To specify the distributions for continuous random variables using probability
density functions .
To calculate and interpret the expectation (mean) , median , mode , variance and
standard deviation for a continuous random variable.
To relate the graph of a probability density function to the values of the
parameters defining that function.
To relate probabilities for intervals to the graph of a probability density function.
To use calculus to calculate probabilities for intervals for a probability density function.
To use technology to calculate probabilities for intervals for a probability density
function.
To apply knowledge of probability density functions for continuous random
variables to solve probability-related problems.
17.1 Continuous random variablesA continuous random variable is one that can take any value in an interval of the real number line. For example, if X is the random variable which takes its values as ‘distance in metres’ thata parachutist lands from a particular marker, then X is a continuous random variable, and herethe values which X may take are the non-negative real numbers.
A continuous random variable has no limit as to the accuracy with which it can bemeasured. For example, let W be the random variable with values ‘a person’s weight inkilograms’, and W i be the random variable with values ‘a person’s weight in kilograms
measured to the ith decimal place’.
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Chapter 17 — Continuous random variables and their probability distributions 595
Then: W 0 =83 implies 82 .5 ≤ W < 83.5W 1 =83 .3 implies 83 .25 ≤ W < 83.35W 2 =83 .28 implies 83 .275 ≤ W < 83.285W 3 =83 .281 implies 83 .2805 ≤ W < 83.2815
and so on. Thus, the random variable W cannot take an exact value, since it is always rounded to the limits imposed by the method of measurement used. Hence, the probability of W beingexactly equal to a particular value is zero, and this is true for all continuous random variables.
That is,
Pr(W =w ) =0 for all w
In practice, considering W taking a particular value is equivalent to W taking a value in anappropriate interval.
Thus, from above:
Pr(W 0
=83)
=Pr(82 .5
≤ W < 83 .5)
To determine the value of this probability you could begin by measuring the weight of alarge number of randomly chosen people, and determine the proportion of the people in thegroup who have weights in this interval.
Suppose after doing this a histogram of weights was obtained as shown.
8382.5 83.5
Weights0
f (w)
From this histogram:
Pr(W 0 =83) =Pr(82 .5 ≤ W < 83.5)
=shaded area from 82.5 to 83.5
total areaIf the histogram is scaled so that the total area under the blocks is one, then:
P(W 0 =83) =Pr(82 .5 ≤ W < 83.5)
= area under block from 82.5 to 83.5
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Suppose that the sample size gets larger and the class interval width gets smaller. If theoretically this process is continued so that the intervals are arbitrarily small, the histogramcan be modelled by a smooth curve, as shown in the following diagram.
w
f (w)
8382.5 83.5
0
The curve obtained here is of great importance for a continuous random variable.
The function f whose graph models the histogram as the number of intervals isincreased is called the probability density function . The probability density function f is used to describe the probability distribution of a continuous random variable, X .
Now, the probability of interest is no longer represented by the area under the histogram, butthe area under the curve. That is:
Pr(W 0
=83)
=Pr(82 .5
≤W < 83.5)
= area under the graph of the function with rule f (w ) from 82.5 to 83.5
= 83.5
82.5 f (w )dw
In general, for the continuous random variable X with density function f :
Pr(a < X < b) = b
a f ( x)dx is given by the area of the shaded region.
x 0 a b
f ( x )
In order to be a probability density function a function must satisfy certain conditions.
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Chapter 17 — Continuous random variables and their probability distributions 597
If the range of the continuous random variable X is [a , b] then the domain of its probabilitydensity function f is [a , b]. The probability density function will satisfy two properties:
1 f ( x) ≥0 for all x ∈[a , b] and 2 b
a f ( x)dx =1
Note that Pr( X < c)
=Pr( X
≤c)
= c
a f ( x)dx
The probability density function does not give probabilities and f ( x) may take valuesgreater than one. The probabilities are given by areas determined by the graph of f .
In many cases the range of X is an unbounded interval, for example [1 , ∞) or indeed R.Therefore, some new notation is necessary.
If the range of X is [1, ∞), then it would be required that the area of the region under the
graph of the probability density function be one. This can be expressed as
∞1
f ( x)dx =1.
This is computed as limk →∞
k
1 f ( x)dx . If the probability density function f has domain R, then
∞
−∞ f ( x)dx =1. This is computed as lim
k →∞ k
−k f ( x)dx .
Any probability density function f with domain [ a , b] (or any other interval) may beextended to a function with domain R by dening f ∗( x) = f ( x) for all x ∈[a , b] and f ∗( x) =0 for all x /
∈[a , b].
This leads to the following:
A probability density function (or its natural extension) satises the following two properties:1 f ( x) ≥0 for all x 2 ∞
−∞ f ( x)dx =1
Note that, since the probability of X taking any exact value is zero, then:
Pr(a < X < b) =Pr(a ≤ X < b) =Pr(a < X ≤b) =Pr(a ≤ X ≤b)
That is, there is no difference between the numerical values of all of these expressions.
