chapter 14 data and reasoning... · chapter 14 calculus [extension] the time between the arrival of...
TRANSCRIPT
CHAPTER
14
Calculus [extension]
The time between the arrival of customers at a counter is a random variable which has an exponential distribution.
424 CALCULUS [EXTENSION]
This chapter is included as part of the calculus requirement of the additional area of study in the extensions of Reasoning and Data.
It includes an extension of the work covered in Chapter 6 to include other continuous probability distributions and applies the differential and integral calculus to mean, mode,
median and variance related to these distributions. A knowledge of differentiation and antidifferentiation of polynomial and exponential functions is assumed. Before considering these, we will consider: a the power series for ex
b antidifferentiation by parts.
14.1 Power series for ex
A power series is an infinite series of ascending powers of x of the form:
ao + 01X + 02X2 + 03X
3 + 04X4 + . . .
and, if there is a series for eX, it will be of this form. We assume that an infinite series can
be differentiated term by term and use the fact that! (ex) = ex .
Set: ex
= a0 + a 1x + a2x2 + a 3x3 + a4x
4 + . . . .. ...................................... (1)
Put: x = 0 in (1), and so e0 = a0 = 1
Differentiate both sides of (1), assuming we can differentiate a series term by term:
ex
= a1 +- 2a2x + 3a3x2 + 4a4x3 + . . . .. ............................................ (2)
Put: x = 0 in (2), and so e0 = a 1 = 1
Differentiate both sides of (2):
ex
= 2 . la2 + 3 . 2a 3x + 4. 3a4x2 +. . . .. ........................................... (3)
Put: x = 0 in (3), and so e0 = 2 . la2 = 1, i.e. a2 = ii
Differentiate both sides of (3):
ex
= 3 . 2 . la 3 + 4. 3 . 2a4x + .................................................... (4)
Put: x = 0 in (4), and so e0 = 3 . 2 . la 3 = 1, i.e. a3 = i
i and so on.
Substituting the values of a0, a1, a2, a3, • • • in (1) yields:
In particular, if x = 1, then: 1 1 1
e = 1 + 1 + 2 ! + 3! + 4! + . . .
""' 2.7183 to four decimal places.
If x is small, then 1 + x is a good linear approximation to ex .
For example:
e0•02 ""' 1 + 0.02 = 1.02 2
1 + x + � is a better approximation.
e0·02
""' 1 + 0.02 + 0.0002 = 1.0202
CALCULUS (EXTENSION) 425
The power series for ex is true for all x. So, if in place of x, we write - x, then:-x x
2 x
3 x
4
e = l - X + 2! - 31
+ 4! • .
.
e2x = 1 + 2x + (�)2
+ (�)3
+ <7;t + ...
1 2x 2x2 4 3 2 4 = + + + 3X + 3X + ...
Example 1 Write the power series for ex and e - x and calculate the values of e - 0
·5 correct to four
decimal places.
In place of x, write - x in this identity. -x x
2 x
3 x
4
So: e = 1 - x + 21 - 31
+ 41 -
0.5 = 1 0 5 + (0.5)2 (0.5)3
+ (0.5)4
+ e + · 2!
+ 3! 4!
· · ·
-0.5 = 1 - 0 5 + (0.5)2 - (0.5)3 (0.5)4
+ e · 2! 3! +
4! · · ·
Positive terms Negative terms
1 0.5
0.125 0.020833
0.002604 0.000260
0.000022 0.000002
1.127626 0.521095
e0•5
""' 1.127626 + 0.521095= 1.648721= 1.6487 (to four decimal places)
e - 0·5
""' 1.127626 - 0.521095 = 0.606531 = 0.6065 (to four decimal places)
These results can be checked using the ex button on your calculator.
Operation sequence
For e0·5
: [)(ID.
For e-0
.5: OlIDlla
Display
1-6487 . . .
D. 6065
The power series gives quite a good approximation if we use a sufficient number of terms.
426 CALCULUS (EXTENSION)
Example 2 By using a power series for e
x and e -zx as far as the term in x2 solve, approximately, the equation 1 + 2e
x = e - zx.
