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S3 Notes (Edexcel)























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Combinations of Random Variables Distribution of linear combinations of independent Normal random variables. If ( )2~ ,x xX N µ σ and ( )2~ ,y yY N µ σ independently, then ( )2 2 2 2~ ,x y x yaX bY N a b a bµ µ σ σ± ± + . No proofs required. The following definitions of the expectation of a random variable X and the variance of X are known from S1.

E( ) i iX p x=∑ and ( ) ( )22 2Var( ) E( )i i i i i iX p x X p x p x= − = −∑ ∑ ∑ The two results which were required in S1 were

E( ) E( )aX b a X b+ = + and )(Var)(Var 2 XabaX =+ Consider two random variables X and Y whose probability distributions are as follows:

x x1 x2 xn y y1 y2 ym ( )P X x= p1 p2 pn ( )P Y y= q1 q2 pm

Suppose ( ) and ij i jr P X x Y y= = = . If X and Y are independent then ( ) ( ) ( ) and i j i jP X x Y y P X x P Y y= = = = × = , that is ij i jr p q= × but in the working shown below X and Y are not assumed to be independent. The probability distribution table is therefore

y1 y2 ym x1 r11 r12 r1m x2 r21 r22 r2m

xn rn1 rn2 rnm Now clearly ( ) 1 2 ...i i i im iP X x r r r p= = + + + = and ( ) 1 2 ...j j j nj jP Y y r r r q= = + + + = . The table gives ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

11 1 1 12 1 2 1 1

21 2 1 22 2 2 2 2

1 1 2 2

E ....

........

....

m m

m m

n n n n nm n m

X Y r x y r x y r x y

r x y r x y r x y

r x y r x y r x y

+ = + + + + + +

+ + + + + + +

+

+ + + + + + +

So ( ) ( ) ( ) ( )

( ) ( ) ( )1 11 12 1 2 21 22 2 1 2

1 11 21 1 2 12 22 2 1 2

E ... ... .... ...

... ... .... ...m m n n n nm

n n m m m nm

X Y x r r r x r r r x r r r

y r r r y r r r y r r r

+ = + + + + + + + + + + + +

+ + + + + + + + + + + + +



But 1 2 ...i i im ir r r p+ + + = and 1 2 ...j j nj jr r r q+ + + = . So ( ) 1 1 2 2

1 1 2 2

E ........

n m

m m

X Y x p x p x py q y q y q

+ = + + +

+ + + +

But ( ) 1 1 2 2E ... n nX p x p x p x= + + + and ( ) 1 1 2 2E ... m mY q y q y q y= + + + . So

( ) ( ) ( )E E EX Y X Y+ = + where X and Y are random variables which may or may not be independent.

Now consider ( )E XY . Using the table that was used for ( )E X Y+ gives

( ) 11 1 1 12 1 2 1 1

21 2 1 22 2 2 2 2

1 1 2 2

E ........

........

m m

m m

n n n n nm n m

XY r x y r x y r x yr x y r x y r x y

r x y r x y r x y

= + + +

+ + + +

++ + + +

Now if it is assumed that X and Y are independent then ( ) ( ) ( ) and i j i jP X x Y y P X x P Y y= = = = × = , that is ij i jr p q= ×

So ( ) 1 1 2 2E ... n nX p x p x p x= + + + and ( ) 1 1 2 2E ... m mY q y q y q y= + + + .

Also 1 2 ... 1np p p+ + + = and 1 2 ... 1mq q q+ + + = . So ( ) 1 1 1 1 1 2 1 2 1 1

2 1 2 1 2 2 2 2 2 2

1 1 2 2

E ........

........

m m

m m

n n n n n m n m

XY p q x y p q x y p q x yp q x y p q x y p q x y

p q x y p q x y p q x y

= + +

+ + +++ + +

So ( ) ( )

( )

( )

1 1 1 1 2 2

2 2 1 1 2 2

1 1 2 2

E ...

.......