Example 1
Suppose that the random variable X has the density function with the rule:
f ( x) =cx 0 ≤ x ≤20 if x > 2 or x < 0
a Find the value of c that makes f a probability density function.b Find Pr( X > 1.5).
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b It is known that f ( x) ≥ 0 for all x as required.
The second condition is that ∞
−∞
f ( x)dx = 1
Now ∞
−∞
2e− 2 xdx = ∞
02e− 2 xdx , since f ( x) = 0 for x ≤ 0
To evaluate this integral, consider limk →∞
k
02e− 2 xdx
= limk →∞
2e− 2 x
− 2
k
0
= limk →∞
− e− 2 x k 0
= limk →∞
(− e− 2k ) − (− e− 0)
= 0 + e− 0
= 1
Thus f satises the requirements of a probability density function.
c Pr( X > 1) = limk →∞
k
12e− 2 xdx
= limk →∞
2e− 2 x
− 2
k
1
= limk →∞
− e− 2 x k 1
= limk →∞
(− e− 2k ) − (− e− 2)
= 0 + e− 2
=1e2
= 0 . 1353 , correct to four decimal places
Using the TI-NspireThis is an application of integration.a The graph is as shown.The hybrid function template, , has been used in this example. Access thetemplate using t , (/ + r on theClickpad)
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Chapter 17 — Continuous random variables and their probability distributions 601
b and c The required integrations are performed to achieve the results.∞ , the symbol for innity, can be found using or / + k (or / + j on theClickpad)
Using the Casio ClassPadThis is an application of integration.a The graph is as shown.b and c The integrations are performed to achieve the results.
Exercise 17A
1 Show that the function f with rule
f ( x) =
24 x3 3 ≤ x ≤ 6
0 x < 3 or x > 6
is a probability density function.
2 Let X be a continuous random variable with the following probability density function:
f ( x) = x2 + kx + 1 0 ≤ x ≤ 20 x < 0 or x > 2
Determine the constant k such that f is a valid probability density function.
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Chapter 17 — Continuous random variables and their probability distributions 603
8 A probability density function has a probability density function given by
f ( x) =k (1 + x) −1 ≤ x ≤0k (1 − x) 0 < x ≤10 x < −1 or x > 1
where k > 0.a Sketch the probability density function. b Evaluate k.c Find the probability that X lies between −0.5 and 0.5.
9 Let X be a continuous random variable with probability density function given by:
f ( x) =3 x2 0 ≤ x ≤10 x < 0 or x > 1
a Sketch the graph of f ( x).b Find Pr(0 .25 < X < 0.75) and sketch this on your graph.
10 A random variable X has a probability density function f with the rule:
f ( x) =
1100
(10 + x) −10 < x ≤0
1100
(10 − x) 0 < x ≤10
0 x ≤ −10 or x > 10
a Sketch the graph of f . b Find Pr(−1 ≤ X < 1).
11 The life in hours, X , of a type of light globe has a probability density function with therule:
f ( x) =k
x2 x > 1000
0 x ≤1000
a Evaluate k . b Calculate the probability the globe will last at least 2000 hours.
12 The weekly demand for petrol, X (in thousands of litres), at a particular service station is arandom variable with probability density function:
f ( x) =2 1
−1
x2 1
≤ x
≤2
0 x < 1 or x > 2
a Determine the probability that more than 1.5 thousand litres are demanded in 1 week.b Determine the probability that the demand for petrol in 1 week is less than
1.8 thousand litres, given than it is more than 1.5 thousand litres.
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13 The length of time, X (in minutes), between the arrival of customers at an autobank is arandom variable with probability density function:
f ( x) =15
e− x5 x ≥0
0 x < 0
a Find the probability that more than 8 minutes elapses between successive customers.b Find the probability that more than 12 minutes elapses between successive customers,
given that more than 8 minutes has passed.
14 A random variable X has density function given by:
f ( x) =0.2 −1 < x ≤00.2 +1.2 x 0 < x ≤10 x ≤ −1 or x > 1
a Find Pr( X
≤0.5). b Hence nd Pr( X > 0.5
| X > 0.1).
15 The continuous random variable X has probability density function f given by:
f ( x) =e− x x ≥00 x < 0
a Sketch the graph of f .b Find:
i Pr( X < 0.5) ii Pr( X ≥1) iii Pr( X ≥1| X > 0.5)
16 The continuous random variable X has probability density function f given by
f ( x) = |ax | −2 ≤ x ≤20 x < −2 or x > 2
where a is a constant.
a Sketch the graph of f. b Hence or otherwise, nd the value of a.
17.2 Cumulative distribution functions ∗Another function of importance in describing a continuous random variable is thecumulative distribution function F , or CDF. For a continuous random variable X , with
probability density function f , the cumulative distribution function dened on an interval[a , b] is given by
f ( x) =Pr( X ≤ x)
= x
a f (t )dt
where f is the probability density function and t is the variable of integration.
∗ Cumulative distribution functions are not mentioned explicitly in the Mathematical Methods Units 3 and 4 CAS studydesign but are useful in our study of continuous probability distributions.