2ex
= 2 ( 1 + X + �)
= 2 + 2x + x2
e -2x = 1 - 2x + (2x)2
2 = 1 - 2x + 2x2
1 + 2ex
= e-Zx
becomes: 3 + 2x + x2
= 1 - 2x + 2x2
x2
- 4x - 2 = 0
- b ± .J b2 - 4ac x =-------,where a= 1, b = -4, c = -2.
2a
4 ± -v'24 2
= 2± v6 """ 4.45 or -0.45
However, on testing, only x = -0.45 gives a reasonable approximation. Why?
Exercises 14a
1 Write the power series for ex and calculate to four decimal places:
a e-o.1s b eo.2s C e-
0.02
Check your results with a calculator.
2 From the series forms of ex and e -x find a series expression for ½<e
x - e -x) and so
calculate its value for x = ½ to four decimal places. 3x -x
3 Express e -x
e in the form e0x - e -ax and find a series expression for it.
e
Evaluate it correct to four decimal places where x = 0.2.
4 By expressing ex in the form of a series as far as the term in x3
, evaluate
11 x2
exdx. -I
5 By expressing e -o.zx in the form of a series as far as the term in x 3
, evaluate correct
1 0.8
to four decimal places xe -o.zx dx. -0.2
6 Write the first three terms in the series expansion for e -xn and use them to evaluate,
to four decimal places, [1
e -xiz dx.
7 Solve for x the equation ex = 1 + e -x by:
a using a substitution, y = ex
b using the series expansions of ex and e -x as far as the terms in x2.
Give your answers correct to two decimal places.
CALCULUS [EXTENSION) 427
8 Givenf(x) = ¼ ex + e - 2x. Find the function g, where g(x) is the quadratic polynomial
obtained by replacing ex and e - 2x with the first three terms of their series expansions.
9 Write the coefficient of x2 in the expansion of ex2
+x.
10 Find the expansion, in increasing powers of x up to and including the term in x3, of
2ex + e - 2x - 3x2
•
11 Evaluate i I e -x212 dx by approximating e - x2
12 by a fourth degree polynomial in x.
14.2 Antidifferentiation by parts The product rule of differentiation is associated with a technique known as antidifferentiation by parts. If u(x) and v(x) are differentiable then the product rule tells us that:
� (uv) = u dv + v dudx dx dx
Transposing this we obtain: dv d du u dx = dx (uv) - v dx"
So: f U : dx = UV - f V : dx
The main value of this rule lies in its application to functions whose antiderivatives cannot be found in easier ways. The integrand on the left-hand side is seen to be a product of two expressions involving x; one of these is denoted by u and the other by:. The choice is made in such a way that the
antiderivative on the right-hand side is one which can be found by normal techniques such as change of variable. We have omitted the arbitrary constant c. This is normally inserted into the solution at the appropriate point.
Example 3 Find an antiderivative of x cos x.
To find f x cos x dx, we put u = x and: = cos x.
With this choice, : = 1 and v = sin x.
Now: f utdx = uv - f v:dx
So: f xcosxdx = xsinx - f sin xdx
= X sin X + cos X + C.
Note that the choice u = cos x and : =· xis not useful since we would have
f x cos x dx = ½x2 cos x - f ½x2 sin x dx, and the second term on the right, namely
f ½x2 sin x dx, is more complicated than f x cos x dx which is what we started with.
428 CALCULUS (EXTENSION]
Example 4
Evaluate 12
xe -x dx.
� Put u = x and t = e -X.
S du 1 d -x o dx = an v = -e
JU : dx = UV - JV : dx
J xe -x dx = - xe -x + J e -x dx
-xe -x - e -x + c= - (x + 1 )e -x + c
So: 1 2
xe -X dx = [ - (x + 1 )e -X r
-3e- 2 - (-2e- 1)
= 2e- 1 -
3e- 2
""'0 .3299 x2 x3 x4 xs
Alternatively, using e-x""' 1 -
x + 21 - 31 + 41 -S! we get:
xe dx ""' x l -x + - - - + - - - dx12 -x 12 (
x2 x3 x4 xs
) I I 2 6 24 120
12
( x3 x4 xs x6
) =
I
X - X
l
+ 2 -6 + 24 -120 dx
[x2 x3 x4 xs x6 x1 ]2
= 2 - 3 + 8 - 30 + 144 -840 I
(2 - i + 2 - !� + i - /;s) - (½ - ½ + ½ - 310 + 1!4 - s!o)= 0.2946 (a fair approximation)
The method of antidifferentiation by parts will be used mostly as a last resort when other known techniques fail . It is used very effectively to find antiderivatives of products of different kinds of expressions, e.g. algebraic and circular such as x cos x or algebraic and logarithmic, such as x log x and the inverse circular functions. Sometimes the operation must be applied more than once.
