribution

The mean is the expected number of successes:

𝐸(𝑋) = 𝑛𝑝

We just multiply the number of trials by the probability of success

The variance:

𝑉𝑎𝑟(𝑋) = 𝑛𝑝𝑞

We just multiply the number of trials by the probability of success by the probability of

failure.

Example 1

The probability that it will be a dry day is 0.4. Find the expected number of dry days in a week and

the standard deviation.

𝐸(𝑋) = 𝑛𝑝 = 7 × 0.4 = 2.8

𝑉𝑎𝑟(𝑋) = 𝑛𝑝𝑞 = 7 × 0.4 × 0.6 = 1.68

But standard deviation is the square root of variance….

𝑆𝐷 = √1.68 = 1.296 … = 1.30 𝑡𝑜 3 𝑠. 𝑓.

Example 2

An examiner is trying to design a multiple choice test. For students answering a test at random, he

requires that the mean score on the test should be 20 and the standard deviation should be at least

4. Assuming that each question has the same number of alternative answers, find how many

questions and how many answers each question should have. The number of alternatives for each

question should be as few as possible.

Let

𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠

𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑛𝑠𝑤𝑒𝑟𝑖𝑛𝑔 𝑎 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛 𝑟𝑖𝑔ℎ𝑡

𝑋 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠 𝑎𝑠𝑤𝑒𝑟𝑒𝑑 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑙𝑦

So, 𝑋~𝐵(𝑛, 𝑝)

Mean must be 20 𝐸(𝑋) = 20 so 𝑛𝑝 = 20

Standard deviation must be at least 4 𝑆𝐷 ≥ 4, 𝑠𝑜 𝑉𝑎𝑟(𝑋) ≥ 42

𝑛𝑝𝑞 ≥ 16

𝑛𝑝(1 − 𝑝) ≥ 16

20(1 − 𝑝) ≥ 16

1 − 𝑝 ≥16

20

1 − 𝑝 ≥ 0.8

0.2 ≥ 𝑝

This means that p, probability of answering a question right, must be at least 0.2 (1 in 5 chance)

So, the examiner should have 5 multiple choice answers for each question.

If 𝐸(𝑋) = 𝑛𝑝 = 20

And 𝑝 = 0.2

Then 0.2𝑛 = 20

So, 𝑛 =20

0.2= 100

So, the examiner should have 100 questions.