DowseyMathsMethodsCh12.fm Page 473 Monday, January 16 ... World Mathematical Methods... ·...

1212

Continuous randomvariables

In the previous chapter we dealt with probability distributions of discrete randomvariables, their associated central values and measures of spread, and their applications.We will now consider the corresponding properties of continuous random variables.

The distinction between discrete variables and continuous variables is that discretevariables can only take particular discrete values, whereas continuous variables cantake any value in an interval of the set of real numbers. For example, the length of afoot is a continuous variable being able to take any value in some interval, but shoesize is a discrete variable being able to take only certain values.

As a rough guide, most variables which are

measured

are continuous; most variableswhich are

counted

are discrete. For example, the waiting time for a train is acontinuous random variable.

DowseyMathsMethodsCh12.fm Page 473 Monday, January 16, 2006 8:09 AM

474

MathsWorld

Mathematical Methods Units 3 & 4

Continuous random variables

We have established previously that for any probability distribution the probability of an event occurring must have a value between zero and 1 inclusive; a probability of zero indicates an event that is impossible, a probability of 1 indicates an event that is certain to occur. This can be written as 0

≤

Pr(

A

)

≤

1 for any event

A

.

A

continuous random variable

is one whose set of possible values is the set of real numbers

R

or a subset of

R

. The observed values of a continuous random variable contain rounding error or measurement error. For example, if

X

cm is the length of the foot of an individual in some population, the set of possible values may be [10, 60] for that population. When we record

X

=

28, it means that the actual foot length is in the vicinity of 28 cm, not that it is

exactly

28 cm. Generally, we would measure a person’s foot length correct to the nearest centimetre, so the recorded value of 28 cm will actually mean any value between 27.5 cm and 28.5 cm.

What is the probability that

X

=

28

exactly

? As there is an infinite number of real numbers in the interval [10, 60], if

c

is any number in this interval, Pr(

X

=

c

) must be 0, otherwise the sum of all the probabilities would be more than 1. This is a very important property of a continuous random variable

X

: for any possible value

x

, Pr(

X

=

x

)

=

0. It only makes sense to consider the probability that

X

lies in an interval. Thus, although Pr(

X

=

28)

=

0, the probability given by Pr(27.5

≤

X

<

28.5) will be greater than zero.

In chapter 11, we saw that the probability distribution of a discrete random variable is described by its probability function

p

(

x

)

=

Pr(

X

=

x

). This function is quite useless for a continuous random variable, as all such probabilities are zero. So the distribution of a continuous random variable is described by a different sort of function.

From our knowledge of calculus, we should recognise that the integral above gives the area below the graph of

y

=

f

(

x

) over the interval [

x

1

,

x

2

], provided

f

(

x

) lies above the

x

-axis. As probabilities cannot be negative, and the total probability is 1, this leads to the following essential conditions for a function to be a probability density function.

What is a probability density function?The probability density function of a continuous random variable X is a function f (x ) with domain R so that the probability that X lies in the interval [x1, x2] is given by

Pr(x1 ≤ X ≤ x2) = f (x ) dx⌠⌡x1

x2

xx1 x2 ba0

y

y = f(x)

Pr(x1 ≤ X ≤ x2)

12 .112 .1



475

chap

ter

12

In many cases, the possible values of

X

lie in a closed interval such as [

a

,

b

]. In such cases, the probability density function

f

(

x

) is taken to be zero for all real numbers outside this

interval. The second property above is then equivalent to

f

(

x

)

dx

=

1, since:

f

(

x

)

dx

=

f

(

x

)

dx

+ f(x) dx + f(x) dx

= 0 dx + f(x) dx + 0 dx

= f(x) dx

Example 1

Consider the function .

a Show that f (x ) satisfies the conditions for it to be the probability density function of a random variable X.

b Find:i Pr(0.2 ≤ X ≤ 0.4). ii Pr(0.2 < X < 0.4). iii Pr(X < 0.5).

Solutiona It is helpful to sketch the graph of y = f (x ).

The graph is shown at the right on the interval [0, 1]; it is zero elsewhere. From the graph, we see that f (x ) ≥ 0 for all real values of x, so the first condition is satisfied.The area below the curve is given by:

f (x ) dx = (1 − x2) dx

=

=

= 1

The second condition is also satisfied, and so f (x ) is the probability density function of a random variable X.

Conditions for a probability density functionFor f (x ) to be a probability density function of a continuous random variable X, we require:

. f (x ) ≥ 0 for all real numbers x

. the total area under the graph of f (x ) is 1, i.e. f (x ) dx = 1⌠⌡−∞

∞

⌠⌡a

b

⌠⌡−∞

∞⌠⌡−∞

a⌠⌡a

b⌠⌡a

∞

⌠⌡−∞

a⌠⌡a

b⌠⌡b

∞

⌠⌡a

b

f x( )32--- 1 x2–( )

0⎩⎪⎨⎪⎧

=0 x 1≤ ≤

otherwise

0.5

0

1

2

1.5

0.2 0.60.4

y

0.8 1 x

y = 3 (1 – x

2), 0 ≤ x ≤ 1

2

The shaded area isone square unit

⌠⌡−∞

∞ ⌠⎮⌡0

132---

3x2

------ x3

2-----–

0

1

32--- 1

2---–

12.1


476

MathsWorld Mathematical Methods Units 3 & 4

b i From the definition of a probability density function, Pr(0.2 ≤ X ≤ 0.4) is found by integrating f (x ) between 0.2 and 0.4, i.e. it is given by the area below the curve between 0.2 and 0.4. Thus:

Pr(0.2 ≤ X ≤ 0.4) = (1 − x2) dx

=

= 0.272ii Pr(0.2 < X < 0.4) = Pr(0.2 ≤ X ≤ 0.4) as Pr(X = 0.2) = Pr(X = 0.4) = 0. Thus

Pr(0.2 < X < 0.4) = 0.272.iii Pr(X < 0.5) is given by the area under the curve to the left of x = 0.5. As f (x ) = 0 for

x < 0, this is the same as the area below the graph of between x = 0 and x = 0.5. Hence:

Pr(X < 0.5) = (1 − x2) dx

=

= 0.6875

Discrete versus continuous random variablesIt is useful to compare the methods used for discrete and continuous random variables. For any probability distribution, the probabilities cannot be negative and the sum of the probabilities of all possible outcomes is 1. Thus we have the following comparison.

Discrete case Continuous case

Probability function p(x ) = Pr(X = x ) Probability density function f (x )p(x ) ≥ 0 f (x ) ≥ 0

∑p(x ) = 1 ∫f (x ) dx = 1Pr(x1 ≤ X ≤ x2) = ∑p(x ) for discrete

values of x between x1 and x2 inclusive.Pr(x1 ≤ X ≤ x2) = f (x ) dx

⌠⎮⌡0.2

0.432---

3x2

------ x3

2-----–

0.2

0.4

y 32--- 1 x2–( )=

⌠⎮⌡0

0.532---

3x2

------ x3

2-----–

0

0.5

GC 5.5CAS 5.5

The probabilities in example 1 part b can be easily found using a graphics calculator, either at the GRAPH screen or at the HOME screen. The screenshots below show the calculations at the HOME screen on the TI-83/84 and TI-89 calculators.

tip

⌠⌡x1

x2

DowseyMathsMethodsCh12.fm 76 Monday, January 16, 2006 8:09 AM


477

chap

ter

12In the discrete case, we find probabilities by summing (∑); in the continuous case, we find

probabilities by integration ( ). The following summarises the key properties of a continuous random variable.