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Chapter 17 — Continuous random variables and their probability distributions 605
For the probability density function dened on R:
f ( x) =Pr( X ≤ x)
= x
−∞ f (t )dt
The graphical representation of the relationship between the probability density function and thecumulative distribution function is shown in thediagram on the right.
f (t )
F ( x)
t x
It is clear that this result is an application of thefundamental theorem of calculus and the derivativeof F ( x) is f ( x), i.e. the derivative of the cumulativedistribution function is the density function.
That is, the function F describes the area under the probability density function between the
lower bound of the domain of f and x (in the diagram, the lower bound is 0). So, to nd Pr( X ≤n) for a particular random variable X , you would substitute the value of n into theexpression for the cumulative distribution function and evaluate.
Thus, the cumulative distributionfunction for a particular value x givesthe probability that the random variable X takes a value less than or equal to thatvalue. As X is a continuous randomvariable, the cumulative distribution
function is also continuous. 0
F ( x )
x
1
There are three important properties of a cumulative distribution function witha probability density function with domain [ a , b].
If F is a cumulative distribution function, with probability density function withdomain [ a , b]:1 The probability that the random variable X takes a value less than a is zero. That is:
F (a ) =0
2 The probability that the random variable X takes a value less than or equal to b is 1.That is:
F (b) =1
3 If x1 and x2 are values of x such that x1 ≤ x2, then:
Pr( X ≤ x1) ≤Pr( X ≤ x2)That is F ( x1) ≤ F ( x2)
The function F is a non-decreasing function.
For a probability density function dened on R, F (−∞) =0 and F (∞) =1
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The time (in seconds) it takes a student to complete a puzzle is a random variable with adensity function with the rule:
f ( x) =5
x2 x ≥5
0 x < 5
Find F ( x), the cumulative distribution function of X .
Solution
F ( x) = x
5 f (t )dt
= x
5
5
t 2
= −5t
x
5
= −5 x +1
That is: F ( x) = 1 −5 x
The importance of the cumulative distribution function is that probabilities for various intervals can be computed directly from F ( x).
Example 5
Use the cumulative distribution function found in Example 4 to nd:a the probability that a student takes less than 12 seconds to complete the puzzleb the probability that a student takes between 8 and 10 seconds to complete the puzzle, given
that he takes less than 12 seconds
Solution
a Pr( X ≤12) can be obtained by substituting 12 in the expression for F ( x).
F (12) = 1 −5
12
=7
12
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Chapter 17 — Continuous random variables and their probability distributions 607
b Pr(8 ≤ X ≤10| X ≤12) =Pr(8 ≤ X ≤10 ∩ X ≤12)
Pr( X ≤12) =Pr(8 ≤ X ≤10)
Pr( X ≤12)
= F (10) − F (8)
F (12)
=12 −
38
712
=3
14
Exercise 17B
1 The probability density function for a random variable X is:
f ( x) =15
0 < x < 5
0 x ≤0 or x ≥5
a Find F ( x), the cumulative distribution function of X . b Hence nd Pr( X ≤3).
2 A random variable X has a cumulative distribution function:
F ( x) =0 x < 01 −e− x2
x ≥0
a Sketch the graph of F ( x). b Find Pr( X
≥2).
c Find Pr( X ≥2| X < 3).
3 The continuous random variable X has cumulative distribution function F given by:
F ( x) =0 x < 0kx2 0 ≤ x ≤61 x > 6
a Determine the value of the constant k . b Calculate Pr 12 ≤ X ≤1 .
17.3 Mean, median and mode for a continuousrandom variableThe centre is an important summaryfeature of a probability distribution.The diagram shows two probability distributions which areidentical except for their centres.
f ( x )
x
More than one measure of centre may be determined for a continuous random variable, and each gives useful information about the random variable under consideration. The most
generally useful measure of centre is the mean.
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MeanIn the same way as the mean or expected value for a discrete random variable is dened, onecan dene the mean or expected value for a continuous random variable.
The mean or expected value of a continuous random variable X is given by:
E( X ) = ∞
−∞ x f ( x) dx
provided the integral exists. As in the case of a discrete random variable the mean isalways denoted by the Greek letter (mu).
If f ( x) =0 for all x /∈
[a , b], then E( X ) = b
a x f ( x)dx
This denition is consistent with the denition of the expected value for a discrete randomvariable. As in the case of a discrete random variable, the mean or expected value of acontinuous random variable is not necessarily the value the random variable takes in any particular experiment, but the long-run average value of the variable. As an example, consider the daily demand for petrol at a service station. The mean of this variable tells us the averagedaily demand for petrol over a very long period of time.
Example 6
Find the expected value of the random variable X which has probability density function withrule:
f ( x) = 0.5 x 0 ≤ x ≤20 x < 0 or x > 2
SolutionBy denition E( X ) = ∞
−∞ x f ( x)dx
= 2
0 x ×0.5 xdx since f ( x) =0 elsewhere
=0.5
2
0
x2dx
=0.5 x3
3
2
0
=43
The mean of a function of X is dened as follows. (In this case the function of X is denoted as g ( X ) and is the composition of the random variable X and the function g .)