...

n n

n n

m m n n

XY q y p x p x p x

q y p x p x p x

q y p x p x p x

= + + +

+ + + +

+

+ + + +



Since ( ) 1 1 2 2E ... n nX p x p x p x= + + + it follows that ( ) ( ) ( ) ( )

( )( )1 1 2 2

1 1 2 2

E E E .... E

E ....m m

m m

XY q y X q y X q y X

X q y q y q y

= + + +

= + + +

Since ( ) 1 1 2 2E ... m mY q y q y q y= + + + it follows that

( ) ( ) ( )E =E EXY X Y if X and Y are independent

( ) ( )( ) ( )( )( ) ( ) ( )( )

22

22 2

Var E E

E 2 E E

X Y X Y X Y

X XY Y X Y

+ = + − +

= + + − +

Since ( ) ( ) ( )E E ES T S T+ = + it follows that ( ) ( ) ( ) ( )2 2 2 2E 2 E E 2 EX XY Y X XY Y+ + = + + . So

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( )( )

22 2

2 22 2

2 22 2

Var E 2E E E E E

E 2E E E E 2E E

E E E E 2 E E E

Var Var 2 E E E

X Y X X Y Y X Y

X XY Y X Y X Y

X X Y Y XY X Y

X Y XY X Y

+ = + + − +

= + + − + +

= − + − + −

= + + −

If X and Y are independent random variables then ( ) ( ) ( )E =E EXY X Y so

( ) ( ) ( )Var Var VarX Y X Y+ = + So, in summary

( ) ( ) ( )( ) ( ) ( )

E E E whether or not and are independent

Var Var Var if and are independent

X Y X Y X Y

X Y X Y X Y

+ = +

+ = +

NB The result that ( ) ( ) ( )Var Var VarX Y X Y+ = + cannot be used to get the result that

( ) ( ) ( ) ( ) ( )Var 2 =Var Var Var =2VarX X X X X X+ = + since ( ) ( ) ( )Var Var VarX Y X Y+ = + can only be used if X and Y are independent. ( ) ( )Var 2 =4VarX X . Combing these results with the results from S1 that ( ) ( )E EaX a X= and that

( ) ( )2Var VaraX a X= , it follows that

2 2

E( ) E( ) E( )Var( ) Var( ) Var( ) if and are independent

aX bY a X b YaX bY a X b Y X Y+ = +

+ = +

Since ( ) ( ) ( )E E EX Y X Y+ = +



If X and Y are independent random variables which are both distributed normally... Suppose 2~ ( , )x xX N µ σ and ),(~ 2

yyNY σµ then

( )( )

2 2 2 2

2 2 2 2

~ ,

~ ,

x y x y

x y x y

aX bY N a b a b

aX bY N a b a b

µ µ σ σ

µ µ σ σ

+ + +

− − +

Example Given the random variables ~ (50, 9)X N and ~ (45, 8)Y N where X and Y are independent, find

( )( )

( )

(a) E

(b) Var

(c) 55 3 2 71

X Y

X Y

P X Y

−

−

< − <

(a) ( ) ( ) ( )E E E 50 45 5X Y X Y− = − = − = (b) ( ) ( ) ( )Var Var Var 9 8 17X Y X Y− = + = + = (c) If 3 2R X Y= − ( )2 2~ 3 50 2 45, 3 9 2 8R N × − × × + × so ( )~ 60,113R N

( ) ( ) ( )

( ) ( )( )( )

( )

55 71 71 55

71 60 55 60113 113

1.03 0.47

0.8485 1 0.47

0.8485 1 0.68080.5293

P R P R P R< < = < − <

− − = Φ −Φ

= Φ −Φ −

= − −Φ

= − −

=

Example The random variable D is defined as 2D A B C= − + where 2~ (18, 3 )A N , 2~ (17, 2 )B N and 2~ (21, 4 )C N and A, B and C are independent. (a) Find ( 40)P D < The random variables B1, B2 and B3 are independent and each has the same distribution as B. The random variable X is defined as

3

15 2i

iX A B C

=

= − −∑

(b) Find P(X > 0).



(a) 2D A B C= − + . ( )2 2 2~ 1 18 1 17 2 21,1 9 1 4 2 16D N × − × + × × + × + × , that is ( )~ 31, 229D N

( )

40 31( 40)229

0.590.7224

P D − < = Φ

= Φ

=

(b) 3

1 2 31

5 2 5 2ii

X A B C A B B B C=

= − − = − − − −∑

( ) ( ) ( ) ( ) ( ) ( )1 2 3E 5E E E 2E5 18 17 17 17 2 21 3

X A B B E B C= − − − −

= × − − − − × = −

( ) ( ) ( ) ( ) ( ) ( )1 2 3Var 25Var Var Var Var 4Var25 9 4 4 4 4 16 301

X A B B B C= + + + +

= × + + + + × =

So ( )~ 3, 301X N − .