Example 2State whether each of the following functions could represent a probability density function.

a b

Solutiona A sketch can assist. From the sketch,

f (x ) ≥ 0 for all real values of x.Also, the shaded area, A, is given byA = 0.2 × 2 + 0.3 × 2 = 1.So f (x ) could represent a probability density function.

b g(x ) ≥ 0 for all real values of x.If A is the area enclosed by the graph of y = g(x ) and the x-axis, thenA = 0.6 × 2 + 0.4 × 1 = 1.6.So g(x ) could not represent a probability density function.

Example 3Show that the following functions are probability density functions.

a , 1 ≤ x ≤ 3

b , 0 ≤ x ≤ loge 5

⌠⌡

Properties of a continuous random variableIf X is a continuous random variable with probability density function f (x ), then:

. Pr(X = x ) = 0 for any real number x

. Pr(x1 ≤ X ≤ x2) = f (x ) dx

. f (x ) ≥ 0 for all real numbers x

. f (x ) dx = 1, or equivalently f (x ) dx = 1 if f (x ) = 0 for all x < a and for all x > b

. Pr(x1 ≤ X ≤ x2) = Pr(x1 < X < x2) = Pr(x1 ≤ X < x2) = Pr(x1 < X ≤ x2).

⌠⌡x1

x2

⌠⌡−∞

∞⌠⌡a

b

f x( )0.20.30⎩

⎪⎨⎪⎧

=

0 x 2≤ ≤2 x 4≤<elsewhere

g x( )0.60.40⎩

⎪⎨⎪⎧

=


0.1

–0.1

0

0.2

0.3

0.5 1.51

y

2 2.5 3 3.5 4 x

tipIt is understood that each function is zero for all other values of x, even if this is not explicitly stated.

f x( ) 16--- 7 2x–( )=

g x( ) 112------e2x=

12.1


478


Solution

a From the graph of y = f (x ), f (x ) ≥ 0 for all real values of x. Also:

(7 − 2x) dx =

=

= 1

So f (x ) is a probability density function.

b g(x ) > 0 for 0 ≤ x ≤ loge 5, as e2x > 0 for any real number x. Hence g(x ) ≥ 0 for all real values of x. Also:

e2x dx =

=

=

=

= 1

So g(x ) is a probability density function.

Example 4Find the value of k if each of the following is a probability density function.

a b

Solutiona First note that f (x ) ≥ 0 provided k ≥ 0.

Use f (x ) dx = 1 to find k.

kx(8 − 3x) dx = k (8x − 3x2) dx

=

= k((16 − 8) − (0))= 8k

Hence 8k = 1, so k = .

0.2

0

0.4

0.8

1

0.6

y

x0.5 1.51 2 2.5 3 3.5 4

y = 1 (7 – 2x)6

⌠⎮⌡1

316--- 1

6--- 7x x2–[ ]1

3

16--- 21 9–( ) 7 1–( )–( )

0

1

2

x

y

loge.5(0, 1

12)

y = 1 e2x

12

⌠⎮⌡0

loge 5112------ 1

24------ e2x[ ]

0loge 5

124------ e2 loge 5 1–( )

124------ e log e 25 1–( )

124------ 25 1–( )

f x( )kx 8 3x–( )0⎩

⎨⎧

=0 x 2≤ ≤otherwise

g x( )k sin x0⎩

⎨⎧

=x 0 π,[ ]∈otherwise

b As sin x ≥ 0 for 0 ≤ x ≤ π, then g(x) ≥ 0.

Use g (x ) dx = 1 to find k.

k sin x dx = = −k(−1 − 1)= 2k

Hence 2k = 1, so k = .

⌠⌡−∞

∞

⌠⌡0

πk cos x–[ ]

0π

12---

⌠⌡−∞

∞

⌠⌡0

2⌠⌡0

2

k 4x2 x3–[ ]02

18---



479

chap

ter

12Example 5Find the value of k if each of the following is a probability density function.

a b

SolutionThese functions cannot be integrated using the techniques in this course, so we use the integration features of a graphics calculator.

a As sin x > 0 and cos x > 0 for ,then f (x ) ≥ 0 provided k ≥ 0.


k sin x cos x dx = k sin x cos x dx

= 0.5k

Hence k = 2.

The integrals in both parts a and b are shown in thescreenshot to the right.

Example 6A continuous random variable X has its probability density function defined by

, and zero elsewhere. Use integration to find:

a Pr(0.5 < X ≤ 1.5) b Pr(X < 1.5) c Pr(X ≥ 0.75) d Pr(X > 2.5)

Solution

a Pr(0.5 < X ≤ 1.5) = (2 − x) dx

=

= 0.5

f x( ) k sin x cos x 0 x π2---<<,= f x( ) k

π 1 x2+( )----------------------- 0 x 1≤ ≤,=

b As the denominator is positive for allvalues of x, f (x ) ≥ 0 provided k ≥ 0.


= k

= 0.25kHence k = 4.

⌠⌡−∞

∞

0

1⌠⎮⌡

kπ 1 x2+( )----------------------- xd

0

1⌠⎮⌡

1π 1 x2+( )-----------------------

0 x π2---< <

⌠⌡−∞

∞

⌠⌡0

π/2⌠⌡0

π/2

GC 5.5CAS 5.5

f x( ) 12--- 2 x–( ) 0 x 2≤ ≤,=

b Pr(X < 1.5) is the same as Pr(0 ≤ X < 1.5), as f (x ) = 0 for x < 0.

Pr(0 ≤ X < 1.5) = (2 − x) dx

=

= 0.9375

12---

⌠⎮⌡0

1.5

12--- 2x x2

2-----–

0

1.5

12---

⌠⎮⌡0.5

1.5

12--- 2x x2

2-----–

0.5

1.5

12.1


480


c Pr(X ≥ 0.75) is the same as Pr(0.75 ≤ X ≤ 2),as f (x ) = 0 for x > 2. (This probability canalso be calculated as 1 − Pr(X < 0.75), if that is easier.)

Pr(0.75 ≤ X ≤ 2) = (2 − x) dx

=

= 0.3906

Example 7For the continuous random variable X with probability density function

, and zero elsewhere find:

a Pr(1.2 < X ≤ 1.9) b Pr(X > 2) c Pr(X ≥ 3)

Solution

a Pr(1.2 < X ≤ 1.9) = x loge x dx

= 0.2311

b Pr(X > 2) is the same as Pr(2 ≤ X ≤ e), as f (x ) = 0 for x > e.

Pr(2 ≤ X ≤ e) = x loge x dx

= 0.6966

c Pr(X ≥ 3) = 0 as 3 > e, and f (x ) = 0 for x > e.

Answers in parts a and b are shown in the screenshot.The function has been stored in Y1.

exercise 12.1Where not explicitly stated, it is understood that the probability density function is zero for all real numbers outside the given interval.

1 State whether each of the following functions could represent a probability density function if the function has rule:

a b

2 Show that the following functions are probability density functions.

a b

d Pr(X > 2.5) = 0, as f (x ) = 0 for x > 2.

tipPhrases such as ‘Use calculus’ or ‘Use integration’ mean you should show your working, e.g. show an appropriate antiderivative. It is not sufficient to just give a calculator-based answer in such cases.

12---

⌠⎮⌡0.75

2

12--- 2x x2

2-----–

0.75

2

f x( ) 4e2 1+---------------x log e x 1 x e≤ ≤,=

tipThis function cannot be integrated using the techniques in this course so we use the integration feature of a graphics calculator.

