Further Statistics

7/21/2019 Further Statistics

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CIE A LEVEL- FURTHER MATHEMATICS [9231]

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1. EXPECTATION ALGEBRA

1.1 Expectation & Variance of a Function of + = + + =

(IS) Ex 6a: Question 12:

The random variable has mean 5 and variance 16.Find two pairs of values for the constants and such

that + = 100 and + = 144

Solution:

Expand expectation equation: + = + = 100 ∴ 5 + = 100

Expand variance equation: + = = 144

16

= 144

= ±3 Use first equation to find two pairs: = 3, = 85 = 3, = 115

1.2 Combinations of Random Variables

Expectations of combinations of random variables: + = +

Variance of combinations of independent random

variables:

+ + = +

± = +

Combinations of identically distributed random variables

having mean and variance 2 = 2 and + = 2 2 = 4 but + = 2

(IS) Ex 6b: Question 3:

It is given that and are independent, and

=

= ,

=

=

Find and (), where = + Solution:

Split the expectation into individual components

12 + = 12 + 12

Substitute given values, hence

12 + = 12 + 12 =

Split the variance into individual components

12 + = 12 + 12

Substitute given values, hence

12 + = 14 + 14 = 12

1.3 Expectation & Variance of Sample Mean( ) = ( ) =

(IS) Ex 6c: Question 5:

The mean weight of a soldier may be taken to be 90kg,

and = 10kg. 250 soldiers are on board an aircraft,

find the expectation and variance of their weight.

Hence find the and of the total weight of soldiers.

Solution:

Let

be the average weight, therefore

( ) = = 90 ( ) = = = 0.4 kg2

To find of total weight, you are calculating + … + = 250 = 22 500kg

To find , first find …+ = 250 = 2500kg = = 25000 ∴ = √ 25000 = 158.1kg

2. POISSON DISTRIBUTION The Poisson distribution is used as a model for the

number, , of events in a given interval of space or

times. It has the probability formula = = − ! = 0, 1, 2, …

Where is equal to the mean number of events in the

given interval.

A Poisson distribution with mean can be noted as ~

2.1 Suitability of a Poisson Distribution Occur randomly in space or time

Occur singly – events cannot occur simultaneously

Occur independently

Occur at a constant rate – mean no. of events in given

time interval proportional to size of interval




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2.2 Expectation & Variance

For a Poisson distribution ~

Mean = = =

Variance = = =

The mean & variance of a Poisson distribution are equal

2.3 Addition of Poisson Distributions If and are independent Poisson random variables,

with parameters and respectively, then + has a

Poisson distribution with parameter +

(IS) Ex 8d: Question 1:

The numbers of emissions per minute from two

radioactive objects and are independent Poisson

variables with mean 0.65 and 0.45 respectively.

Find the probabilities that:

i.

In a period of three minutes there are at least three

emissions from .ii.

In a period of two minutes there is a total of less

than four emissions from and together.

Solution:

Part (i):

Write the distribution using the correct notation ~0.65×3 = ~1.95

Use the limits given in the question to find probability ≥ 3 = 1 < 3

= 1 1.95−.

2! +1.95−.

1! +1.95−.

0!

= 10.690 = 0.310

Part (ii):

Write the distribution using the correct notation + ~(20.65+0.45) = + ~2.2

Use the limits given in the question to find probability

< 4 = −. 2.23! + 2.22! + 2.21!+ 2.2

0!

= 0.819

2.4 Relationship of Inequalities

< = ≤ 1

= = ≤ ≤ 1

> = 1 ≤

≥ = 1 ≤ 1

2.5 Poisson Approximation of a Binomial

Distribution

To approximate a binomial distribution given by: ~,

If

> 50 and

> 5

Then we can use a Poisson distribution given by:

~

(IS) Ex 8d: Question 8:

A randomly chosen doctor in general practice sees, on

average, one case of a broken nose per year and each

case is independent of the other similar cases.

i.