( ) ( )

( )

0 1 0

0 31301

1 0.171 0.56750.4325

P X P X> = − <

− − = −Φ

= −Φ

= −=

Example Two children each select a bag of sweets. The mass of the bags in the counter from which Lydia selects her bag are normally distributed with mean 40g and standard deviation 4g. The mass of the bags in the counter from which Joshua selects his bag are normally distributed with mean 35g and standard deviation 3g. Find the probability that: (a) The sum of the masses of Lydia’s and Joshua’s bags of sweets is greater than 81g. (b) Lydia’s bag of sweets is heavier than Joshua’s. (a) If L is the distribution of mass of the bags on the counter from which Lydia chooses her

sweets then ( )2~ 40, 4L N . If J is the distribution of mass of the bags on the counter from which Joshua chooses his sweets then ( )2~ 35, 3J N .

So ( )2 2~ 40 35, 4 3L J N+ + + . That is ( )~ 75, 25L J N+ . The probability that the sum of the masses of Lydia’s and Joshua’s bags of sweets is greater than 81g is represented by ( )81P L J+ > .



( ) ( )

( )

81 1 81

81 75125

81 75125

1 1.21 0.88490.1151

P L J P L J+ > = − + <

− = −Φ

− = −Φ

= −Φ

= −=

(b) ( )2 2~ 40 35, 4 3L J N− − + . That is ( )~ 5, 25L J N− . The probability that Lydia’s bag of sweets is heavier than Joshua’s is represented by ( )0P L J− > .

.

( ) ( )

( )( )

0 1 0

0 5125

1 1

10.8413

P L J P L J− > = − − <

− = −Φ

= −Φ −

= Φ

=

If n items are chosen from a distribution with mean ( )2~ ,x xX N µ σ and m items are chosen from a

distribution with mean ( )2~ ,y yY N µ σ

then their sum has distribution ( )2 2~ ,x y x yS N n m n mµ µ σ σ+ + .



Sampling Methods for collecting data. Simple random sampling. Use of random numbers for sampling. Other methods of sampling: stratified, systematic, quota. The circumstances in which they might be used. Their advantages and disadvantages. The S2 course dealt with the idea of taking a census or a sample. Sometimes it is possible to investigate the entire population. This would be called a census. A census is when every member of the population is investigated. January 2008, S2, Q1 Census If the IQ of pupils in a school was being investigated then it would be possible to take a census, since every pupil’s IQ could be measured. If the voting intentions of voters in the UK were being investigated then a census would be (a) extremely costly and (b) unnecessary since the level of accuracy needed does not justify asking every member of the population. So a census is used only if • The size of the population is small enough to enable the census to be taken. • A very high level of accuracy is required. NB There is a census taken in the UK every ten years where every member of the population is required to fill in a long form about all sorts of things. The obvious advantage of a census over a sample survey is that a census is completely accurate since it avoids the uncertainty associated with sampling The obvious disadvantages of a census are: • It is very costly and takes a long time • It produces vast quantities of data which takes ages to investigate • It cannot be used in certain cases, such as the ability of a certain car to cope with an impact.

Clearly sampling is greatly advantageous since a census would involve ruining every model of that car.

Sample In the example mentioned above about investigating the IQ of pupils in a school, whilst a census is possible it would be very time-consuming. It would be much easier (though less reliable) to take a sample. As was mentioned in S2, any sample should be a true (unbiased) representation of the population from which it is being drawn. The sampling method to be used must be chosen carefully – using a telephone directory when doing a survey on homelessness would not be very useful. Clearly it is true that as the sample gets larger so the accuracy of the results increases but it is also true that as the sample gets larger, the amount of time and resources required also increases. The sample chosen must be random so that each member of the population has an equal chance of being selected.