GC 5.5CAS 5.5

4e2 1+---------------

⌠⎮⌡1.2

1.9

4e2 1+---------------

⌠⎮⌡2

e

f x( )0.4

0.6

0⎩⎪⎨⎪⎧

=


g x( )0.6

0.4

0⎩⎪⎨⎪⎧

=


f x( ) 110------ 9 4x–( ) x 0 2,[ ]∈,= g x( ) 5

242---------e5x 0 x log e 3< <,=



481

chap

ter

12c d

e f

g h

3 Which of the following could be a probability density function? Give reasons for your answer in each case.

a b

c

4 Find the value of k if each of the following is a probability density function.

a b

c d

e f

g h

5 Find the value of k if each of the following is a probability density function. You will need to use a graphics calculator for each part of this question.

a b

c d

e f

6 A continuous random variable X has its probability density function defined by

. Use integration to find:

a Pr(0.6 < X ≤ 1.7) b Pr(X ≤ 2.5) c Pr(X > 0.6) d Pr(X > 3.2)

7 For the continuous random variable X with probability density function

, find:

a Pr(2 < X ≤ 4) b Pr(X ≤ 5.4) c Pr(X ≥ 4.6) d


, find:

a Pr(2.4 ≤ X < 4.2) b Pr(X < 4.4) c Pr(X > 3.2) d

9 A continuous random variable X has probability density function defined by

. Find:

a Pr(2.5 < X ≤ 4.5) b Pr(X > 3.2) c Pr(X ≤ 1.9) d Pr(X ≤ 4.8)

f x( ) 136------ 2x 3+( ) 1 x 5≤ ≤,= f x( ) 2 cos 2x 0 x

π6---< <,=

f x( ) 374------x x 1+( ) 2 x 4≤ ≤,= f x( ) 2 sin x

π4--- x

π2---≤ ≤,=

f x( ) 641------ x2 3x–( ) 4 x 5≤ ≤,= g x( ) 24

7------e 3x– 0 x log e 2≤ ≤,=

f x( )6x 2–

0⎩⎨⎧

=0 x 1< <elsewhere

f x( )6x 2–

0⎩⎨⎧

=1 x 2< <elsewhere

f x( )2 6x–

0⎩⎨⎧

=13---– x 0< <

elsewhere

f x( ) k 12 5x–( ) 0 x 2≤ ≤,= g x( ) k cos 2x 0 xπ4---≤ ≤,=

f x( ) kx 9 2x–( ) x 1, 4)(∈,= g x( ) ke2x 0 x log e 7≤ ≤,=

f x( ) kx 4x 3+( ) x 0 5,( )∈,= g x( ) k sin 3x 0 xπ6---≤ ≤,=

f x( ) k 3x2 2x 1+ +( ) x 0 2,[ ]∈,= g x( ) ke 4x– 0 x log e 2≤ ≤,=

f x( ) k sin 2x cos 2x 0 xπ4---≤ ≤,= f x( ) k

π 9 x2+( )----------------------- 0 x 3≤ ≤,=

f x( ) k log e x x 1 e,( )∈,= f x( ) k

3 cos2 x------------------------- 0 x

π6---≤ ≤,=

f x( ) kx sin x 0 xπ2---< <,= f x( ) k sin x

cos3 x---------------- 0 x

π3---≤ ≤,=

f x( ) 112------ 7 2x–( ) 0 x 3≤ ≤,=

f x( ) 165------ 27 4x–( ) 1 x 6≤ ≤,=

Pr X 37<( )

f x( ) 293------ 3x 5+( ) 2 x 5≤ ≤,=

Pr X 3≥( )

f x( ) 151------x 16 3x–( ) 2 x 5≤ ≤,=

12.1


482



, find the exact value of:

a b c



a b c


. Find the exact value of:

a Pr(loge 2 < X ≤ loge 3) b Pr(X > 1) c Pr(X ≤ 0.5)



a b


. Find:

a Pr(X ≤ 1.8) b Pr(X > 0.8)

15 For the continuous random variable X with probability density function, find:

a Pr(1.7 ≤ X < 2.7) b Pr(X ≥ 1.5) c


. Find:

a Pr(0.1 < X ≤ 0.5) b Pr(X > 0.16)

Example 8For the function f (x ) = ce−0.2x for x ≥ 0, and zero elsewhere, find the value of c for which f (x ) is a probability density function.

Solution

As e−0.2x > 0 for all values of x, then f (x ) ≥ 0 if c > 0. Use ce−0.2x dx = 1 to find c.

ce−0.2x dx =

= −5c(e−∞ − e0)= −5c(0 − 1)= 5c

Hence 5c = 1, so c = 0.2.

f x( ) 12--- sin x 0 x π≤ ≤,=

Prπ6--- X

π2---<≤⎝ ⎠

⎛ ⎞ Pr X3π4

------<⎝ ⎠⎛ ⎞ Pr X

π3---≥⎝ ⎠

⎛ ⎞

f x( ) cos x 0 xπ2---≤ ≤,=

Prπ6--- X

π4---<≤⎝ ⎠

⎛ ⎞ Pr Xπ3---≥⎝ ⎠

⎛ ⎞ Pr Xπ6---<⎝ ⎠

⎛ ⎞

f x( ) 13---ex 0 x log e 4≤ ≤,=

f x( ) 2

x2----- 1 x 2≤ ≤,=

Pr X 112---≤⎝ ⎠

⎛ ⎞ Pr X 114---≤⎝ ⎠

⎛ ⎞

f x( ) 1π--- x sin x 0 x π≤ ≤,=

f x( ) log e x x 1 e,( )∈,=

Pr X e<( )

f x( ) 3

3 cos2 x------------------------- 0 x

π6---≤ ≤,=

⌠⌡0

∞

⌠⌡0

∞ c0.2–

-----------e 0.2x–

0

∞



483

chap

ter

12The second and third lines in the solution need some explanation. When we substitute ∞ (think of this as an immeasurably large number) theindex becomes −∞ and then e−∞ = 0. Alternatively, think of this step as asking about the behaviourof e−0.2x as x → ∞. The graph of y = e−0.2x hasthe x-axis as horizontal asymptote, i.e. asx → ∞, e−0.2x → 0, so the limit is 0.

Example 9A continuous random variable X has probability density function defined by

, and zero elsewhere. Find:

a b c

SolutionThe screenshot shows the results for the first two parts. Hence the answers are:

a

b

c =

= (since )

= 0.6340

CAS 5.5tipThe TI-89 can help with an ‘infinite’ integral. The screenshot below showsthe result for the integral in example 8.

f x( ) 13 sin2 x

-------------------------π6--- x π

2---≤ ≤,=

Pr X π4---<⎝ ⎠

⎛ ⎞ Pr X π3---≤⎝ ⎠

⎛ ⎞ Pr X π4---< X π

3---≤⎝ ⎠

⎛ ⎞⎝ ⎠⎛ ⎞

GC 5.5CAS 5.5

Pr X π4---<⎝ ⎠

⎛ ⎞ 0.4226=

Pr X π3---≤⎝ ⎠

⎛ ⎞ 23---=

Pr X π4---< X π

3---≤⎝ ⎠

⎛ ⎞Pr X π

4---< X π

3---≤∩⎝ ⎠

⎛ ⎞

Pr X π3---≤⎝ ⎠

⎛ ⎞----------------------------------------------

Pr X π4---<⎝ ⎠

⎛ ⎞

Pr X π3---≤⎝ ⎠

⎛ ⎞------------------------ X π

4---<⎝ ⎠

⎛ ⎞ X π3---≤⎝ ⎠

⎛ ⎞⊂

tipNote that the calculation in part c was made using all of the digits given by the calculator for part a rather than the answer given. If that answer was used we would get 0.6339. This is an important principle—you should always use the most accurate answer available in subsequent parts of the question even if the earlier answer is requested to be given to a particular number of decimal places.