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Chapter 17 — Continuous random variables and their probability distributions 609
The expected value of g ( X ) is
E[ g ( X )] = ∞
−∞ g ( x) f ( x) dx
provided the integral exists.
Generally, as in the case of a discrete random variable, the expected value of a function of X is not equal to that function of the expected value of X . That is:
E{ g ( x)}= g {E( X )}Example 7
If X is the random variable with probability density function f :
f ( x)
=0.5 x 0
≤ x
≤2
0 x < 0 or x > 2
Find:a the expected value of X 2 b the expected value of e x (correct to four decimal places)
Solution
a E[ X 2] = ∞
−∞ x2 f ( x)dx
= 2
0 x2
×0.5 xdx since f ( x) =0 elsewhere
=0.5 2
0 x3dx
=0.5 x4
4
2
0
=2
b E[e x] = ∞
−∞e x f ( x)dx
= 2
0e x f ( x)dx since f ( x) =0 elsewhere
= 2
0e x ×0.5 xdx
=4.195 (correct to three decimal places)
A case where the equality does hold is where g is a linear function of X . Then
E(a X +b) = a E( X ) +b (a , b constant)
Percentiles and the medianAnother value of interest is the value of X which bounds a particular area under a probabilitydensity function. For example, a teacher may wish determine the mark ( p) below which lie
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75% of all students’ marks. This is called the 75th percentile of the population, and is found bysolving:
p
−∞ f ( x)dx =0.75
This can be stated more generally.
The value p of X which is the solution of an equation of the form
p
−∞ f ( x)dx =q
is called a percentile of the distribution. Here q is the required percentile. (For theexample above q =75% =0.75.)
Example 8
The duration of telephone calls to the order department of a large company is a randomvariable X minutes with probability density function:
f ( x) =13
e− x3 x > 0
0 x ≤0
Find the value of a such that 90% of phone calls last less than a minutes.
Solution
To nd the value of a , solve the equation:
a
0
13
e− x3 dx =0.90
That is, −e− x3
a
0 = 0.90
∴ 1 −e−a3 =0.90
∴ −a3 = loge 0.10
∴ a =3log e (10)
= 6.908 (correct to three decimal places)
So 90% of the calls to this company last less than 6.908 minutes.
As for discrete probability distributions, a percentile of special interest is the median, or 50th percentile. The median is the middle value of the distribution. That is, the probability of X taking a value below the median is 0.5, and the probability of X taking a value above themedian is 0.5. Thus, if m is the median value of the distribution, then:
Pr( X
≤m)
=Pr( X > m)
= 0.5
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Chapter 17 — Continuous random variables and their probability distributions 611
The median may be determined readily from the cumulative distribution function, as it is the50th percentile of the distribution. Graphically, the median is the value of the random variablewhich divides the area under the probability density function in half.
The median is another measure of centre of
a continuous probability distribution.The median, m, of a probability distributionfunction is the value of X such that:
m
−∞ f ( x)dx =0.5
f ( x )
x m
0.5
Example 9
Suppose the probability density function of weekly sales of topsoil, X (in tonnes), is given bythe rule:
f ( x) =2(1 − x) 0 ≤ x ≤10 x < 0 or x > 1
Find the median value of X , and interpret.
Solution
The solution here is the value of m such that:
m
02(1 − x) =0.5
∴ 2 x − x2
2
m
0 =0.5
∴ 2m −m2 =0.5∴ m2
−2m
+0.5
=0
m =0.293 or 1 .707
But since 0 ≤ x ≤1 the median m =0.293 tonnes. That is, in the long run, 50% of weekly sales will be less than 0.293 tonnes, and 50% will be more.
ModeAnother value of interest is the mode, M , of the distribution. Although not actually a measureof centre the mode does tell us the most likely or most probable value of X . Thus, the modeoccurs at the maximum value of f ( x), and may be located through considering the graph of
the probability density function.
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Chapter 17 — Continuous random variables and their probability distributions 613
Example 11
If the probability density function of weekly sales of topsoil, X (in tonnes), is as given inExample 9, nd the most likely weekly sales.
Solution
The most likely weekly sales correspond tothe mode of f ( x). Since the probability densityfunction is a linear function of x, and moreimportantly strictly decreasing, calculus willnot help, so the solution will need to bedetermined graphically.
f ( x )
x
2
10Consideration of the graph of f shows that the
maximum value of f ( x) occurs when x =0.
Exercise 17C
1 Find the mean, E( X ), of the continuous random variables with the following probabilitydensity functions:
a f ( x) =2 x, 0 < x < 1 b f ( x) =1
2√ x , 0 < x < 1
c f ( x) =6 x(1 − x), 0 < x < 1 d f ( x) = 1 x2 , x ≥1
2 Use your calculator to rstly check that each of the following is a probability densityfunction, and then to nd the mean, E( X ), of the continuous random variables with thefollowing probability density functions:
a f ( x) = sin x, 0 < x <2
b f ( x) = loge x , 1 < x < e
c f ( x) =1
sin2 x,
4 < x <
2d f ( x) = −4 x loge x , 0 < x < 1
3 A continuous random variable X has the probability density function given by:
f ( x) =2 x3 − x +1 0 ≤ x ≤10 x < 0 or x > 1
a Find the mean value of X , .b Find the probability that X takes a value less than or equal to the mean.
