© Boardworks Ltd 20061 of 61 © Boardworks Ltd 2006 1 of 61 A2-Level Maths: Statistics 2 for Edexcel S2.2 Continuous random variables This icon indicates

© Boardworks Ltd 20061 of 61 © Boardworks Ltd 20061 of 61

A2-Level Maths: Statistics 2for Edexcel

S2.2 Continuous random variables

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© Boardworks Ltd 20062 of 61

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Relative frequency histograms


Probability density functions

Mode

Cumulative distribution functions

Median and quartiles

Expectation

Variance


A histogram has the important property that the area of each bar is in proportion to the corresponding frequency.

A histogram with unequal widths can be drawn by plotting the frequency density on the vertical axis, where:

frequencyfrequency density =

class width


frequency densityrelative frequency density =

total frequency

A relative frequency histogram is one in which the area of a bar corresponds to the proportion of the data falling into the corresponding interval.

The relative frequency is defined as:


Example: The relative frequency histogram below shows the distribution of ages of the UK population.


The area of each bar corresponds to the proportion of the population with ages in that interval.

The total area of all the bars is 1.


The distribution of the ages can be modelled by a curve.


This curve is called a probability density function.


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Mode



Expectation

Variance


A probability density function (or p.d.f.) is a curve that models the shape of the distribution corresponding to a continuous random variable.


A probability density function has several important properties.


If f(x) is the p.d.f corresponding to a continuous random variable X and if f(x) is defined for a ≤ x ≤ b then the following properties must hold:

f ( )b

a

x dx 1

i.e. the total area under a p.d.f. is 1.


1.


If f(x) is the p.d.f. corresponding to a continuous random variable X and if f(x) is defined for a ≤ x ≤ b then the following properties must hold:

2.


f ( ) x a x b 0 for

i.e. the graph of the p.d.f. never dips below the x-axis.


If f(x) is the p.d.f. corresponding to a continuous random variable X and if f(x) is defined for a ≤ x ≤ b then the following properties must hold:

3.


( ) f ( )x

x

P x X x x dx 2

1

1 2

i.e. probabilities correspond to areas under the curve.


Example: Sketch the graph of each of the following functions. Decide in each case whether it could be the equation of a probability density function:

f ( )x

x xx

3

11

0 1


f ( )x x x

x

2124 3 36 2 6

320 otherwise

f ( )x x x

x

2 4 3 0 4

0 otherwise3.

2.

1.



The function is non-negative everywhere.

If f represents a p.d.f., thenall

f ( )x

x dx 1

allf ( ) ( )

xx dx x x dx

62

2

124 3 36

32

( ) ( )

1 432 216 216 48 8 72 132

So f(x) could represent a probability density function.

1. f ( )x x x

x

2124 3 36 2 6

320 otherwise

x x x 62 3

2

112 36

32


The function is non-negative everywhere.

For f to represent a p.d.f. we need to check that

But:


dx x dx xx

3 2

31 1 1

1 12

all

f ( )x

x dx 1

So f(x) could not represent a probability density function.

2. f ( )x

x xx

3

11

0 1

1 10

2 2



The function is clearly negative for some values of x.

Consequently f(x) cannot represent a probability density function.

f ( )x x x

x

2 4 3 0 4

0 otherwise3.




Question 1: A continuous random variable X is defined by the probability density function

( )f ( )

k x xx

5 0 5

0 otherwise

Question 1

a) Sketch the probability density function.

b) Find the value of the constant k.

c) Find P(1 ≤ X ≤ 3).


a)

The diagram shows the probability density function.

Question 1

( )k x dx k x x

55 2

0 0

15 5

2

2

25k Therefore, 12.5k = 1

b) To find k, we can use the property that

( )f ( )

k x xx

5 0 5

0 otherwise

all

f ( )x

x dx 1

( . )k k 25 12 5 0 12.5


c) P(1 ≤ X ≤ 3)

( )

323

11

2 25 5

25 25 2

xx dx x

Question 1

= 0.48

( . ) ( . )2

15 4 5 5 0 525


Examination-style question 1: A continuous random variable X is defined by the probability density function

Examination-style question 1

( )

f ( ) ( )( )

k x x

x k x x x

1 1 3

5 2 3 5

0 otherwise



c) Find P(X > 2).



a) ( )

f ( ) ( )( )

k x x

x k x x x

1 1 3

5 2 3 5

0 otherwise


b) To find k, we can use the property that

Note that (5 – x)(x – 2) = 7x – 10 – x2


3 52 2 3

1 3

1 7 110 1

2 2 3k x x k x x x

( )3 5 2

1 31 7 10 1k x dx k x x dx

1 1 1 11 4 7 1

2 2 6 2k k

15 1

3k Therefore

all

f ( )x

x dx 1

So,

1i.e.

3

6k


c) P(X > 2) =


f ( )x dx5

2

x x x x x

3 52 2 3

2 3

3 1 3 7 110

16 2 16 2 3

( )3 5 2

2 3

3 31 7 10

16 16x dx x x dx

3 1 3 1 11 0 4 7

16 2 16 6 2

29

32 An alternative method would be to utilise P(X > 2) = 1 – P(X ≤ 2)


Examination-style question 2: The life, T hours, of an electrical component is modelled by the probability density function



c) Find P(1500 ≤ T ≤ 2000).