12.1


484


exercise 12.1

17 A continuous random variable X has probability density function:

a Find the value of k.

b Find exactly:

i Pr(X ≤ 2) ii Pr(X ≥ 2 | X = 3)


and zero elsewhere. Find the exact value of:

a Pr(X ≥ 5) b Pr(X > 3) c Pr(X ≥ 5 | X > 3)


and zero elsewhere, find the exact value of:

a Pr(X ≤ 4) b Pr(X < 3) c Pr(X < 3 | X ≤ 4)

20 A continuous random variable x has probability density function defined by

and zero elsewhere. Find, correct to 4 decimal

places, the values of:

a Pr(0.5 < X ≤ 1) b Pr(0.3 < X ≤ 1.2) c Pr(0.5 < X ≤ 1 | 0.3 < X ≤ 1.2)

21 A continuous random variable X has probability density function:

a Verify that f(x) is a valid probability density function.

b Find the exact values of:

i Pr(X > 1) ii Pr(−2 < X < 2) iii Pr(X < loge 2)

continued

f x( )k

x2-----

0⎩⎪⎨⎪⎧

=x 1≥

elsewhere

f x( ) 156------ 15 2x–( ) 0 x 7≤ ≤,=

f x( ) 1144---------x 20 3x–( ) 0 x 6≤ ≤,=

f x( ) 14 log e 4 3–----------------------------xex 0 x log e 4≤ ≤,=

f x( ) 2e 2x–

0⎩⎨⎧

=x 0≥elsewhere

Numbersense with the spence

96For five and a half years, Australia's first prime time soap opera, Number 96, commanded

the attention of TV viewers with its unique blend of sex, suspense and sit-com. Set in a block of flats in Paddington, Sydney, it purported to trace "the lives,loves and emotions of ordinary people". We’re not sure why the number 96 was used.



485

chap

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12Statistics of continuous random variables

The important statistics for any random variable are the mean, the median, the mode, the variance and the standard deviation. The mean, median and mode provide a picture of the distribution of the random variable by giving measures of the central tendency of the distribution. The standard deviation, which is the square root of the variance, is a measure of the spread of the distribution.

Mean, median and modeThe meanThe mean, μ, is defined by μ = E(X), the expected value of X. Recall that the expected value of X for a discrete random variable is defined by:

μ = E(X) = Σx Pr(X = x) = ∑xp(x).

The equivalent statement for a continuous random variable is as follows:

The medianThe median is the halfway mark of the population. For a continuous random variable,

the median is the value m such that Pr(X ≤ m) = .

The modeThe mode is the most common value. For a continuous random variable, this will be the value M such that f(M) ≥ f(x) for all values of x.

There can be more than one mode. The mode occurs at the value(s) of x for which the probability density function, f(x), is a maximum. This can be found by using the derivative of f(x) (by solving f ′(x) = 0) or by graphing y = f(x) and using the graph to find the maximum. The method used will depend on the probability density function and the wording of the question.

Variance and standard deviationVariance, σ 2 = var(X), is a measure of the deviation from the mean or average. Recall that the variance of X for a discrete random variable is defined by:

var(X) = E[(X − μ)2] = E(X2) − μ2 = ∑x2p(x) − μ2.

μ = E(X) = xf (x ) dx

In general, E [g (X )] = g(x )f (x ) dx

⌠⌡−∞

∞

⌠⌡−∞

∞

12---

To find m solve:

f (x ) dx = , where a is the minimum value of X.⌠⌡a

m 12---

12.212 .212 .2


486


The equivalent statement for a continuous probability distribution is:

The standard deviation of X is given by .

Example 1For the continuous random variable with probability density function

, and zero elsewhere, find the:

a mean. b median. c mode. d variance. e standard deviation.

Solution

a μ = x

=

=

b For the median m, we require f (x ) dx = .

=

=

=

= 0

= 0

Using the quadratic formula, we get . However, since 2 < m < 4, the solution is , correct to 3 decimal places.(Note: In many cases the equation involving m will need to be solved numerically; an exact value may not be possible to find. See the tip at the end of this example on how to do this with a graphics calculator.)

c For the mode, we require the value Mthat gives the maximum value of f(x).As the graph of f(x) is a straight line and therefore has no stationary points, the maximum value occurs at an endpoint. Hence M = 2.

var(X ) = σ 2 = E[(X − μ)2] = (x − μ)2f (x ) dx

= x2f (x ) dx − μ2

⌠⌡−∞

∞

⌠⌡−∞

∞

sd X( ) σ var X( )= =

f x( ) 2 12---x– 2 x 4≤ ≤,=

⌠⎮⌡2

4

2 12---x–⎝ ⎠

⎛ ⎞ xd

x2 16---x3–

2

4

83---

GC 2.4, 5.5CAS 2.4, 5.5

tipAn approximate value for the median can be found graphically as follows.

[2, 4] by [0, 10]

⌠⌡2

m 12---

⌠⎮⌡2

m

2 12---x–⎝ ⎠

⎛ ⎞ dx 12---

2x 14---x2–

2

m 12---

2m 14---m2– 4 1–( )–

12---

14---m2 2m– 31

2---+

m2 8m– 14+

m 4 ± 2=m 4 2– 2.586= =

0

0.5

y

x1 2 3 4

y = 2 – 1 x, 2 ≤ x ≤ 42

1



487

chap

ter

12d σ 2 = (using x2f (x ) dx − μ2)

=

=

=

=

=

e sd(X) =

=

Example 2

A continuous random variable has probability density function and zero elsewhere. Find the:a mean and variance. b exact value of the median. c exact value of the mode.

⌠⌡2

4x2 2 1

2---x–⎝ ⎠

⎛ ⎞ x 83---⎝ ⎠

⎛ ⎞2

–d ⌠⌡−∞

∞

⌠⌡2

42x2 1

2---x3–⎝ ⎠

⎛ ⎞ x 649

------–d

23---x3 1

8---x4–

2

4 649

------–

1283

---------- 32–⎝ ⎠⎛ ⎞ 16

3------ 2–⎝ ⎠

⎛ ⎞– 649

------–

223

------ 649

------–

29---

var X( )23

--------

GC 5.5, 7.1CAS 5.5, 7.1, 10.2

The integrals in parts a and d of example 1 can be done using a graphics calculator. Finding the median m in part b can also be done on a graphics calculator. On a TI-83/84, with the function entered as Y1, we can use a combination of the fnInt command and the Solver. Enter the equation as shown in the screenshot at the left (as the equation in the Solver must

be set to zero, we subtract from the integral). Press ENTER, guess a value for M (e.g. 3)

and then SOLVE. The screenshot on the right shows that the median is 2.586, correct to 3 decimal places, as before.

On a TI-89, the exact value of m can be found using a combination of the integrate and solve commands as shown in the screenshot at the right. We still have tochoose the correct value for m, however.

12---

tip

f x( ) sin x 0 x π2---≤ ≤,=

12.2


488


Solutiona We use a graphics calculator as shown in the tip below.