4 Consider the probability density function given by:
f ( x) =1
2(1 +cos x), − ≤ x ≤
Find the expected value of X .
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Chapter 17 — Continuous random variables and their probability distributions 615
11 The cumulative distribution function ∗ of a continuous random variable X , F ( x), is given by:
F ( x) =04 x3 −3 x4
1
x < 00 ≤ x ≤1 x > 1
Find:a the probability density function of X b the mode, M , of X
12 A random variable X has a probability density function:
f ( x) =12
sin x 0 < x <
0 x ≤0 or x ≥Find the mode, M , of X .
13 A continuous random variable X has the probability density function given by:
f ( x) = x 0 ≤ x < 12 − x 1 ≤ x < 20 x < 0 or x ≥2
a Find , the expected value of X . b Find m, the median value of X .c Find M , the mode of X .
14 Let the probability density function of X be:
f ( x) =30 x4(1
− x) 0 < x < 1
0 x < 0 or x ≥1
a Find the expected value of X , .b Find the median value of X , m, and hence show that the mean is less than the median.
15 A probability model for the mass x kg of a 2-year-old child is given by:
f ( x) =1
20sin
110
( x −7) , 7 ≤ x ≤17
a Find the mode of X , M . b Find the median value of X , m.
16 A random variable X has density function given by:
f ( x) =0.2 −1 < x ≤00.2 +1.2 x 0 < x ≤10 x < −1 or x > 1
a Find , the expected value of X . b Find m, the median value of X .c Find M , the mode of X .
∗ Cumulative distribution functions are not mentioned explicitly in the Mathematical Methods Units 3 and 4 CAS studydesign but are useful in our study of continuous probability distributions.
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17 The exponential probability distribution describes the distribution of the time betweenrandom events, such as phone calls. The general form of the exponential distribution with parameter is:
f ( x) =1
e− x
x ≥0
0 x < 0a Differentiate kxe−kx and hence nd an antiderivative of kxe−kx.b Show that the mean of an exponential random variable is .c On the same axes, sketch the graph of an exponential random variable for:
i =12
ii =1 iii =2
d Describe the effect of the variation of the parameter on the graph of the distribution.
17.4 Measures of spreadAnother important summary feature of a distribution is variation or spread. The followingdiagram shows two distributions that are identical except for their spreads.
x
f ( x )
As in the case of centre, there is more than one measure of spread. The most commonlyused is the variance, together with its companion measure, the standard deviation. Others thatyou may be familiar with are the range and the interquartile range.
Variance and standard deviationThe variance of the random variable X is a measure of the spread of a probability distribution
about its mean or expected value . It is dened as follows.
Var( X ) =E[( X − )2]
= ∞
−∞( x − )2 f ( x)dx
As for discrete random variables the variance is usually denoted 2, where is thelower case of the Greek letter sigma.
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Chapter 17 — Continuous random variables and their probability distributions 617
Variance may be considered as the long-run average value of the square of the distance fromthe values of X to . Since the variance is determined by squaring the distance from the valuesof X to it is no longer in the units of measurement of the original random variable. Ameasure of spread in the appropriate unit is found by taking the square root of the variance.
The standard deviation of X is dened as:sd( X ) = Var( X )
The standard deviation is usually denoted .
As in the case of discrete random variables, an alternative (computational) formula for variance is generally used.
To calculate variance, use:
Var( X )
=E( X 2)
−2
The computational form of the expression for variance is derived as follows:
Var( X ) = ∞
−∞( x − )2 f ( x)dx
= ∞
−∞( x2
−2 x +2) f ( x)dx
= ∞
−∞ x2 f ( x)dx − ∞
−∞2 x f ( x)dx + ∞
−∞2 f ( x)dx
=E( X 2
) −2 ∞
−∞ x f ( x)dx +2
∞
−∞ f ( x)dx
Since ∞
−∞ x f ( x)dx = and ∞
−∞ f ( x)dx =1
Var( X ) =E( X 2) −2 2+
2
=E( X 2) − 2
Example 12
Find the variance and standard deviation of the random variable X which has the probabilitydensity function f with rule:
f ( x) =0.5 x 0 ≤ x ≤20 x < 0 or x > 2
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Use the computational formula Var( X ) =E( X 2) − 2
First, evaluate E( X 2):
E( X 2) =
∞
−∞
x2 f ( x)dx
= 2
0 x2×0.5 xdx
=0.5 2
0 x3dx
=0.5 x4
4
2
0
=0.5 ×4
=2
Now E( X ) = 43
(from Example 6)
so Var( X ) =2 −43
2
=29
and sd( X ) = 29
=
√ 23
or, correct to three decimal places , 0.471
It helps to make the standard deviation more meaningful to give it an interpretation whichrelates to the probability distribution. As already stated for discrete random variables, it is alsothe case for many continuous random variables that about 95% of the distribution lies withintwo standard deviations either side of the mean.