Different types of Samples 1. Simple Random Sample It is as if each member of the population has a numbered ticket and these tickets are then put into a bowl and mixed thoroughly. If a sample of 50 was required then 50 tickets would be chosen. Sampling without replacement means that the ticket is not replaced after being chosen. If simple random sampling is done with replacement – so a ticket could be chosen more than once, it is said to be unrestricted random sampling. Alternatively a random number sample could be taken. In this, every member of the pupil would be allocated a number. So in a school of 800 each pupil would be allocated a different number from 1 to 800. It is worth noting that each number must have the same number of digits so the pupils would be given the numbers 001, 002,003,……,010, 011, 012,…..,100, 101,…..,799, 800 On page 32 of the formula book there is a page with Random Numbers. This has been created so that each digit 0, 1, 2,..9 appears with the same frequency.

In the example of the school with 800 pupils, suppose that a sample of 50 was taken. The numbers would be taken in groups of 3, so 861 is the first number then 384 etc.

In this example any number greater than 800 is ignored. So the first 5 numbers would be 384, 100, 730, 390, 597. If the school had 500 pupils then instead of giving each pupil a number between 1 and 500 and then ignoring every number over 500, each pupil could be given 2 numbers. For example the first pupil could be given the numbers 001 and 501, the next 002 and 502 etc. 2. Systematic Sample In the example of the school of 800, if every tenth pupil was chosen from a list of all pupils then this would create a sample of 80. This method of sampling is called a systematic sample. To overcome the weakness of this sample that the first pupil on the list is always chosen, a random number between 1 and 10 could first of all be chosen. Suppose that was 3. Then the third pupil would be in the sample along with every tenth pupil on from there. So the 3rd, 13th, 23rd etc. pupil would be chosen.

NB Start in a random place on the page.



If a sample of n is being chosen from a population of N then every kth member of the population

could be chosen where Nkn

≤ . The process begins by first of all choosing a random number

between 1 and k and then choosing every kth member from there. Systematic sampling can be used when the population is too large for simple random sampling. Example Explain briefly how a researcher could select a sample of 50 employees from 550 employees using a systematic sample

S3, June 2008, Q1 • Label employees (1-550) or obtain an ordered list • Select first using random numbers (from 1 - 11) • Then select every 11th person from the list 3. Stratified Sample In a stratified sample the population is divided into groups, or strata, which have certain characteristics (e.g. smokers, car owners, colour of hair etc.) and then simple random samples are taken from each group. The number we choose from each group is proportional to the size of the group. The strata are mutually exclusive, for example, smokers and non-smokers. For example if an investigation was being made into lung cancer then whilst it would seem strange to split the population into those who wear glasses and those who do not, it would be logical to split the population into those who smoke and those who do not. So if it was established that 25% of the population smoke and a sample of 200 people was being taken then, in a stratified sample, 25% of these 200 people, that is 50 would be smokers and 150 would be non-smokers. The 50 smokers and 150 non-smokers would be chosen by simple random samples. Example State two reasons why stratified sampling might be chosen as a method of sampling when carrying out a statistical survey. • Population divides into mutually exclusive groups or strata. [Seems like one reason to me but

that is what the mark scheme says] S3, June 2005, Q1

Example Explain how to obtain a sample from a population using stratified sampling Give one advantage and one disadvantage of [stratified sampling]. S3, June 2003, Q1 Take a (simple) random sample from (mutually exclusive) groups of the population Sample sizes within strata in strict proportion to numbers in each strata in the population Advantage: [Any one] More accurate estimate of variance of population mean Individual estimates for strata available Disadvantage [Any one] Difficult if strata are large / Definition of strata problematic (may overlap) 4. Quota Sample