Regarding a month as a twelfth part of a year,

a.

Show that the probability that, between them,

three such doctors see no cases of a broken

nose in a period of one month is 0.779

b.

Find the variance of the number of cases seen

by three such doctors in a period of six monthsii.

Find the probability that, between them, three

such doctors see at least three cases in one year.

iii.

Find the probability that, of three such doctors,

one sees three cases and the other two see no

cases in one year.

Solution:

Part (i)(a):

Write down the information we know and need1 doctor = 1 nose per year = noses per month

3 doctors= = noses per monthWrite the distribution using the correct notation ~0.25

Use the limits given in the question to find probability

= 0 = 0.25−.0! = 0.779

Part (i)(b):

Use the rules of a Poisson distribution = =

Calculate

in this scenario:

= 6 × ℎ = 6 × 0.25 = 1.5 ∴ = 1.5

Part (ii):

Calculate in this scenario: = 12 × ℎ = 12 × 0.25 = 3

Use the limits given in the question to find probability ≥ 3 = 1 ≤ 2 = 1 − 32! + 31! + 30! = 10.423 = 0.577




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Part (iii):

We will need two different s in this scenario: = 1 ℎ = 2 × 1 = 2

For the first doctor:

= 3 = − 13!

For the two other doctors:

= 0 = − 10!

Considering that any of the three could be the first

= − 13! × − 10! × = 0.025

2.6 Normal Approximation of a Poisson

Distribution

To approximate a Poisson distribution given by: ~

If > 15

Then we can use a normal distribution given by: ~,

Apply continuity correction to limits:

Binomial Normal = 6 5.5≤≤6.5

> 6

≥6.5

≥ 6 ≥5.5 < 6 ≤5.5 ≤ 6 ≤6.5

(IS) Ex 10h: Question 11:

The no. of flaws in a length of cloth, m long has a

Poisson distribution with mean 0.04

i.

Find the probability that a 10m length of cloth has

fewer than 2 flaws.

ii.

Find an approximate value for the probability that a

1000m length of cloth has at least 46 flaws.

iii.

Given that the cost of rectifying flaws in a 1000mlength of cloth is pence, find the expected cost.

Solution:

Part (i):

Form the parameters of Poisson distribution = 10 and = 0.04 ∴ = 0.4

Write down our distribution using correct notation ~0.4

Write the probability required by the question < 2

From earlier equations:

< 2 = −. 0.4

0! + 0.4

1! = 0.938

Part (ii):

Using information from question form the parameters

of Poisson distribution = 10 and = 0.04 ∴ = 40 > 15

Thus we can use the normal approximation

Write down our distribution using correct notation ~40 → ~40,40

Write the probability required by the question

≥ 46

Apply continuity correction for the normal distribution

≥ 45.5

Evaluate the probability ≥ 45.5 = 1 Φ 45.540√ 40 = 0.192

Part (iii):

Using the variance formula = ( )

For a Poisson distribution

= = and

= 40

Substitute into equation and solve for the unknown

∴ 40 = 40 = 1640 pence = ₤16.40

Expected cost for rectifying cloth is ₤16.40

3. CONTINUOUS R ANDOM V ARIABLE

3.1 Probability Density Functions (pdf)

Function whose area under its graph represents

probability used for continuous random variables

Represented by




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Conditions:

Total area always = 1 ∫ = 1

Cannot have –ve probabilities ∴ graph cannot dip below

-axis;

≥ 0

Probability that lies between and is the area from to < < = ∫

Outside given interval = 0; show on a sketch

= always equals 0 as there is no area

Notes:

o

< = ≤ as no extra area added

o

The mode of a pdf is its maximum (stationary point)

(IS) Ex 9a: Question 6:Given that: = 60 2 < < 5otherwise

i.

Find the value of

ii.

Find the mode,

iii.