μ = x sin x dx

= 1For the variance, we can use the formula var(X ) = (x − μ)2f (x ) dx rather than the more

usual computational formula, since the integration feature of a graphics calculator can handle this as easily.

var(X ) = (x − μ)2f (x ) dx

= (x − 1)2 sin x dx

= 0.1416 correct to 4 decimal places

b For the median m, we require f (x ) dx = 0.5.

sin x dx = 0.5

= 0.5−cos m + cos 0 = 0.5

cos m = 0.5 (as cos 0 = 1)m =

c For the mode, we require the value M that gives the maximum value of f (x ). As sin x is amaximum at , then .

Example 3For the continuous random variable with probability density function

, and zero elsewhere, find:

a the mean. b the median. c the mode.d the exact value of the standard deviation.

Solutiona We can use integration as previously, to find the mean μ:

μ = x(2 − x ) dx

= 1

⌠⌡0

π/2

⌠⌡−∞

∞

⌠⌡−∞

∞

⌠⌡0

π/2

GC 5.5CAS 5.5

The integrals required cannot be found using techniques in this course, so they are evaluated using the integration feature of a graphics calculator. Notice that on the TI-89, the exact value of the variance can be found.

tip

⌠⌡0

m

⌠⌡0

m

cos x–[ ]0m

π3---

x π2---= M π

2---=

f x( ) 34---x 2 x–( ) 0 x 2≤ ≤,=

34---

⌠⎮⌡0

2



489

chap

ter

12However, note that the graph of y = f (x ) for 0 ≤ x ≤ 2 is part of a parabola, symmetrical about x = 1. If a probability distribution function is symmetrical, as in this case, it can be shown that the mean μ is thex-value corresponding to the line of symmetry. So we can say: ‘By symmetry, μ = 1’.

b The median can also be found using the symmetry of the parabola and is m = 1.c The value M that gives the maximum value of f (x ) can also be found using the symmetry

of the parabola and so M = 1.

d σ 2 = x3(2 − x) dx − 12 (using σ2 = x2f (x ) dx − μ2)

= (2x3 − x4) dx − 1

=

=

=

σ =

Example 4A continuous random variable has its probability density function defined by

, and zero elsewhere.

a Find the exact value of the mean.b Find the exact value of the mode.c Without calculating the median, show that it lies between the mean and the mode.d Find the median correct to 4 decimal places.

Solution

a μ = x2(9 − x2) dx

=

=

=

=

= 1.6

0

y

x1 2

34

34---

⌠⎮⌡0

2

⌠⌡−∞

∞

34---

⌠⎮⌡0

2

34--- x4

2----- x5

5-----–

0

2

1–

34--- 8 32

5------–⎝ ⎠

⎛ ⎞ 1–

15---

15

--------

f x( ) 481------x 9 x2–( ) 0 x 3≤ ≤,=

b The value M that gives the maximum value off (x ) can be found using calculus.

f (x ) =

f ′(x ) =

9 − 3x2 = 0 (f ′ (x ) = 0 at a maximum)x2 = 3x = ±

But x > 0, so x = .Hence the mode .

481------ 9x x3–( )

481------ 9 3x2–( )

33

M 3=

481------

⌠⎮⌡0

3

481------ 3x3 1

5---x5–

0

3

481------ 81 243

5----------–⎝ ⎠

⎛ ⎞

4 1 35---–⎝ ⎠

⎛ ⎞

85---

12.2


490


c First observe that the mean μ is 1.6 and the mode M is , so the mean is less than the mode. As the median m is the halfway mark, then for μ < m < M to be true, the probability that X is below the mean would be less than 0.5, while the probability that X is below the mode would be more than 0.5. So we calculate these probabilities:

Pr(X < μ) = x(9 − x2) dx

= 0.488 correct to 3 decimal places

Pr(X < M) = x(9 − x2) dx

= 0.556 correct to 3 decimal placesHence the median is indeed between the mean and the median.

d To find the median, we need to solve for m:

x(9 − x2) dx = 0.5

We could integrate this and simplify the result to give the following equation in m:m4 − 18m2 + 40.5 = 0

This equation could be solved graphically or with a solver.Alternatively, we can use a combination of the integration and solving features of a graphics calculator to find an approximate value of m in one step as shown in the tip after example 1 earlier. The screenshots below show this approach on the TI-83/84 and the TI-89.

Example 5Find the exact values of the mode and the median for the random variable X with probability density function:

SolutionThe graph of y = f (x ) is part of a truncus, which approaches the x-axis as x approaches ∞. The maximum point on the graph is at (0, 1) as shown in the screenshot at the right,so the mode M = 0.

GC 5.5CAS 5.5

3 1.732≈

481------

⌠⎮⌡0

8/5

481------

⌠⎮⌡0

3

481------

⌠⎮⌡0

m

tipNote that in the TI-89 screenshot on the right, adding the condition 0 < m < 3 means that

just one solution for m is provided. If the condition is omitted, additional values of m are given, and you must choose the correct one.

f x( )1

x 1+( )2-------------------

0⎩⎪⎨⎪⎧

=x 0≥

elsewhere

[0, 5] by [0, 1.5]

GC 2.1CAS 2.1



491

chap

ter

12For the median, we need to solve for m:

=

=

=

=

m = 1So the median m = 1.

exercise 12.2Where necessary, assume that the given probability density function is zero for all real numbers outside the given interval.

1 For the continuous random variable X with probability density function , find the exact value of the:

a mean. b median. c mode.

d variance. e standard deviation.


. Find the:

a mean. b median. c mode. d variance.


, find the exact values of the:

a mean. b median. c mode. d standard deviation.


. Find the:



, find the:



, find the exact values of the:


1 2 3 4x

y

0.2

0.4

0.6

0.8

1

mode median

Area = 0.5

Area = 0.5

0

y = , x ≥ 0(x + 1)

2

1

⌠⎮⌡0

m1

x 1+( )2------------------- xd 1

2---

1x 1+------------–

0

m 12---

1m 1+--------------– 1+

12---

1m 1+-------------- 1

2---

f x( ) 2x 0 x 1≤ ≤,=

f x( ) 221------ 6 x–( ) 1 x 4≤ ≤,=

f x( ) 112------ 2x 1+( ) 0 x 3≤ ≤,=

f x( ) cos x 0 xπ2---≤ ≤,=

f x( ) 12--- sin x 0 x π≤ ≤,=

f x( ) 332------x 4 x–( ) 0 x 4≤ ≤,=

12.2


492


7 A continuous random variable X has the probability density function defined by

. Find the exact values of the:


8 A continuous random variable X has the probability density function defined by

. Find the :

a exact value of the mean. b median.

c exact value of the mode. d exact value of the standard deviation.


, find the exact value of the:

a mean. b median. c mode.


. Find the:


11 For the continuous random variable X with probability density function , find the:

a exact value of the mode. b exact value of the median.


. Find the:


13 A continuous random variable has probability density function

a Find k.

b Find the exact value of the mean.

c Find the exact value of the mode.

d Show that the median lies between the mean and the mode.

e Find the median, correct to 3 decimal places.

f Find the exact value of the standard deviation.