Exercises 14b
1 Find the antiderivatives in the following:
a J xsin xdx
c f xcos 2xdx
e Jx�dx
g J xl�gexdx
b J xsin2xdx
d J xsec 2 xdx
f jx..Jx+l dx
h J xe2x dx
CALCULUS [EXTENSION) 429
2 Evaluate:
a 13
te1 dt b f 2 X loge X dx
1�,2 i�/3c
O
xsinxdx d (x + 1) cos 2.x dx �/6
3 Show that f xnex dx = xnex - n f xn - 1ex dx and find the antiderivative of:
a xex b x3ex
4 Find the derivative of xn loge x and deduce, to three decimal places, the value of: a f 2 X2 loge X dx b f 2 x3 loge X dx
14.3 Other density functions It was once thought that the normal distribution would fit all sets of measurements occurring in nature. However, this theory was later dispelled because of the existence of so many distributions which were obviously non-normal. We consider now other probability density functions which have properties similar to the normal density function but whose equation defining the function is such that areas under the curve can be calculated by simple arithmetical means or with the aid of the integral calculus. The properties common to all probability density functions are: a The variable is continuous and can assume, theoretically at least, all real number values. b The function values are non-negative for all x in the domain. c The total area under the curve is unity.
i.e. J: f(x) dx = 1
Usually, however:
lb
f(x) dx = 1
the function values being zero for all x >band all x < a. Some examples of non-normal distributions are:
(i) Rectangular or uniform distributionRectangular or uniform distribution is defined by:
J(x) = { b � a'
0,
Figure 14-1
y
1 b-a
6
for a:,;; x:,;; b
for all other values of x.
a b X
430 CALCULUS [EXTENSION)
(ii) Triangular distributionTriangular distribution is defined by:
f(x) = 1
{ :2
(x + a), for - a :;;;; x :;;;; 0
2 ( - x + a), for 0 ::;;;
x :;;;; a
g, for all other values of x.
y
-a 0
Figure 14-2
Example 5
a X
To perform a certain task , a student spends up to five minutes. What is the probability that the task is performed in less than two.minutes, on the assumption that the distribution of times taken is: a uniform? b triangular , the triangle being isosceles? State a rule which defines the probability density function in each case. Can we give a reasonable justification of why the distribution should be triangular or uniform as the case may be?
a Since the distribution of times is uniform over the five minute interval we can assign equal probabilities to any two time intervals of equal width. If the probabilities are to be measured by areas, we require a rectangle of dimensions
5 by ½ since the total area is unity.
y
0.2
0 2 5 X
Figure 14-3
The shaded area represents the probability that the task is performed in less than two minutes. The required probability is 0.4.
f(x) = [ 00·2
• ffor
Ot,h::;;; x ::;;;l
5 f defines the probability density function., or o er va ues o x
CALCULUS (EXTENSION) 431
Alternatively, the required probability can be evaluated by integration. If X denotes the random variable, then:
Pr(X � x) = lx
f(x) dx for 0 � x � 5
= lx
0.2dx
The function Fwhere:
{
0,x<0 F(x) = �, 0 � X � 5
1,x>5
is called the cumulative distributionfunction, the graph of which is shown in Figure 14-4. To find the probability that X < 2, we use
. 2 F(2) = 5
.
0.4
0
Pr (X..:x)
2
Figure 14-4
b Since the distribution of times is triangular, we require an isosceles triangle whose base measurement is 5 and altitude 0.4 as shown in Figure 14-5. Why?
Figure 14-5
f(x)
0.4
0 2 2.5 5 X
X
The shaded area in Figure 14-5 represents the probability that the task is performed in less than two minutes. By simple calculation of the area of a triangle, the required
b b·1· . 8 pro a 1 1ty 1s 25.The equations of the lines forming the triangle are y = 11 for O � x � 2.5 andy = - 2� (x - 5) for 2.5 � x � 5. So the function/where:
{ 11, for O �'x � 2.5
f(x) = 4 f - 25 (x - 5), or 2.5 � x � 50, for other values of x.
defines the probability density function in this case.