Find <

Solution:

Part (i):

Total area must equal 1 hence

∫ 6 = 3 3 = 1

= 75 1253 12 + 83 = 24 = 1

∴ = 124

Part (ii):

Mode is the value which has the greatest probability

hence we are looking for the max point on the pdf

6 = 6 2

Finding max point hence stationary point

6 2 = 0

= 6 2 = 3

∴ mode = 3

Part (iii): < can be interpreted as ∞ < <

∫ 6 − = ∫ 6

= 3 3

= 124 33 3

3 32 + 2

3 = 1336

3.2 Cumulative Distribution Function (cdf)

Gives the probability that the value is less than < or ≤

Represented by

It is the integral of = ∫ −

Median: the value of

for which

= 0.5

(apply analogy to quartiles/percentages)

Notes:

o

Since it is always impossible to have a value of

smaller than ∞ or larger than ∞:∞ = 0 ∞ = 1

o

As increase, either increase or remains

constant, but never decreases.

o

is a continuous function even if is discontinuous

Useful relations:

o < < =

o

> = 1

(IS) Ex 9b: Question 9:

Given that:

= 40 0 < < 11 < < 3otherwise

i.

Find the value of

ii.

Find

iii.

Find the difference between the median and the

fifth percentile of




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Solution:

Part (i):

Total area must equal 1 hence∫ +∫ 4

= + 4 = 1

= 0 + 12 4 = 9 = 1

∴ = 19

Part (ii):

Integrate each case separately from its ∞ to

For the first interval 0 ≤ ≤ 1 = ∫ 19 = [19 ]

= 19

We must split next interval 0 ≤ ≤ 3 as = ≤ 3 = ≤ 1 + 1 ≤ ≤ 3

and ≤ 1 = 1 = 1 9⁄

∴ = 19 + ∫ 4 × 19

= 19 + [4 × 19 ] = 49 39

Writing in correct notation and fixing intervals (adding

equal sign to inequalities)

={

019 49 391

≤ 019 0 ≤ ≤ 1 1919 1 ≤ ≤ 3 39 ≥ 3

Part (iii):

Finding the median, you must check in which interval itlies. Do this by substituting the maximum value for in

the first case 19 × 1 = 19 < 12

This means the median does not lie in this interval ∴ 49 39 = 0.5

= 158

The fifth percentile lies in the first interval as

<

so

19 = 120 = 920

Find the difference 158 920 = 5740

3.3 Expectation and Variance

To calculate expectation = ∫ −

To calculate variance:

o

First calculate

as above

o

The calculate by = ∫ −

o

Substitute information and calculate using =

3.4 Obtaining f(x) from F(x)

As is obtained by integrating , then can be obtained

by differentiating

(IS) Ex 9d: Example 13:The random variable has cdf given by

= 0 181 ≤ 111 1 ≤ ≤ 3 11 ≥ 3

Find the form of the pdf of

Solution: is unchanging for < 1 and for > 3, therefore is equal to 0. Hence we must find differentiate in

the interval

1 < < 3

= ′ = 18 = 38 1

Hence:

= 38 10 1 < < 3otherwise

3.5 Distribution of a Function of a Random

Variable

We can deduce the distribution of a simple function of

either increasing or decreasing with this procedure:f X → FX → FY → f Y

(IS) Ex 9e: Example 15:

The random variable has pdf given by, = 10 2 < < 3otherwise

The random variable is given by = 2 + 3.

Determine the pdf and cdf of .




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Solution:

First step is to find and suppose we do,

FX = ≤ = 0 21 ≤ 22 ≤ ≤ 3 ≥ 3

Find the ranges for

2 × 2 + 3 ≤ ≤ 3 × 2 + 3 7 ≤ ≤ 9

Convert cdf from to using relationship givenFY = ≤ = 2 + 3 ≤ = ≤ 1 2⁄ 3

Now substitute 1 2⁄ 3 for in cdf function1 2⁄ 3 2 ⟹ 1 2⁄ 7

Expressing cdf of with ranges worked out

F = ≤ = 01 2⁄ 71 ≤ 77 ≤ ≤ 9 ≥ 9

Differentiate function to find pdf

= 1 2⁄0 7 < < 9otherwise

Method can be used for both increasing and decreasing

functions as well functions with powers (e.g. = )