14 The probability density function of the continuous random variable X is given by .

a Sketch the graph of y = f(x) and hence find the value of k.

b Find without using calculus.

c Write down the values of the

i mean ii median iii mode.

d Find the exact value of the standard deviation.

f x( ) 29---x 3 x–( ) 0 x 3≤ ≤,=

f x( ) 1108---------x2 6 x–( ) 0 x 6≤ ≤,=

f x( ) 1x--- 1 x e≤ ≤,=

f x( ) 12---ex 0 x log e 3≤ ≤,=

f x( ) 0.5e 0.5x– x 0≥,=

f x( ) 3

π 1 x2+( )----------------------- 0 x 3≤ ≤,=

f x( ) kx2 1 x–( )0⎩

⎨⎧

= 0 x 1≤ ≤elsewhere

f x( ) k 1 x–( ) 1 x 1≤ ≤–,=

Pr X12--->⎝ ⎠

⎛ ⎞



493

chap

ter

12

12.2Size does matter

CD

SAC

analysis task

12.1Time for a sleep

CD

SAC

analysis task

The time, X seconds, for a fast-acting anaesthetic to put a patient to sleep is a continuous random variable with a probability density function , and zero elsewhere.a Use calculus to show that .

b Find the average time in seconds which the anaesthetic takes to put a patient to sleep.c Find the median of the probability density function correct to 4 decimal places.d Find the most common time, in seconds, that the anaesthetic takes to put a patient to

sleep.e Calculate the standard deviation.f Find the probability, correct to 3 decimal places, that a patient will be put to sleep

within one standard deviation of the mean.

f x( ) k 8x x2–( ) 1 x 8≤ ≤,=

k 3245---------=

analysis task 1—time for a sleep

SAC

The length, X centimetres, of the wings of a Goliath moth has been found to have a probability density function , and zero elsewhere.

Part 1 a Use calculus to show that .

b Find the average length of the wings of a Goliath moth in centimetres.

Part 2 c Fifty per cent of Goliath moths have wings longer than m centimetres.

Find the value of m, giving your answer correct to 4 decimal places.

d Find the most common length of the wings of a Goliath moth.

Part 3e Calculate the exact value of the variance of the probability density function.

f Calculate the exact value of the standard deviation in the form .

ExtensionThe speed with which Goliath moths can fly is determined by the length of their wings. If the length of their wings is less than 3 centimetres, they are unable to fly quickly enough to avoid being eaten by the David sparrow, a predator that has recently been introduced to their region.g What percentage of the Goliath moth population will survive the introduction of the

David sparrow?h What percentage of the Goliath moth population that survives the introduction of the

David sparrow has wings longer than 5 centimetres?i What percentage of the Goliath moth population that survives the introduction of the

David sparrow has wings shorter than 4 centimetres?

f x( ) k 6x2 x3–( ) 2 x 6≤ ≤,=

k 196------=

a bc

------------

analysis task 2—size does matter

SAC

12.2


494


Applications of continuous random variables

There are numerous situations where a continuous random variable can be used to calculate probabilities, to make predictions or plan for future eventualities. These predictions may relate to situations involving various quantities such as mass, length, area, volume, and time. We can use the techniques we have learned with regard to properties of continuous random variables to determine the probabilities of specified events occurring or to find important statistics such as the mean and the standard deviation.

Example 1The duration, t minutes, of telephone calls from a particular phone is a continuous random

variable with probability density function , and zero elsewhere.

Find, correct to 4 decimal places, the probability that a telephone call has a duration ofa exactly 5 minutes. b between 2 and 5 minutes. c at least 8 minutes.

Solutiona Pr(T = 5) = 0 as Pr(T = 5) = (t ) dt = 0

b Pr(2 < T < 5) =

= 0.3021

c Pr(T ≥ 8) = Pr(8 ≤ T ≤ 12), since f (t ) = 0 for t > 12. So:

Pr(T ≥ 8) =

= 0.2593

Example 2The temperature, X degrees Celsius, inside a refrigerator has been found to have aprobability density function , and zero elsewhere.a Find the value of k.b Find the probability that the temperature inside the refrigerator is above 5°C.c Calculate the average temperature inside the refrigerator correct to 2 decimal places.

Solutiona We know that .

We use a graphics calculator to evaluate:

So k = 8.

f t( ) 1288----------t 12 t–( ) 0 t 12≤ ≤,=

GC 5.5CAS 5.5

tipThe probabilities in parts b and c can be evaluated with a graphics calculator as shown.

⌠⌡5

5

⌠⎮⌡2

51

288----------t 12 t–( ) td

⌠⎮⌡8

121

288----------t 12 t–( ) td

f x( ) 1k π 2–( )--------------------x cos x

4--- 0 x 2π≤ ≤,=

GC 5.5CAS 5.5

1k π 2–( )--------------------

⌠⎮⌡0

2π

x cos x4--- xd 1=

1π 2–( )

-----------------⌠⎮⌡0

2π

x cos x4--- xd 8=

12 .312 .3



495

chap

ter

12b Pr(X > 5) =

= 0.1213

c μ =

= 3.28The average temperature is 3.28°C.

With the function entered as Y1, the calculations in parts b and c are shown in the screenshot on the right.

Example 3The waiting time, X hours, for the next customer to arrive in a shop is known to have a probability density function f (x ) = αe−αx, x ≥ 0, and zero elsewhere, where α is a constant.a Use calculus to show that f (x ) is a probability density function for any value of α > 0.

It can be shown that for this random variable X, and .b If α = 10, calculate:

i the probability that the waiting time is longer than 12 minutes.ii the average waiting time in minutes.iii the standard deviation of the waiting time in minutes.iv the probability that the waiting time is within two standard deviations of the mean.

Solutiona If f (x ) is a probability density function, we require f (x ) ≥ 0 and f (x ) dx = 1. As e−αx > 0

for all real values of x, it follows that α must be positive. Now check the integral:

αe−αx dx =

= −e−∞ − (−1) (for any value of α > 0 )= 0 − (−1)= 1

b i First convert 12 minutes to 0.2 hours.

Pr(X > 0.2) = 10e−10x dx

= = 0 + e−2

= 0.1353Alternatively, we could calculate

Pr(X > 0.2) = 1 − Pr(X < 0.2)

= 1 − 10e−10x dx

=

= 1 − (−e−2 + 1)= 0.1353

18 π 2–( )---------------------

⌠⎮⌡5

2π

x cos x4--- xd

18 π 2–( )---------------------

⌠⎮⌡0

2π

x2 cos x4--- xd

E X( ) 1α---= E X2( ) 2

α2------=

⌠⌡−∞

∞

⌠⌡0

∞e αx––[ ]0

∞

ii

So the average waiting time is hour, or 6 minutes.

iii σ 2 = E(X2) − μ2

=

=

So σ = .

The standard deviation of the waitingtime is hour, or 6 minutes.

μ E X( ) 1α--- 1

10------= = =

110------

2102--------- 1

102---------–

1102---------

110------

110------

⌠⌡0.2

∞

e 10x––[ ]0.2

∞

⌠⌡0

0.2

1 e 10x––[ ]00.2

–

12.3


496


iv Pr(μ − 2σ ≤ X ≤ μ + 2σ ) = Pr(−0.1 ≤ X ≤ 0.3)= Pr(0 ≤ X ≤ 0.3) (as f (x ) = 0 for x < 0)

= 10e−10x dx

= = −e−3 + 1= 0.9502

Example 4The length X cm of a particular brand of candle is a continuous random variable with

probability density function and zero elsewhere.

Find the probability that a randomly chosen candle has a length: a greater than 13 cm, given that it is at least 12.5 cm.b at most 12.8 cm, given that it is less than 13.2 cm.

Solutiona Pr(X > 13 | X ≥ 12.5) =

= (since (X > 13) ⊂ (X ≥ 12.5))

= (since f (x ) = 0 for x > 13.5)

= 0.2565

b Pr(X ≤ 12.8 | X < 13.2) =

=

= 0.8073

The screenshot at right shows the calculations in each part,using the fnInt command, with the function in Y1.

exercise 12.31 The volume X millilitres per bottle of a particular perfume is a continuous random

variable with probability density function and zero elsewhere.