432 CALCULUS [EXTENSION]
The required probability can likewise be found by integration.
Pr(X < 2) = 12
il dx
8 25 (as before)
Pr(X � x) = lx
f(x) dx for O � x � 2.5
= lx
il dx
-2x2
-25
and: Pr(X � x) = 1-"
f(x) dx + J2
·5 il dx for 2.5 � x � 5
.5 0
4 l
x 1 - - (x - 5) dx + -25 2.5 2
[ 2x2 4X] X 1 - -25 + 5 2.5 + 2
= _..!_ (2x2
- 20x + 25) 25
The function F where:
F(x) =
0, 2x225' - 1... (2x2 - 20x + 25) 25 ' 1,
x<O
0 � X � 2.5
2.5 � X � 5
x>5
is the cumulative distribution function, the graph of which is shown in Figure 1
4
-6.
Pr ( X < x)
0.92
0.5 -------------------------------
0.32 -----------------------
0 2 2.5 4 5 X
Figure 14-6
CALCULUS [EXTENSION] 433
So: Pr (X � 2) = F(2) = 0. 32
Pr (X � 4) = F(4) = -15
(32 - 80 + 25) = 0.92
(iii) Polynomial distribution
Example 6
y
4A
A continuous random variable, X, can assume values only between 0 and 2, and its pdf is defined by f(x) = A (2 - x)2 where Ais a constant. Calculate:
3A
2A
a the numerical value of A A
b the probability that Xis greater than 1.5 c the probability that Xlies between 0.5 and 1.
0
a Since the total area under the curve is unity, then:
b
C
A l 2 (2 - x) 2 dx = l
A 1 2
(2 - x) 2 dx = A 12
(4 - 4x + x 2) dx
[ x3 ] 2
= A 4x - 2x2 + 3 o
=A[s-s+U
= 8A = l3
A= l8
Pr(X> 1.5) = if 2 (4 - 4x +
x 2) dx1.5
= - 4x - 2x + -3 [
2 x 3] 2
8 3 u
1 64
Pr(0.5 < X < 1.0) = i [ 4x - 2x 2 + �3
] �-5
19 64
Figure 14-7
= i [ ( 4 - 2 + ½) - ( 2 - ½ + i4 ) ]
1964
1 64
2 X
434 CALCULUS (EXTENSION]
(iv) Exponential distributionExponential distribution is defined by:
f(x) = [ 'J,.e->.x, for all� � 0
0, otherwise f(x)
0
Figure 14-8
The cumulative distribution function (cdf) is defined by:
So:
F(x) = lx
'J,.e->.x dx for all x � 0
1 - e->.x
P ( ) F( [ 1 - e->.x, X � 0.
r X � x = x) =
0 0, x< .
Pr ( X � x)
0
Figure 14-9
X
X
The exponential distribution is applicable in situations where we are concerned with the length of time between events. The variable is time which, theoretically at least, can assume any non-negative value, however large. The proprietor of a service station may be concerned with the time lapse between successive cars entering the driveway of the service station. The manager of a departmental store is concerned with the time lapse between the arrival of customers. The manufacturer of electric light globes is concerned with the time a globe will burn from the moment it is switched on. (In Chapter 4.4, mention was made about this distribution in connection with the Poisson distribution.)
CALCULUS [EXTENSION) 435
Example 7 The time, X minutes, between cars passing a checkpoint is a random variable defined bythe probability functionf(x) = 0.2e -0·2X, x � 0. Calculate:
a Pr (X � 5) b Pr(2 � X � 4)
f(x)
0.2 f (x) = o.2e·02•
Figure 14-1 O 0 2 4 5 X
a The required probability is measured by the area under the curve to the right of X = 5 in Figure 14-10.
Pr(X � 5) = 1 - Pr(X < 5)
= 1 - 0.2 l s e-o.2xdx
= 1 _ [ -e-o.2xJ:
= 1 - [ -e - 1 + I]
= 0.3679 b The required probability is measured by the area under the curve between
X = 2 and X = 4.
