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Transcript of © Boardworks Ltd 20061 of 61 © Boardworks Ltd 2006 1 of 61 A2-Level Maths: Statistics 2 for...
© 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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For more detailed instructions, see the Getting Started presentation.
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Relative frequency histograms
Relative frequency histograms
Probability density functions
Mode
Cumulative distribution functions
Median and quartiles
Expectation
Variance
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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
Relative frequency histograms
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:
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Example: The relative frequency histogram below shows the distribution of ages of the UK population.
Relative frequency histograms
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.
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The distribution of the ages can be modelled by a curve.
Relative frequency histograms
This curve is called a probability density function.
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Probability density functions
Relative frequency histograms
Probability density functions
Mode
Cumulative distribution functions
Median and quartiles
Expectation
Variance
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A probability density function (or p.d.f.) is a curve that models the shape of the distribution corresponding to a continuous random variable.
Probability density functions
A probability density function has several important properties.
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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.
Probability density functions
1.
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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.
Probability density functions
f ( ) x a x b 0 for
i.e. the graph of the p.d.f. never dips below the x-axis.
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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.
Probability density functions
( ) f ( )x
x
P x X x x dx 2
1
1 2
i.e. probabilities correspond to areas under the curve.
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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
Probability density functions
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.
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Probability density functions
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
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The function is non-negative everywhere.
For f to represent a p.d.f. we need to check that
But:
Probability density functions
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
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Probability density functions
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.
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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).
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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
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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
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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
a) Sketch the probability density function.
b) Find the value of the constant k.
c) Find P(X > 2).
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Examination-style question 1
a) ( )
f ( ) ( )( )
k x x
x k x x x
1 1 3
5 2 3 5
0 otherwise
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b) To find k, we can use the property that
Note that (5 – x)(x – 2) = 7x – 10 – x2
Examination-style question 1
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
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c) P(X > 2) =
Examination-style question 1
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)
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Examination-style question 2: The life, T hours, of an electrical component is modelled by the probability density function
a) Sketch the probability density function.
b) Find the value of the constant k.
c) Find P(1500 ≤ T ≤ 2000).
.f ( )
tke tt
0 001 10000 otherwise
Examination-style question 2
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b) To find k, we use the fact that f ( )t
t dt all
1
So:
Examination-style question 2
. 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
0 001 10000 otherwise
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c) P(1500 ≤ T ≤ 2000) = .f ( ) tt dt k e dt
2000 20000 001
1500 1500
Examination-style question 2
. tk e 20000 001
15001000
.ee e 2 1 51000 1000
1000
= 0.239 (3 s.f.)
.f ( )
tke tt
0 001 10000 otherwise
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Mode
Relative frequency histograms
Probability density functions
Mode
Cumulative distribution functions
Median and quartiles
Expectation
Variance
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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.
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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
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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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Cumulative distribution functions
Relative frequency histograms
Probability density functions
Mode
Cumulative distribution functions
Median and quartiles
Expectation
Variance
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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..
Example: A random variable X has p.d.f. f(x), where
f ( )x x
x
311 0 2
60 otherwise
Find the c.d.f. and find P(X < 1).
Cumulative distribution functions
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Solution: The c.d.f., F(x) is given by:
F( ) f ( )x x dx x dx 31 1
6 6
Cumulative distribution functions
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.
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Cumulative distribution functions
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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Median and quartiles
Relative frequency histograms
Probability density functions
Mode
Cumulative distribution functions
Median and quartiles
Expectation
Variance
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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.
Median and quartiles
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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).
Median and quartiles
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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):
Median and quartiles
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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
Median and quartiles
( )1 1 4 1 18
2m
2 18 0m m
.m m 4 77 or 3.77
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Median and quartiles
c) P(3 ≤ X ≤ 4) = F(4) – F(3)
2 21 124 244 4 6 3 3 6
13
7 112 4
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Example: A random variable X has p.d.f. f(x), where
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.
Median and quartiles
0 otherwise
x x 3 32 8 2 4
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f ( ) F( )x x x x x x x c 2 3 23 3 3 3 314 4 4 4 42For ,
Median and quartiles
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
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Median and quartiles
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.
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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.
Median and quartiles
24 576 4 3 40
6m
m = 5.63 or m = 2.37
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Expectation
Relative frequency histograms
Probability density functions
Mode
Cumulative distribution functions
Median and quartiles
Expectation
Variance
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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
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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
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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
Relative frequency histograms
Probability density functions
Mode
Cumulative distribution functions
Median and quartiles
Expectation
Variance
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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.
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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].
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[ ] 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
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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
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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)
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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.)
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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
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To find k we use ( )30
3
0
30 1kx x dx
Examination-style question
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
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.
( )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
Examination-style question
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) =
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( )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
Examination-style question
124300000 0
1215000
20
305 6
0
1 16
1215000 6x x
c)