Find the probability that a randomly chosen bottle has a volume of:a exactly 8.6 millilitres. b between 8.55 and 8.75 millilitres.c more than 8.6 millilitres.

2 The length, X millimetres, of a particular brand of matchstick is a continuous random

variable with probability density function and zero

elsewhere. Find the probability that a randomly chosen matchstick has a length of:a at least 34 millimetres. b between 31 and 32.5 millimetres.c at most 32 millimetres.

⌠⌡0

0.3

e 10x––[ ]00.3

f x( ) 8225----------x 27 2x–( ) 12 x 13.5≤ ≤,=

Pr X 13>( ) X 12.5≥( )∩( )Pr X 12.5≥( )

------------------------------------------------------------------ using Pr A B( ) Pr A B∩( )Pr B( )

--------------------------=⎝ ⎠⎛ ⎞

Pr X 13>( )Pr X 12.5≥( )--------------------------------

Pr 13 X 13.5≤ ≤( )Pr 12.5 X 13.5≤ ≤( )--------------------------------------------------

Pr X 12.8≤( ) X 13.2<( )∩( )Pr X 13.2<( )

-----------------------------------------------------------------------

Pr X 12.8≤( )Pr X 13.2<( )--------------------------------

GC 5.5CAS 5.5

12.3Waiting tim

e

CD

TAI

f x( ) 85--- 10 x–( ) 8.5 x 9≤ ≤,=

f x( ) 2279---------x 33 x–( ) 30 x 33≤ ≤,=



497

chap

ter

123 The duration, T minutes, of commercial breaks in prime time on a particular television

station is a continuous random variable with probability density function

and zero elsewhere. Find the probability that a

commercial break has a duration of:

a more than 5 minutes. b between 4 and 5 minutes. c exactly 4.5 minutes.

4 The area, X cm2, of a particular brand of tile is a continuous random variable with

probability density function and zero elsewhere. Find the

probability that a randomly chosen tile has an area of:

a at least 25.1 cm2. b between 25.5 and 26.8 cm2. c less than 25 cm2.

5 The volume, X millilitres, of liqueur in the small bottles of a particular brand is labelled as 30 mL and is known to have a probability density function f(x) = k, 29.4 ≤ x ≤ 30.9, and zero elsewhere.


b Find the probability that a bottle has a volume not less than that shown on the label.

c State the value of the median.

d Calculate the average volume of liqueur in the bottles.

e Calculate the exact value of the standard deviation.

f Find the exact probability that the volume is within one standard deviation of the mean.

6 The mass, X kilograms, of sugar sold in bags labelled as 2 kg is known to have a probability density function f(x) = k(5x + 3), 1.95 ≤ x ≤ 2.15 and zero elsewhere.


b Calculate the percentage of bags of sugar with a mass not less than that shown on the label. Give your answer correct to 1 decimal place.

c Find the value of the median correct to the nearest gram.

d Calculate the average mass of the bags of sugar, correct to the nearest gram.

e Calculate the value of the standard deviation.

7 The length, X millimetres, of the whiskers of a type of rat has been found to have a

probability density function and zero elsewhere.


b Find the probability that a rat has whiskers shorter than 1.5 mm long.

c Calculate the average length of the whiskers correct to 2 decimal places.

d Calculate the standard deviation correct to 4 decimal places.

e Find the probability that the length of the whiskers is within two standard deviations of the mean.

8 The daily rainfall, X millilitres, in a particular town has been found to have

a probability density function and zero elsewhere.


b Find the most common daily rainfall correct to 2 decimal places.

f t( ) 81455------------ 3t2 4t+( ) 3 t 5.5< <,=

f x( ) 3154--------- 27x x2–( ) 25 x 27< <,=

f x( ) kπ 2–------------x cos

x2--- 0 x π≤ ≤,=

f x( ) 1kπ------x sin

x6--- 0 x 6π≤ ≤,=

12.3


498


c Calculate the average daily rainfall correct to 2 decimal places.

d Calculate the standard deviation correct to 4 decimal places.

e Find the probability that the daily rainfall is within two standard deviations of the mean.

9 The waiting time T hours in a doctor’s surgery has been found to have a probabilitydensity function f(t) = 6e−6t, t ≥ 0, and zero elsewhere.

a Find the average waiting time in minutes (see example 3 in this section).

b Calculate the probability that the waiting time is shorter than 5 minutes.

c Calculate the exact value of the median waiting time in the form .

d Find the standard deviation of the waiting time (see example 3 in this section).

e Find the probability that the waiting time is within two standard deviations of the mean.

10 The mass X kilograms of prawns caught in a net by a fishing trawler has been found to

have a probability density function , and zero elsewhere.


b Calculate the probability that the mass of prawns caught is greater than 2 kilograms.

c Find the exact average mass of prawns caught.

d Use calculus to calculate exactly the most common value for the mass of prawns caught.

e Calculate the exact value of the standard deviation in the form .

f Find the probability that the mass of prawns caught is within two standard deviations of the mean.

11 The height, X m, of a particular brand of basketball pole is a continuous random

variable with probability density function , and zero

elsewhere. Find the probability that a randomly chosen pole has a length:

a greater than 3.3 m, given that it is at least 3.2 m.

b at most 3.25 m, given that it is less than 3.36 m.

12 The mass, X kg, of bags of a particular brand of sugar is a continuous random variable

with probability density function , and zero

elsewhere. A bag is rejected if its mass is less than 1 kg. What is the probability that a randomly chosen bag is accepted if its mass is greater than 0.98 kg?

13 The volume, X millilitres, per bottle of a particular brand of liqueur is a continuous

random variable with probability density function ,

and zero elsewhere. A bottle is rejected if its volume is less than 18.8 millilitres. What is the probability that a randomly chosen bottle is accepted, if its volume is greater than 18.6 millilitres?

14 The time, T seconds, for an electrical component to respond to a stimulus is a continuous

random variable with probability density function ,

and zero elsewhere. A component is rejected if it takes more than 0.25 seconds to respond. What is the probability that a randomly chosen component is rejected, if it has a response time of less than 0.27 seconds?

1a--- log e b

f x( ) 1k--- 16x x3–( ) 0 x 4≤ ≤,=

a bc

------------

f x( ) 56--- 17 4x–( ) 3.1 x 3.4≤ ≤,=

f x( ) 809

------x 11 10x–( ) 0.95 x 1.1≤ ≤,=

f x( ) 8171--------- 20x x2–( ) 18.5 x 20≤ ≤,=

f t( ) 500253--------- 3t2 2t 2+ +( ) 0.1 t 0.3< <,=



499

chap

ter

1215 The time T days between breakdowns of a sensitive piece of electronic equipment is a

continuous random variable with probability density function f(t) = ce−ct, t ≥ 0, and zero otherwise, where c is a constant.

a If Pr(T ≤ loge 4) = , find c.

b Find, correct to 3 decimal places, the probability that after the equipment is repaired it does not break down again for:

i at least 5 days.

ii at least 10 days if it does break down in the first 5 days.

16 The response time T minutes for a person to react to a stimulus has been found to have a probability density function f(t) = kte−2t, t ≥ 0, and zero elsewhere.


b Calculate the probability that the response time is greater than 2 minutes.

c Find the average response time μ minutes.

d Use calculus to calculate the most common value for the response time.

e Find the standard deviation σ minutes.

f Find the probability that the response time is greater than (μ + 2σ) minutes.