Pr(2 � X � 4) = 0.2 i4 e-0-2x dx
= _ e -o.s + e -o.4= 0.2210
Exercises 14c
1 Verify that each of these functions may be a probability density function (pdf) and sketch its graph:
a f(x) = { 0.5, for O � x � 20, elsewhere
b f(x) = { r· for 0 � X � 1elsewhere
c f(x) = { lxl , for - I � x � 10, elsewhere 0l ,
436 CALCULUS (EXTENSION)
d f(x) = { ½ (x + 1), for -1 :,;; x :,;; 1
0, elsewhere
e f(x) = { sinx,0,
for 0:,;; X:,;; ! elsewhere
f f(x) = { e0,-x, for all x � O elsewhere
2 In each of the following questions find the value of A so that the functions may be probability density functions.
{ Ax, for 0 :,;; x :,;; 4
a f(x) = A (8 - x), for 4 :,;; x :,;; 8 0, elsewhere
b f(x) = { A(2 - x),
0, for -2:,;; X:,;; 2 elsewhere
c f(x) = { Ax(6 - x), 0,
for 0:,;; X:,;; 6 elsewhere
d f(x) = {Ae-x15, for allx � 0
0, elsewhere
3 Xis a random variable denoting the number of minutes in excess of two hours which a person takes to travel from one town to another. The probability density function is defined by:
{ k(l0 + x),
f(x) = k(l0 - x), 0,
a Find the value of k.
b Sketch the graph off.
c Find the probability that: (i) Xis less than 5
-10:,;;x:,;;o0:,;; X:,;; 10
elsewhere
(ii) Xis less than 0, · given that Xis less than 5(iii) -2 :,;; X < 3.
4 The probability density function of a random variable, X, is defined by:
f(x) = { kx2(1 - x), 0,
0:,;; X:,;; 1 elsewhere
a Determine k.
b Find the probability that Xis less than l
c Find the probability that Xis less than½, given that Xis less than l 5 A continuous random variable, X, can assume values only between 0 and 3, and its pdf
is defined by f(x) = Ax2(3 - x), where A is a constant such that the area under the curve is unity. Calculate: a the numerical value of Ab the probability that Xis less than 2.
CALCULUS [EXTENSION) 437
6 a Show that the function defined by:
{ 0.le- 0·1\ X � 0
f(x) = 0, otherwise can be a probability density function.
b The number of days, X, that a certain drug, A, takes to cure a disease, B, is a random variable with probability density function defined above. Sketch the graph of the function.
c Find the probability that: (i) X< 5
(ii) X> 10 (iii) 5 <X < 10 (iv) X>10IX>5
7 A continuous random variable, X, can assume values only between O and 2 and its pdf is defined by f(x) = kx(2 - x) 3
• Determine: a the numerical value of k b the probability thatX � xwherex = 0, 0.5, 1, 1.5, 2.Plot the cumulative graph for the probability distribution of X.
8 The probability density function, f, of a random variable, X, is given by:
f(x) = [ 6 - 2x, 2 � X � 30, elsewhere
a Obtain a function F for the probability that Xis less than any specified value x.
b Calculate the probability that: (i) X<2.2
(ii) X> 2.5(iii) 2.2 <X < 2.5.Plot the cumulative graph for the probability distribution of X.
9 Y is a continuous random variable with probability density function:
f(y) = [ �.-y· �ti Find: a Pr(Y<2) b Pr(Y < 3 I Y> 1).
10 The random variable, X has pdf such that:
f ) [ k, 0 � X � 1(x = 0, otherwise
a Find the value of k. b Find the cumulative distribution function of X, i.e. the function F, such that
F(x) = Pr (X � x).
11 The time, X seconds, between arrival of particles at a radiation counter has been found to have the cumulative distribution function:
F(x) = [ 0, for x < 0 1 - e - X, for x � 0
a Sketch the graph of y = F(x).
b Find: (i) Pr(X � 1)(ii) Pr(l < X � 2)
438 CALCULUS [EXTENSION]
12 The number of hours, X, that an electric light globe will burn from the moment it is first switched on is a random variable defined by:
f(x) = 'i,.e->-x, x;:, 0 and 'A= 1�00Find the probability that a randomly selected globe: a will burn for more than 1000 hours b will burn between 100 and 1000 hours.