4. GEOMETRIC & EXPONENTIAL DISTRIBUTION

4.1 Geometric DistributionConditions for a Geometric Distribution:

Only two possible outcomes: success or failure

Probability of success, , is constant

Each event is independent

The geometric distribution is used to find the number of

trials required to obtain the first success = = 1 − = 1, 2, 3, …

Where is the probability of success, 1 is the

probability of failure and

is the number of trials

A geometric distribution with probability of success can be noted as ~

The distribution is called geometric because successive

probabilities, , 1 , 1 … form a geometric

progression with first term and common ratio 1

4.2 Cumulative Probabilities

Calculating cumulative probabilities ≤ = 1 1 ≥ = 1 − < = 1 1 − > = 1 Example:

In the village of Nanakuli, about 80% of the residentsare of Hawaiian ancestry. Suppose you fly to Hawaii and

visit Nanakuli.

i.

What is the probability that the fifth villager you

meet is Hawaiian?

ii.

What is the probability that you do not meet a

Hawaiian until the third villager?

Solution:

Part (i):

Using the formula

= 5 = 10.80−

0.80 = 0.00128

Part (ii):

Not meeting until third means the probability > 3

Using relationships above > 3 = 10.80 = 0.008

4.3 Mean & Variance of a Geometric

Distribution

The expectation (mean) of a geometric distribution: = 1

The variance of a geometric distribution: = 1

4.4 Exponential Distribution

Used for modeling duration of events < = 1 − > = − < < = − −

Where is the average no. of events in 1 unit of time

and

is the duration

An exponential distribution with average can be noted ~

The exponential distribution is memory-less > + | > = >

o

e.g. if a motor has been running for 3 hours and you

are asked to calculate the probability of it running for

more than 4 hours, you only need to find the

probability of it running for the next hour as the

previous condition does not affect the probability




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4.5 Mean & Variance of an Exponential

Distribution

The expectation (mean) of an exponential distribution: = 1

The variance of an exponential distribution:

= 1

Example:

Calls arrive at an average rate of 12 per hour. Find the

probability that a call will occur in the next 5 minutes

given that you have already waited 10 minutes.

Solution:

Interpreting the information, = 12 per hour = 0.2 per minute

We are being asked to calculate

≤ 15| > 10 As the exponential distribution is memory-less; the

previous condition does not affect it hence we are

simply being asked to find ≤ 5 ≤ 5 = 1 −.× = 0.63

5. S AMPLING & CENTRAL LIMIT THEOREM

5.1 Central Limit TheoremIf

,

, … ,

is a random sample of size

drawn from

any population with mean

and variance

then the

sample has:

Expected mean,

Expected variance,

It forms a normal distribution: ~ ,

(IS) Ex 10f: Question 12:

The weights of the trout at a trout farm are normally

distributed with mean 1kg & standard deviation 0.25kg

a.

Find, to 4 decimal places, the probability that a trout

chosen at random weighs more than 1.25kg.

b.

If kg represents mean weight of a sample of 10

trout chosen at random, state the distribution of :

evaluate the mean and variance.

Find the probability that the mean weight of a

sample of 10 trout will be less than 0.9kg

Solution:

Part (a):

Write down distribution ~1,0.25

Write down the probability they want

> 1.25 = 1 < 1.25

Standardize and evaluate1 < 1.25 10.25 = 0.1587

Part (b):

Write down initial distribution ~1,0.25

For sample, mean remains equal but variance changes

Find new variance

Variance of sample = = . = 0.00625

Write down distribution of sample

~1,0.00625

Write down the probability they want < 0.9

Standardize and evaluate

Standardized probability is negative so do 1 minus < 0 .9 10.00625 = 1 < 0.10.00625 = 0.103

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