In general, for many continuous random variables X :
Pr( −2 ≤ X ≤ +2 ) ≈ 0.95
Example 13
The life of a certain brand of battery, X , is a continuous random variable with mean 50 hoursand variance 16 hours. Find an (approximate) interval for the time period for which 95% of the batteries would be expected to last.
Solution
Pr( −2 ≤ X ≤ +2 ) ≈ 0.95. In this case =50 and =√ 16 =4, so onecould expect 95% of the batteries to last between 42 hours and 58 hours.
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Chapter 17 — Continuous random variables and their probability distributions 619
RangeThe range of a random variable is the difference between the smallest and the largest value thevariable may take, and is clearly determined from the domain of the probability densityfunction.
Example 14
Determine the range of the random variable X which has the probability density function:
f ( x) =19
(4 x − x2) 0 ≤ x ≤3
0 x < 0 or x > 3
Solution
From the probability density function X may take values from 0 to 3. Thus:
range of X =3 −0
=3
Interquartile rangeThe interquartile range is the range of the middle 50% of the distribution. It is the difference between the 75th percentile, also known as Q3, and the 25th percentile, also known as Q1.
Example 15
Determine the interquartile range of the random variable X which has the probability densityfunction:
f ( x) =2 x 0 ≤ x ≤10 x < 0 or x > 1
Solution
To nd the 25th percentile a , solve:
a
02 xdx =0.25
⇒[ x2]a
0 =0.25⇒
a 2 =0.25⇒
a =√ 0.25 =0.5
To nd the 75th percentile b, solve:
b
02 xdx =0.75
⇒[ x2]b
0 =0.75⇒
b2=0.75
⇒b =√ 0.75 =0.866 ,
correct to three decimal places
Thus the interquartile range is 0.866 −0.5 =0.366, correct to three decimal places. Note that the negative solutions to each of these equations were not appropriate, as
0 ≤ x ≤1.
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The validity of this statement can be readily demonstrated.
E(aX +b) = ∞
−∞(ax +b) f ( x)dx
=
∞
−∞
ax f ( x)dx +
∞
−∞
b f ( x)dx
= a ∞−∞
x f ( x)dx +b ∞−∞
f ( x)dx
= a E( X ) +b , since ∞
−∞ f ( x)dx =1
Consider the variance of a linear function of X , Var( aX +b).
Var( a X +b) =E(aX +b)2−[E(a X +b)]2
Now [E(a X +b)]2= [a E( X ) +b]2
= [a +b]2= a 2 2 +2ab +b2
and E(a X +b)2=E(a 2 X 2 +2ab X +b2)
= a 2E( X 2) +2ab +b2
Thus Var( a X +b) = a 2E( X 2) +2ab +b2 −a 2 2 −2ab −b2
= a 2E( X 2) −a 2 2
= a 2Var( X )
That is:
For any random variable X :
Var( ax +b) = a 2 Var( X )
Although initially the absence of b in the result may seem surprising, on reection it makessense that adding a constant to X merely translates the probability density function of X , buthas no effect on the spread of the resultant distribution. If the random variable X has a probability density function with the rule f ( x), then the probability distribution with randomvariable X +b has a corresponding probability density function with the rule f ( x −b).
Similarly, multiplying by a is similar to a dilation by a factor of a , and this is consistent withthe answer obtained. However, there has to be an adjustment to determine the probabilitydensity function of the random variable aX as the transformation must be area-preserving. This
is chosen as the function with the rule 1
a f
x
a.
Thus if f ( x) is the probability density function for the random variable X , then the
probability density function of the random variable a X +b is 1a
f x −b
a. The
transformation is described by ( x, y) → ax +b, ya
. That is, if a and b are positive, a
dilation of factor a from the y-axis and 1a
from the x-axis and a translation of b units in the positive direction of the x-axis.
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Chapter 17 — Continuous random variables and their probability distributions 623
Example 16
Suppose X is a continuous random variable with mean =10 and variance 2 =2.a Find E(2 X +1).b Find Var(1
−3 X ).
c If X has a probability density function with rule f ( x), describe the rule for a probabilitydensity function g , in terms of f for 2 X +1.
Solution
a E(2 X +1) =2E( X ) +1 =2 ×10 +1 =21b Var(1 −3 X ) = (−3)2Var( X ) =9 ×2 =18
c The rule is g ( x) =1a
f x −b
awhere a =2 and b =1. Therefore
g ( x) =1
2 f
x
−1
2
Sums of independent random variables *
In previous work the idea of independent events has been introduced. The term ‘independent’is also applied to random variables. While a formal denition of independent random variablesis beyond the scope of this course, basically two random variables are independent if their joint probability density function is a product of their individual probability density functions. Thisis also true for their cumulative distribution functions. Knowing random variablesare independent allows us to make certain conclusions concerning their mean and variance.
Suppose X and Y are independent random variables; then:
E( X +Y ) =E( X ) +E(Y )Var( X +Y ) =Var( X ) +Var( Y )
Note that in fact E( X +Y ) =E( X ) +E(Y ) whether or not X and Y are independent, but thesame cannot be said for variance.