13 The length of time, in minutes, that a person speaks on the telephone is a random variable with probability density function given by:
f(x) = 0.2e - 0·2X, x ;:: 0
What is the probability that the length of a call is: a more than 10 minutes? b less than five minutes? c between five and 10 minutes?
14 The time, Xminutes, between haircuts at a ladies' salon is a random variable defined by:
f(x) = 0.05e - 0·05X, x ;:, O
What is the probability that a haircut: a exceeds 20 minutes? b takes between 20 and 30 minutes?
14.4 Measures of location for probability distributions
The mean, median and mode are measures of central tendency and are often called measures of location because they locate some central value of a probability distribution.
Mean
If Xis a continuous random variable with a probability distribution function (pdf), the mean of Xis defined by:
µ = E(X) = J: xf(x) dx.
In a number of situations, the values of the variable lie between X = a and X = b and
are zero elsewhere, and so J: xf(x) dx = lb
xf(x) dx. In this case:
I µ � E(X) � f xf(x)dx, I Compare this definition of the mean of a continuous random variable with the mean of a discrete random variable:µ = �x Pr(X = x).
Recall that just as� Pr(X = x) = 1 for a probability distribution of a discrete variable, so the total area enclosed by a pdf and the X-axis is equal to 1.
i.e. ( J:J(x) dx = 1 for a continuous variable.
CALCULUS [EXTENSION] 439
Example 8 A random variable, X, has a pdf defined by:
a
b
f(X) = { �' 0 � X � 2
0, elsewhere
f(x) = { sin x, 0 � x � !0, elsewhere
Find, in each case, the mean value of X.
a µ, = E(X) = .f X • � dx
= [ ¼x3]:
f (x)
0 1 µ X
Figure 1 4-11
Note also that the area (A) enclosed between the X-axis and the graph of f(x) is:
as required for f(x) to denote a pdf.
b r .. ,2 µ, = E(X) = Jo xsinxdx
Figure 14-1 2
[ ]
.-12 = - x cos x + sin x O
using antidifferentiation by parts.
f (x)
0 µ
X
r .. ,2 Note also that Jc sin x dx =is 1 unit. 0
[ ].-,2- cos x
O
= 1, i.e. the total area under the curve
440 CALCULUS [EXTENSION)
Median The median value of Xis that value, m, such that there is an equal probability that X willbe greater than or smaller than m.
1i.e. Pr(X ,s;;; m) = Pr(X;;., m) = 2Pr(X ,s;;; m) = f:f(x) dx = ½
or, if in reality the values of X lie between X = a and X = b and are zero elsewhere, then:
Example 9
lm
1Pr(X ,s;;; m) = 0
f(x) dx = 2
Find the median value of Xfor each pdf defined as in Example 8.
a
[ ¼x2J: = ½
.!m2 = .! 4 2
m = -J2The median is -./2.
The median can be well illustrated by considering the graph of the cumulative distribution function ( cdf).
Pr { X <:. x)
0.5
0 1 ../2
Figure 14-13
The pdf defined by f(x) = � has cdf defined by:
{o, x<O
x2
F(x) = 4
, 0 ,s;;; X � 2
1, x>2
the graph of which is shown in Figure 14-13.
X
b 1 111
1
0
sin xdx = 2
[ - COS X t = ½-cos m + 1 = l2
cos m = ½
m = JI.3
The median is f.
The pdf defined by f(x) = sin x has cdf defined by:
{o,
F(x) = 1
1,
x<O
COS X, 0 � X � ! x>Ii
2
the graph of which is shown in Figure 14-14.
Pr ( X ,;;; x)
CALCULUS [EXTENSION] 441
1 ----------------------------
0.5
0 X
Figure 1 4-1 4
Mode
A mode of a continuous random variable, X, with a pdf is a number, M, such that F(M) > f(x) for all x in a neighbourhood of M. It is a value of X for whichf(x) has its maximum. It is possible to have more than one mode. The mode of the two probability
density functions, f(x) = � andf(x) = sin x, of Example 9 are 2 and! respectively.
Differential calculus was not necessary in this example.
Example 10 A random variable, X, has a pdfdefined by:
f(x) = p2x2 (1 - x), 0 � x � 1lo, elsewhere
Find the mode of X.