In each of the previous samples, the population has been known and the sample has been random. A quota sample is a non-random sample in which the researcher asks people with certain characteristics. The proportion of people chosen in each group (called the quota) reflects the proportion of people in each group of the population. So in an example on attitudes to school meals, pupils from each year group would be interviewed but within that it might be decided to separate vegetarians and non-vegetarians since this might affect the view of the meals. Year group Vegetarian / Non Vegetarian Number Year 9 Vegetarian 2 Non Vegetarian 19 Total 21 Year 10 Vegetarian 2 Non Vegetarian 20 Total 22 Year 11 Vegetarian 2 Non Vegetarian 19 Total 21 Year 12 Vegetarian 2 Non Vegetarian 17 Total 19 Year 13 Vegetarian 1 Non Vegetarian 16 Total 17 Total 100 The quotas are the numbers by each category. As the researcher asks someone, he first of all establishes which group the pupil is in. If the pupil refuses to answer or if the relevant quota is full then the researcher ignores that pupil and asks the next one. Example Explain how to obtain a sample from a population using quota sampling. Give one advantage and one disadvantage of each sampling method. S3, June 2003, Q1 Non-random sampling from groups of the population Advantage [Any one (not quick)] Representative sample can be achieved with small sample size Cheap (costs kept to a minimum) Administration relatively easy Disadvantage [Any one] Not possible to estimate sampling errors due to lack of randomness Judgment of interviewer can affect choice of sample – ‘bias’ is sufficient Non-response not recorded Difficulties of defining controls e.g. social class



Summary of the Different Types of Samples Sample Simple Random

Sample

Systematic Sample

Stratified Sample

Quota Sample

Use when… the population is quite small.

the population is large.

(i) the sample is large (ii) the population divides into mutually exclusive, relevant strata.

the one doing the survey is asked to interview a certain number of people with specific characteristics.

Advantage (i) Random process so possible to estimate sampling error. (ii) It is not biased (June 06) (iii) It is simple to use (iv) Each member of the population is equally likely to be chosen since each number is equally likely to be chosen.

(i) Simple or easy to use not “quick” or “cheap” or “efficient” (Jun 2009) (ii) It can be used for large samples (not population).

(i) Can give accurate estimates as it is a random process, (ii) Gives estimates for each course (iii) representative of [BUT not “proportional” to] the whole population. (o.e.) [June 2013]

(i) can be done quickly, (ii) administering is easy, (iii) costs are kept to a minimum (cheap), (iv) OK for large populations or sampling frame not required (o.e.) [June 2013]

Disadvantage (i) It cannot be used when the sample size is large. (ii) Sampling frame required which may not exist or may be difficult to construct for a large population. (June 06)

(i) It only works if the ordered list is truly random. (ii) Requires a list of the population or must assign a number to each member of the pop.

(i) Sampling frame required (ii) strata may not be clear [as some students overlap courses] (iii) not suitable for large populations. (o.e.) [June 2013]

(i) Non-random process (ii) not possible to estimate the sampling (iii) or non response not recorded (iv) interviewer can introduce bias in sample choice. (o.e.) [June 2013]

For example if a wheel with 10 spokes in a machine was faulty so that it every tenth item produced was faulty then a systematic sample using every tenth item would give either all being faulty or all being fine.



Example A motel has 60 rooms of which 10 are classified as Platinum, 20 Gold and 30 as Silver. The owner wants to get information about the use of rooms in the motel by taking a 20% sample of the rooms. (a) Suggest a suitable sampling method. (b) Explain in detail how the manager should obtain the sample.

(a) Stratified (b) 20% of 60 is 12. So the sample is 12 rooms. 16

of rooms in the motel are Platinum, so 16

of rooms in the sample should be Platinum. So there

are 2 Platinum rooms in the sample. 13

of rooms in the motel are Gold, so 13

of rooms in the sample should be Gold. So there are 4

Gold rooms in the sample. 12

of rooms in the motel are Silver, so 12

of rooms in the sample should be Silver. So there are 6

Silver rooms in the sample. Label Platinum rooms 1-10 and use random numbers in range 1-10 and select 2 rooms. Label Gold rooms 1-20 and use random numbers in range 1-20 and select 4 rooms. Label Silver rooms 1-30 and use random numbers in range 1-30 and select 6 rooms.

Example There are 75 girls and 125 boys in a school. Explain briefly how you could take a random sample of 24 pupils using (a) a simple random sample, (b) a stratified sample. (a) • Allocate a number between 1 and 200 to each pupil. • Use random number tables, computer or calculator to select 24 different numbers between 1

and 200 • Pupils corresponding to these numbers become the sample. (b)

• Select 24 75 9200

× = girls and 24 125 15200

× = boys.

• Allocate numbers 1 – 75 to girls and 1 – 125 to boys. • Select 9 random numbers between 1-75 for girls • Select 15 random numbers between 1-125 for boys.

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