12---

12.4Giving is a bloody good thing to do

CD

SAC

analysis task

When people donate blood to the Blood Bank, 450 millilitres of blood is taken from each person. The rate at which people bleed varies and it takes between 10 and 30 minutes for the desired volume to be collected. The time, T minutes, for a person to donate blood has

been found to have a probability density function

a Use calculus to show that .

b Evaluate the average time, in minutes, for a person to donate blood.

c Find the median time for a person to donate blood, correct to the nearest second.

d Find the most common time, in minutes, that it takes for a person to donate blood.


f What percentage of people will take longer than 25 minutes to donate blood?

g What is the probability that a person takes between 15 and 20 minutes to donate blood?

h What is the probability that a person takes less than 17 minutes, given that the person takes less than 22 minutes to donate blood?

i Find the probability that the time taken to donate blood is within two standard deviations of the mean.

f t( ) k 50t t2– 400–( )0⎩

⎨⎧

=10 t 30≤ ≤elsewhere

k 310 000 -----------------=

a b

analysis task 3—

giving is a bloody good thing to do

SAC

12.3


Chapter review

500

MathsWorld

Mathematical Methods Units 3 & 4

Summary


Let

X

be a continuous random variable with probability density function (pdf)

f

(

x

).

.

f

(

x

)

≥

0 for all real values of

x

.

f

(

x

)

dx

=

1

.

Pr(

X

=

x

)

=

0 so Pr(

x

1

≤

X

≤

x

2

)

=

Pr(

x

1

<

X

<

x

2

)

=

Pr(

x

1

≤

X

<

x

2

)

=

Pr(

x

1

<

X

≤

x

2

)

.

Pr(

x

1

≤

X

≤ x2) = f(x) dx

Statistics of continuous random variables

. The mean μ = E(X) = f(x) dx.

. The median, m, is the value such that Pr(X ≤ m) = , so we can calculate the median by

solving f(x) dx = .

. The mode M occurs at the maximum value of the probability density function, f(x), which can be found by solving f ′(x) = 0 or by using the graph of f(x) to find the maximum. There can be more than one mode.

. The variance var(X) = σ2 = (x − μ)2f(x) dx = x2f(x) dx − μ2; the standard deviation

.

Revision questionsShort answer1 State whether each of the following functions could represent a probability density

function if the function has rule:

a b

2 Sketch the graphs of the following functions and show that they are probability density functions.

a b

3 Find the value of k in each of the following probability density functions.

a b

⌠⌡−∞

∞

⌠⌡x1

x2

⌠⌡−∞

∞

12---

⌠⌡−∞

m 12---

⌠⌡−∞

∞⌠⌡−∞

∞

σ var X( )=

f x( )0.3

0.4

0⎩⎪⎨⎪⎧

=


g x( )0.3

0.4

0⎩⎪⎨⎪⎧

=


f x( )239------ 11 3x–( )

0⎩⎪⎨⎪⎧

=0 x 3≤ ≤

elsewhereg x( )

121------e3x

0⎩⎪⎨⎪⎧

=0 x log e 4≤ ≤

elsewhere

f x( ) kx 5x 2+( ) x 0 3,( )∈,= g x( ) k sin 4x 0 xπ8---≤ ≤,=


Continuous random variables 12chap

ter

501

4 Define the probability density functions with the following graphs and verify that they are indeed probability density functions.a b

5 The length, X centimetres, of a particular brand of candle is a continuous random variable

with probability density function , and zero elsewhere.

Find the probability that a randomly chosen candle has a length of:

a exactly 27 centimetres. b between 26.4 and 27.6 centimetres.

c at most 27 centimetres.


. Find the:

a exact value of the mean. b exact value of the median.

c exact value of the mode. d variance.


. Find the:

a mean. b exact value of the median.

c mode. d standard deviation.


. Find the exact value of the:

a mean. b mode.


, and zero elsewhere, find the:

a mean. b median. c exact value of the mode.

10 For the continuous random variable with probability density function

, and zero elsewhere, find:

a Pr(1.3 ≤ X < 3.5) b Pr(X ≤ 2.8) c Pr(X > 1.7) d

11 For the continuous random variable X with probability density functionf(x) = 3x2, 0 ≤ x ≤ 1, and zero elsewhere, find the probability that X lies between themean and the mode.

12 Consider the function f(x) = kxk − 1, 0 ≤ x ≤ 1, and zero elsewhere.

a Show that f is the probability density function of a continuous random variable X for any k > 0.

b If E(X) = 0.8, find k.

0

1

2

y

x10

1

2

y

x1 2

(1, )43

(2, )23

f x( ) 3160---------x 28 x–( ) 26 x 28≤ ≤,=

f x( ) 160------ 4x 3+( ) 1 x 5≤ ≤,=

f x( ) 12--- cos

x2--- 0 x π≤ ≤,=

f x( ) 328------x2 8 3x–( ) 0 x 2≤ ≤,=

f x( ) 3

π 1 x2–----------------------- 0 x

32

--------≤ ≤,=

f x( ) 124------ 2x 3+( ) 1 x 4≤ ≤,=

Pr X 19≤( )


502


Extended response1 A continuous random variable X has probability density function

, and zero elsewhere.

a Find the value of k. b Find the exact value of the mean.c Calculate the value of the median, correct to 3 decimal places.d Find the exact value of the mode. e Find the exact value of Pr(0.5 < X ≤ 1.5).f Calculate Pr(X < 1.5 | X > 0.5).

2 A swimming pool has a pipe with an intermittent leak. The volume, X litres, of water lost during a period that the pipe is leaking has been found to have probability density

function , and zero elsewhere.

a Find the value of k.b Determine the probability that less than four litres of water are lost during a period of

leaking.c Find the probability that during a period of leaking less than 2 litres of water are lost

given that less than 4 litres of water are lost.d Calculate the average volume of water lost per period, correct to the nearest millilitre.e Calculate the standard deviation correct to 4 decimal places.f Find the probability that the volume of water lost in a period of leaking is within two

standard deviations of the mean.

3 The length, X centimetres, of the fur of a type of otter is known to have a probability density function f(x) = k(6x2 − x3 − 8x), 2 ≤ x ≤ 4, and zero elsewherea Find the value of k.b Find the probability that an otter has fur longer than 3.5 cm.c Find the probability that an otter has fur longer than 3.5 cm, given that it has fur longer

than 3 cm.d Calculate exactly the average length of the fur of this type of otter.


f Find the probability that the length of the fur is within two standard deviations of the mean.

4 The time, T days, it takes for a seed to germinate after being planted is known to have a probability density function f(t) = ke−0.5t, t ≥ 0, and zero elsewhere.a Show that k = 0.5.b Find the probability that a seed takes longer than 1 week to germinate.c Find the probability that a seed takes longer than 1 week to germinate given that it takes

longer than 3 days.

If a random variable T has probability density function given by f(x) = αe−αt, t ≥ 0, and zero

elsewhere, then it is known that and .

d Use the results above to find:i the average time for a seed to germinate.ii the standard deviation of the time for a seed to germinate.

e Find the probability that the time taken for a seed to germinate is within two standard deviations of the mean.

f Calculate the exact value of the median time for a seed to germinate.

f x( ) 1k--- 14x 3x2–( ) 0 x 2≤ ≤,=

f x( ) 1k---x sin

x4--- 0 x 2π≤ ≤,=

a bc

------------

E T( ) 1α---= E T2( ) 2

α2------=

CD12.5–12.7A

trio of testsC

hapter tests


DowseyMathsMethodsCh12.fm Page 473 Monday, January 16 ... World Mathematical Methods... ·...

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