.f ( )

tke tt

0 001 10000 otherwise




Solution:

a)


b) To find k, we use the fact that f ( )t

t dt all

1

So:


. tke dt

0 001

1000

1

. tk e

0 001

10001000 1

( )k e 10 1000 1

.e

ke

1

1

10

000100000272Therefore

.f ( )

tke tt



c) P(1500 ≤ T ≤ 2000) = .f ( ) tt dt k e dt

2000 20000 001

1500 1500


. tk e 20000 001

15001000

.ee e 2 1 51000 1000

1000

= 0.239 (3 s.f.)

.f ( )

tke tt



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Mode



Mode



Expectation

Variance


Suppose that a random variable X is defined by the probability density function f(x) for a ≤ x ≤ b.

The mode of X is the value of x that produces the largest value for f(x) in the interval a ≤ x ≤ b.

A sketch of the probability density function can be very helpful when determining the mode.

Mode

Differentiationcould be usedto find the mode here.


Example: A random variable X has p.d.f. f(x), where

Find the mode.

( )f ( ) x x xx

2 2 0 20 otherwise

Sketch of f(x):

Mode


Mode

The mode can be found using differentiation:

f ( ) f ( )x x x x x x 2 3 22 4 3

To find a turning point, we solve f ( )x 0

( )4 3 0x x Factorize:

f ( ) f ( )x x 434 6 4 8 4 0

Check that gives the maximum value: 43x

So the mode is . 43x

24 3 0x x

or 430x x


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Mode



Expectation

Variance


The cumulative distribution function (c.d.f.) F(x) for a continuous random variable X is defined as

F(x) = P(X ≤ x).

Therefore, the c.d.f. is found by integrating the p.d.f..


f ( )x x

x

311 0 2

60 otherwise

Find the c.d.f. and find P(X < 1).



Solution: The c.d.f., F(x) is given by:

F( ) f ( )x x dx x dx 31 1

6 6


41 1

24 6x x c

To find the constant c we can use the fact that P(X ≤ 0) = 0

(because the random variable X is only non-zero between 0 and 2)

Therefore F(0) = 0, i.e. c = 0.



So the c.d.f. is F( )

x

x x x x

x

41 124 6

0 0

0 21 2

P(X < 1) = F(1) =1 1

24 6

5

24


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Mode



Expectation

Variance


The median, m, of a random variable X is defined to be the value such that

F(m) = P(X ≤ m) = 0.5

where F is the cumulative distribution function of X.

Likewise the lower quartile is the solution to the equation

F(x) = 0.25

and the upper quartile is the solution to

F(x) = 0.75.





Example: A random variable X is defined by the cumulative distribution function:

F( ) 2124

0 2

6 2 5

1 5

x

x x x x

x

a) Calculate and sketch the probability density function.

b) Find the median value.

c) Work out P(3 ≤ X ≤ 4).



a) We can get the p.d.f. by differentiating the c.d.f.

f ( ) F ( )x x x 1 112 24

So the p.d.f. is

f ( )x x

x

1 112 24 2 5

0 otherwise

Sketch of f(x):



The median must be 3.77 (as the p.d.f. is only non-zero for values in the interval [2, 5]).

2 6 12m m

b) The median, m, satisfies F(m) = 0.5.

Therefore .2124 6 0 5m m


( )1 1 4 1 18

2m

2 18 0m m

.m m 4 77 or 3.77



c) P(3 ≤ X ≤ 4) = F(4) – F(3)

2 21 124 244 4 6 3 3 6

13

7 112 4



f ( )x x x

x

23 3 34 2 4 1 2

a) Calculate the cumulative distribution function and verify that the lower quartile is at x = 2.

b) Work out the median value of X.


0 otherwise

x x 3 32 8 2 4


f ( ) F( )x x x x x x x c 2 3 23 3 3 3 314 4 4 4 42For ,


c 3 314 4 4 0

f ( ) , F( )x x x x x c 23 3 3 32 8 2 16For

We know that P(X ≤ 1) = 0, i.e., that F(1) = 0.

23 32 164 4 1c

F( )x x x x 3 23 31 14 4 4 4Therefore

We know that P(X ≤ 4) = 1, i.e. that F(4) = 1.

F( )x x x 23 32 16Therefore 2

So,

So 2c

a)

14c



F( )3 2

2

3 31 14 4 4 4

3 32 16

0 1

1 2

2 2 4

1 4

x

x x x xx

x x x

x

So

To verify that the lower quartile is 2, we simply need to check that F(2) = 0.25:

F( ) . 3 21 3 3 12 2 2 2 0 25

4 4 4 4

Therefore the lower quartile is 2.


b) The median, m, must lie in the interval [2, 4] because F(2) = 0.25.

To find the median we must solve F(m) = 0.5:

F( )m m m 23 3 1i.e. 2

2 16 2This can be rearranged to give the quadratic equation:

23 24 40 0m m

Using the quadratic formula,

As 5.63 does not lie in the interval [2, 4], the median must be 2.37.