Example 17
A manufacturing process involves two stages. The time taken to complete stage 1, X hours, isa continuous random variable with mean =4 and standard deviation =1.5. The timetaken to complete stage 2, Y hours, is a continuous random variable with mean =7and standard deviation =1. Find the mean and standard deviation of the total time takento complete the manufacturing process, if the times taken at each stage areindependent.
∗ This section covers material not included in the Mathematical Methods Units 3 and 4 CAS study design.
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The total processing time is given by X +Y . To nd the mean total processing time:
E( X +Y ) =E( X ) +E(Y )
= 4 +7
= 11Since X and Y are independent:
Var( X +Y ) =Var( X ) +Var( Y )
= (1.5)2 +(1)2
=2.25 +1
=3.25
Hence the standard deviation of the total processing time is:
sd( X +Y ) =√ 3.25
=1.803
Exercise 17E
1 The amount of our used each day in a bakery is a continuous random variable X withmean of 4 tonnes. The cost of the our is C =300 X +100. Find E( C ).
2 For certain glass ornaments the proportion of impurities per ornament, X , is a randomvariable with a density function given by:
f ( x) =3 x2
2 + x 0 ≤ x ≤1
0 x < 0 or x > 1
The value of each ornament (in dollars) is V =100 −1.5 X
a Find E( X ) and Var( X ). b Hence nd the mean and standard deviation of V .
3 Let X be a random variable with probability density function:
f ( x)
=
3 x2
2 −1 ≤ x ≤1
0 x < −1 or x > 1
Find:
a E(3 X ) and Var(3 X ) b E(3 − X ) and Var(3 − X )
c the rule of a probability density function for 3 X
4 The time spent waiting in line at a fast-food outlet is a continuous random variable X withmean =2.5 minutes and standard deviation =1.5 minutes. The time taken for the fastfood to be prepared and served is a continuous random variable Y with mean =5minutes and standard deviation
=2.5 minutes.
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Chapter 17 — Continuous random variables and their probability distributions 625
a Find the mean and variance of the total time the customer waits for the order.b Find the time interval within which about 95% of customers would wait.
5 The length of time Joan waits for her train to work each morning is a continuous randomvariable with mean =7 minutes and standard deviation =2 minutes. If the time
waited each day is independent of the time waited any other day, nd the mean and thestandard deviation of the total time waited each week (assume 5 working days per week).
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A continuous random variable is one that can take any value in an interval of the realnumber line.A continuous random variable can be described by a probability density function f . Thereare many different probability density functions with different shapes and properties.However, they all have the following fundamental properties: For any value of x, the value of f ( x) is positive. That is:
f ( x) ≥0 for all x
The total area enclosed by the graph of f ( x) and the x-axis is equal to 1. That is:
∞
−∞ f ( x)dx =1
The probability of X taking a value in the interval ( a , b) is found by determining thearea under the probability density curve between a and b. That is:
Pr(a < X < b) = b
a f ( x)dx
The mean or expected value of a continuous random variable X which has probabilitydensity function f ( x) given by
E( X ) = ∞
−∞ x f ( x)dx
provided the integral exists.The mean is always denoted by the Greek letter (mu).If g( X ) is a function of X , then the mean or expected value of g ( X ) is
E[ g ( X )] = ∞
−∞ g ( x) f ( x)dx
provided the integral exists.In general:
E[ g ( X )] = g [E( X )]
The median for a continuous random variable X is m such that:
m
−∞ f ( x)dx =0.5
If f ( x) is the probability density function of X , then the mode of M is the value of X suchthat:
f ( M ) ≥ f ( x) for all other x
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Chapter 17 — Continuous random variables and their probability distributions 627
For a continuous random variable X with probability density function f ( x) the variance isgiven by
Var( X ) =E[( X − )2]
= ∞
−∞( x − )2
f ( x)dx
provided the integral exists.The variance is usually denoted 2, where is the lower case of the Greek letter sigma.The standard deviation of X is dened as:
sd( X ) = Var( X )
To calculate variance, use:
Var( X ) =E( X 2) − 2
In general, for most continuous random variables X :
Pr( −2 ≤ X ≤ +2 ) ≈ 0.95
The interquartile range of X is
IQR =b −a
where a and b are such that
a
−∞ f ( x)dx
=0.25 and
b
−∞ f ( x)dx
=0.75
and where f ( x) is the probability density function of X .For any random variable X :
E(a X +b) = a E( X ) +b
For any random variable X :
Var( a X +b) = a 2 Var( X )
Multiple-choice questions1 Which of the following graphs could not represent a probability density function f ?