442 CALCULUS [EXTENSION]
f(x) = 12x2 - 12x3, 0 :::;; x ::;;; 1
f'(x) = 24x - 36x2
= 12x(2 - 3x) 2= Owhenx = Oor3.
Whenx < j,f'(x) > 0
When x > j,f'(x) < 0So f has a maximum value when x = l The mode of Xis l
If a distribution is symmetrical, the mean, median and mode will coincide. This is true for the normal pistribution, the rectangular distribution and the isosceles triangular distribution. You may have observed in Examples 9 and 10 that the median lies between the mean and the mode.
Exercises 14d
1 For each of the following probability density functions find: (i) the mean(ii) the median
(iii) the mode.Also check that eachf(x) satisfies the area condition for a pdf.
a f(x) = [0.5, for O :::;; x ::;;; 2 0, elsewhere
b f(x) = 1 � X, 0 ::,;; X ::,;; 1 { 1 + X, - 1 ::,;; X ::,;; 0
0, elsewhere
c f(x) = [io (5 - x), -2 :::;; x::;;; 20, elsewhere
df() [e-x, x �O x = 0, elsewhere
e f(x) = [6x(l - x), 0 :::;; x ::;;; 1 (delete median)0, elsewhere
f f(x) = { � (6 - x), 0 :::;; x ::;;; 6
0, elsewhere
f( ) [2X, 0 ::,;; X ::,;; 1g x = 0, elsewhere
2 The probability density function of a random variable, X, is defined by:
f(x) = [12x2(1 - x), 0 :::;; x ::;;; 10, elsewhere
Calculate: a the mean b the mode.
CALCULUS [EXTENSION] 443
3 The cumulative distribution function of a continuous random variable, X,
F(X) = Pr(X:::;; x), is given by: x<O
F(x) = 4x3 - 3x4 , 0 :::;; x:::;; 1
{o, 1, x> 1
a Find the probability density function of X. b Find the mode of X and show that it is less than the median.
4 The cumulative distribution function of Xis F(x) = 1 - e - xn for x ;;;: 0. Calculate the mean of X.
5 a Find the value of k for whichf(x) represents the pdf of a random variable, X, where:
f(x) =
rkx(X - 1 )2, 0:::;; X:::;; 2 lo, elsewhere.
b Find the mode of X. c Find whether the median of Xis greater than, less than, or equal to 1.
6 If X has cumulative distribution function F given by:
F( ) = [½e\ X :::;; 0 X 1 I -x > 0- 2e , X
a sketch the graph of y = F(x) b state the median of Xc find the interquartile range, i.e. b - a, where a and bare such that Pr(X::,;; a) = 0.25
and Pr(X:::;; b) = 0.75.
7 The time, X seconds, between arrivals of particles at a radiation counter has been found to have the cumulative distribution function:
ro, x<O F(x) = U - e-x, x;;;: 0
a Sketch the graph of y = F(x). b Find the median of X.
c Find the mean of X.
8 A random variable X has density function:
f(X) = ra + bx, 0 � X:::;; 1lo, elsewhere.
Given that the mean of Xis i, find the values of a and b.
9 The pdf of a continuous random variable, X, is defined by:
f(x) = r¼ (1 - x)J,lo,
Calculate: a the median of Xb the mode of Xc the mean of X.
-1 :::;; X:::;; lelsewhere .
444 CALCULUS (EXTENSION]
14.5 The mean [expected value) of g[X) Suppose that Xis a continuous random variable defined on the interval [a, b] and zero elsewhere and that Y is another continuous random variable such that, when X has value x, then Yhas value g(x).
We define the expected value of Yby: E(Y) = E(g(X)) = l
b
g(x)f(x) dx ...................................................... (1)
For example, if f(x) = 2x and g(X) = x2, then g(x) = x2 and so:
E(g(X)) = lb
x2• 2.xdx
If: g(X) = cxX + (3
then: E(aX + (3) = lb
(ax + (3)f(x) dx
= lb
axf(x) dx + lb
(3f(x) dx
= a [b
xf(x) dx + (3 lb
f(x) dx
= aE(X) + (3 since lb
f(x) dx = 1 If: g(X) = X - µ
then: E(X - µ) = lb
(x - µ)f(x)dx
= lb
xf(x) dx - lb
µf(x) dx
= E(X) - µ= 0 since µ = E(X)
14.6 Variance and standard deviation The variance and standard deviation are the measures of spread most frequently used. IfXis a continuous random variable with meanµ, then the variance of X, Var(X), frequently denoted by a2, is defined as the expected value of (X - µ)2. The standard deviation of X,
SD (X), frequently denoted by a, is defined as the positive square root. of the variance.