24 576 4 3 40

6m

m = 5.63 or m = 2.37


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Expectation



Mode



Expectation

Variance


If X is a continuous random variable defined by the probability density function f(x) over the domain a ≤ x ≤ b, then the mean or expectation of X is given by

E[X] is the value you would expect to get, on average.

This mean value of X is also sometimes denoted μ.

[Note: if the p.d.f. is symmetrical, then the expected value of X will be the value corresponding to the line of symmetry].

We can also find the expected value of g(X), i.e. any function of X:

Expectation

[g( )] g( )f ( )b

a

X x x dxE

[ ] f ( )b

a

X x x dxE


Example: A random variable X is defined by the probability density function

f ( )x

x x

3

21

0 otherwise

Calculate the value of E[X] and E[1/X].

[ ] f ( ) .x

X x x dx x dxx

3all 1

2E

21

2dx

x

Expectation

2 1

11

2 2x dx x

( ) ( )0 2 2


f ( ) .x

x dx dxX x x x

3all 1

1 1 21E

4

1

2x dx

Expectation

3

1

2

3x

( )2

03

So, [ ]1E X 2

3

2

3


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Variance



Mode



Expectation

Variance


If X is a continuous random variable defined by the probability density function f(x) over the domain a ≤ x ≤ b, then the variance of X is given by

[ ]22Var[ ] E EX X X

or

Variance

[ ] f ( )b

a

X x x dx μ 2 2Var

The standard deviation of X is the square root of the variance.

The standard deviation is sometimes denoted by the symbol σ.


Example: A continuous random variable Y has a probability density function f(y) where

( )f ( )

332 4 0 4

0 otherwise

y y yy

Sketch of f(y):

The p.d.f. is symmetrical. Therefore E[Y] = 2.

Variance

Calculate the value of Var[Y].


[ ] f ( ) . ( )4 4

2 2 2

0 0

3E 4

32Y y y dy y y y dy

Variance

( )4

3 4

0

34

32y y dy

( )4 53 14 4 0

32 5

44

5

44 5

0

3 1

32 5y y

Therefore Var[Y] = 24

4 25

4

5


Example: A continuous random variable x has a probability density function f(x) where

f ( )

14

38 16

0 2

2 6

0 otherwise

x

x

x x

Calculate

a) the mean value, μ.

b) the standard deviation, σ.

Variance


Variance

[ ] . .2 6

0 2

1 3E

4 8 16

xX x dx x dx

2 6 2

0 2

3

4 8 16

x x xdx dx

2 62 2 3

0 2

3

8 16 48

x x x

4 1 70 2

8 4 12

f ( )

14

38 16

0 2

2 6

0 otherwise

x

x

x x

1

26

a)


Variance

[ ] . .2 6

2 2 2

0 2

1 3E

4 8 16

xX x dx x dx

2 62 2 3

0 2

3

4 8 16

x x xdx dx

2 63 3 4

0 212 8 64

x x x

2 3 30 6

3 4 4 , [ ]

22 1 35

So Var 6 2 13 6 36

X

f ( )

14

38 16

0 2

2 6

0 otherwise

x

x

x x

b)

26

3

Therefore σ = 7136 = 1.40 (3 s.f.)


Examination-style question: The mass, X kg, of luggage taken on board an aircraft by a passenger can be modelled by the probability density function

( )f ( )

3 30 0 30

0 otherwise

kx x xx

a) Sketch the probability density function and find the value of k.

b) Verify that the median weight of luggage is about 20.586 kg.

c) Find the mean and the variance of X.

Examination-style question


To find k we use ( )30

3

0

30 1kx x dx


1215000 0 1k 1

1215000k

( )30

3 4

0

30 1k x x dx 30

4 5

0

30 11

4 5k x x

a)

( )f ( )

3 30 0 30

0 otherwise

kx x xx


.

( )20 586

3

0

30kx x dx

To verify that the median is about 20.586, we need to check that P(X ≤ 20.586) = 0.5


1607525 0

1215000

.0 500

.20 5864 5

0

1 30 1

1215000 4 5x x

b) ( )f ( )

3 30 0 30

0 otherwise

kx x xx

P(X ≤ 20.586) =


( )f ( )

3 30 0 30

0 otherwise

kx x xx

[ ] . ( ) ( )30 30

3 4 5

0 0

E 30 30X x kx x dx k x x dx


124300000 0

1215000

20

305 6

0

1 16

1215000 6x x

c)


[ ] . ( ) ( )30 30

2 2 3 5 6

0 0

E 30 30X x kx x dx k x x dx


.428 5714

[ ] . ( ).2Var 428 5714 20 to428 s.f.57X

306 7

0

1 15

1215000 7x x

c)

Therefore, .

( )f ( )

3 30 0 30

0 otherwise

kx x xx

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© Boardworks Ltd 20061 of 61 © Boardworks Ltd 2006 1 of 61 A2-Level Maths: Statistics 2 for Edexcel S2.2 Continuous random variables This icon indicates