a
0
f ( x )
x
b
0
f ( x )
x
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2 If the function f ( x) =4 x represents a probability density function, then which of thefollowing could be the domain of f ?A 0 < x < 0.25 B 0 < x < 0.5 C 0 < x < 1
D 0 < x <1
√ 2 E1
√ 2 < x <2
√ 23 If a random variable X has probability density function
f ( x) =
1
2 sin x 0 < x < k
0 x ≥k or x ≤0
then k is equal to:A 1 B
2 C 2 D E 2
The following information relates to questions 4, 5 and 6 .A random variable X has probability density function:
f ( x) =34
( x2−1) 1 < x < 2
0 x
≤1 or x
≥2
4 Pr( X ≤1.3) is closest to:A 0.0743 B 0.4258 C 0.3 D 2.25 E 0.9258
5 The mean of X , E( X ), is equal to:
A 1 B32
C94
D2732
E2716
6 The variance of X is:
A2716
B67
1280C
8116
D81
256E
729256
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3 The probability density function of X is given by:
f ( x) =sin x
2 0 ≤ x ≤
0 x > or x < 0
Find the median and mode of X .4 The cumulative distribution function of X is given by:
F ( x) =1 x ≥514
( x −1) 1 ≤ x < 5
0 x < 1
a Find Pr(1 < X < 3). b Find Pr( X > 2|1 < X < 3). c Find Pr( X > 4| X > 2).
5 Consider the random variable X having the probability density function given by:
f ( x) = 12 x2(1 − x) 0 ≤ x ≤10 otherwise
a Sketch the graph of y = f ( x)b Find Pr( X < 0.5) and illustrate this probability on your sketch graph.
6 The probability density function of a random variable X is:
f ( x) =kx2(1 − x) 0 ≤ x ≤10 otherwise
a Determine k . b Find the mode of X .
c Find the probability that X is less than 23
.
d Find the probability that X is less than 13
, given that X is less than 23
.
7 Y is a continuous random variable with probability density function:
f ( y) =e− y, y ≥00, y < 0
Find:a Pr(Y < 2) b Pr(Y < 3
|Y > 1)
8 The cumulative distribution function of X is given by:
F ( x) =loge x 1 ≤ x ≤e0 x > e or x < 1
a Find the median value of X . b Find the interquartile range of X .c Find the probability density function of X , f ( x).
9 The amount of uid, X , in a can of soft drink is a continuous random variable with mean330 mL and standard deviation 5 mL. Find an (approximate) interval for the amount of softdrink contained in 95% of the cans.
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Chapter 17 — Continuous random variables and their probability distributions 631
10 The weight, X , of cereal in a packet is a continuous random variable with mean 250 g and variance 4 g. Find an (approximate) interval for the weight of cereal contained in 95% of the packets.
Extended-response questions
1 The distribution of X , the life of a certain electronic component, in hours, is described bythe following probability density function:
f ( x) = a100
1 − x
100 100 < x < 1000
0 x ≤100 and x ≥1000
a What is the value of a? b Find the expected value of the life of the components.c Find the cumulative distribution function, F ( x), for this probability density function.
2 The continuous random variable X has cumulative distribution function given by:
F ( x) =0 x < 02 x − x2 0 ≤ x ≤11 x > 1
a Find Pr( X > 0.5).b Find a such that Pr( X < a ) =0.8.
c Find E( X ) and E(√ X ).
3 The probability density function of X is given by:
f ( x) = 20 cos
10( x −6) 1 ≤ x ≤11
0 x < 1 or x > 11
a Find the median and interquartile range of X .b Find the mean and the variance of X.
4 A hardware shop sells a certain size nail either in a small packet at $1.00 per packet, or loose at $4.00 per kilogram. On any shopping day the number, X , of packets sold is a binomial random variable with number of trials equal to 8 and probability of success of 0.6,and the weight, Y kilograms, of nails sold loose is a continuous random variable with
probability density function f given by:
f ( y) =2( y −1)
25 1 ≤ y ≤6
0 y < 1 or y > 6
a Find the probability that the weight of nails sold loose on any shopping day will be between 4 kg and 5 kg.
b Calculate the expected money received on any shopping day from the sale of this sizenail in the store.
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5 The continuous random variable X has the probability density function f , where:
f ( x) =( x −2)
2 2 ≤ x ≤4
0 x < 2 or x > 4
By rst expanding ( X −c)2, or otherwise, nd two values of c such that:
E[( X −c)2] =23
6 A wholesaler sells material offcuts in 5 m lengths. The amount, in metres, of unusablematerial are the values of a random variable, X , with probability density function:
f ( x) =k ( x −1)(3 − x) 1 ≤ x ≤30 x < 1 or x > 3
a Show that k
=0.75.
b Find the mean and variance of X .c Find the probability that X is greater than 2.5 m.
7 The continuous random variable X has probability density function f , where:
f ( x) =k
12( x +1)3 0 ≤ x ≤4
0 x < 0 or x > 4
a Find k .b Evaluate E( X
+1). Hence, nd the mean of X .
c Use your calculator to verify your answer to part b .d Find the value of c > 0 for which Pr( X ≤c) = c.
8 The yield of a variety of corn has probability density function:
f ( x) =kx 0 ≤ x < 2k (4 − x) 2 ≤ x < 40 x < 0 or x > 4
a Find k .b Find the expected value, , and variance of the yield of corn.
c Find the probability Pr( | X − | < 1).d Find the value of a such that Pr( X > a ) =0.6, giving your answer correct to one