Var(X) = E(X - µ)2
Example 11
= E(X2 - 2µX + µ 2)
= E(X2) - E(2µX) + E(µ 2)
= E(X2 ) - 2µE(X) + µ2 sinceE(µ2 ) = µ2
= E(X2 ) - 2µ2 + µ2
= E(X2 ) - µ2
= E(X2 ) - (E(X) )2 . . . . .. . . .. . .. .. .. .. . . . . .. .. . . .. .. . . .. . . .. .. . . . .. . .. .. . . .. .. .. .. (2)
Let X be a continuous random variable with uniform distribution defined by:
f( ) [¼, 2 � X � 6x
= 0, elsewhere.
Find the variance of X and the standard deviation of X.
CALCULUS [EXTENSION) 445
Figure 14-15
By symmetry:
f (x)
1 4
0
µ=4
2
Using (1):b
E(g(X)) = i g(x)/(x) dx
Var(X) = E(X - µ)2
I 6 2 l = 2 (x - µ) 4dx
4
I6 1 = 2
(x - 4)2
4dxsinceµ = 4
= - - x2 + 4x [x3 ]6
12 2
2 = (18 - 36 + 24) - <3 - 4 + 8)
=1½
SD(X) = ..Jf
=�
Alternatively, using (2): Var(X) = E(X2) - µ2
= I 6
x2• 1 dx - 16
2 4
= [I;]: - 16
= 1 ½ as before
Example 12
6
Let Xbe a continuous random variable with triangular distribution defined by:
{X, 0 � X � 1
/(X) = 2 - X, 1 � X � 2 0, elsewhere.
Find the variance and standard deviation of X.
X
446 CALCULUS [EXTENSION]
Figure 14-16
By symmetry:
f (x)
0
µ, = 1
Var (X) = 12
(x -µ,)2 f(x) dx
2
= 1 1
(x-1)2 x dx+ f2
(x-1)2(2-x)dx
X
[ x4 2 3 x2]1 [ x4 4 3 5 2 ]2
= 4 - 3x + 2 o + -4 + 3x -2x + 2x 1
1 1 = 12 + 121
SD(X) = -v"f
Exercises 14e
_1 For each of the following probability density functions find the variance of X:
ffi\J( ) [l , 0 � X � 1 \J x
= 0, elsewhere ·
b f(x) = [x, 0 � x � 1 2 -X, 1 � X � 2
FA f( ) [3x2, 0 � x � 1 {) x = 0, elsewhere
d f(x) = [6x(l -x), o_� x � 10, elsewhere
e f(x) = [2x, 0 � x � 1 0, elsewhere
f f(x) = [½ (4 -x), 0 � x � 4 0, elsewhere
{ 3: (2 - X), 0 � X � 2
g f(x) = 0, elsewhere
\.\ f(x) = [1 + X,
- 1 � X � 0� 1 -X, 0 � X � 1
2 The exponential distribution is defined by the pdf:
f(X) - I ' "" [A.e=>o:-- X >.l 0
- 0, _ elsewhere Show that: a E(X)-= _!:--
'A \
b E(X2) = _1_ ; 'A
2
l Var(X) - ;,
3 If .Xhas pdf:
f( [½e-x, x;;;:0x) = i x x<0 ,_e,
find the mean and variance of X.
4 If Xhas pdf defined by:
f(x) = I ½, 2 ::;; x :,;; 4 l 0, elsewhere
find: a E(X)b E(X3) c E(Y) where Y = 2X - 1 d Var (Y).
5 If X has pdf defined by:
f(x) = [t: find: a E(X) b E(X 2
)
c E(3X - 2) d E(X2 + 1).
1 ::,; X:,;; 2 elsewhere
CALCULUS [EXTENSION) 447