Download The continuous probability density function

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Continuous
probability
distributions
12.1
Kick off with CAS
12.2 Continuous random variables and probability functions
12.3 The continuous probability density function
12.4 Measures of centre and spread
12.5 Linear transformations
12.6 Review
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12.1 Kick off with CAS
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To come
Please refer to the Resources tab in the Prelims section of your eBookPlUs for a comprehensive
step-by-step guide on how to use your CAS technology.
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12.2
Continuous random variables
and ­probability functions
Continuous random variables
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Discrete data is data that is finite or countable, such as the number of soft-centred
chocolates in a box of soft- and hard-centred chocolates.
A continuous random variable assumes an uncountable or infinite number of possible
outcomes between two values. That is, the variable can assume any value within a
given range. For example, the birth weights of babies and the number of millimetres
of rain that falls in a night are continuous random variables. In these examples, the
measurements come from an interval of possible outcomes. If a newborn boy is
weighed at 4.46 kilograms, that is just what the weight scale’s output said. In reality,
he may have weighed 4.463 279 . . . kilograms. Therefore, a possible range of outcomes
is valid, within an interval that depends on the precision of the scale.
Consider an Australian health study that was conducted. The study targeted young
people aged 5 to 17 years old. They were asked to estimate the average number of
hours of physical activity they participated in each week. The results of this study are
shown in the following histogram.
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Frequency
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Physical activity
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y
400
350
300
250
200
150
100
50
0
364
347
156
54
0
1
2
3
4
Hours
32
10
5
7
6
7
x
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Remember, continuous data has no limit to the accuracy with which it is measured.
In this case, for example, 0 ≤ x < 1 means from 0 seconds to 59 minutes and
59 seconds, and so on, because x is not restricted to integer values. In the physical
activity study, x taking on a particular value is equivalent to x taking on a value in an
appropriate interval. For instance,
Pr(X = 0.5) = Pr(0 ≤ X < 1)
Pr(X = 1.5) = Pr(1 ≤ X < 2)
and so on. From the histogram,
Pr(X = 2.5) = Pr(2 ≤ X < 3)
=
156
(364 + 347 + 156 + 54 + 32 + 10 + 7)
= 156
970
In another study, the nose lengths, X millimetres, of 75 adults were measured. This
data is continuous because the results are measurements. The result of the study is
shown in the table and accompanying histogram.
454 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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27.5 < X ≤ 32.5
2
32.5 < X ≤ 37.5
5
37.5 < X ≤ 42.5
17
42.5 < X ≤ 47.5
21
47.5 < X ≤ 52.5
11
52.5 < X ≤ 57.5
7
57.5 < X ≤ 62.5
6
62.5 < X ≤ 67.5
5
67.5 < X ≤ 72.5
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Nose length
27.5 32.5 37.5 42.5 47.5 52.5 57.5 62.5 67.5 72.5
Length in mm
x
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35
30
25
20
15
10
5
0
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Frequency
Frequency
Nose length
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It is possible to use the histogram to find the number of people who have a nose
length of less than 47.5 mm.
Pr(nose length is < 47.5) =
2 + 5 + 17 + 21
75
= 45
75
= 35
It is worth noting that we cannot find the probability that a person has a nose length
which is less than 45 mm, as this is not the end point of any interval. However, if we
had a mathematical formula to approximate the shape of the graph, then the formula
could give us the answer to this important question.
In the histogram, the midpoints at the top of each bar have been connected by line
segments. If the class intervals were much smaller, say 1 mm or even less, these line
segments would take on the appearance of a smooth curve. This smooth curve is of
considerable importance for continuous random variables, because it represents the
probability density function for the continuous data.
This problem for a continuous random variable can be addressed by using calculus.
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 455
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For any continuous random variable, X, the probability density function is
such that
Pr(a < X < b) = 3 f(x)dx
b
a
which is the area under the curve from x = a to x = b.
0
a
b
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f (x)
x
• 3 f(x)dx = 1; this is absolutely critical.
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A probability density function must satisfy the following conditions:
• f(x) ≥ 0 for all x ∈ [a, b]
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a
Other properties are:
• Pr(X = x) = 0, where x ∈ [a, b]
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• Pr(a < X < b) = P(a ≤ X < b) = Pr(a < X ≤ b) = Pr(a ≤ X ≤ b) = 3 f(x)dx
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• Pr(X < c) = Pr(X ≤ c) = 3 f(x)dx when x ∈ a, b and a < c < b.
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Concept 1
a
Probability density functions
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AOS 4
Topic 3
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Units 3 & 4
c
b
In theory, the domain of a continuous probability density function is R, so that
∞
3 f(x)dx = 1.
Probability
density functions
Concept summary
Practice questions
−∞
However, if we must address the condition that
3 f(x)dx = 1,
b
Interactivity
Probability density
functions
int-6434
456 a
then the function must be zero everywhere else.
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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Sketch the graph of each of the following functions and state whether each
function is a probability density function.
b f(x) = e
c f(x) = e
tHinK
2(x − 1), 1 ≤ x ≤ 2
0,
elsewhere
0.5, 2 ≤ x ≤ 4
0,
elsewhere
2e−x, 0 ≤ x ≤ 2
0,
elsewhere
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a 1 Sketch the graph of f(x) = 2(x − 1)
a
f(x)
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f(x) = 2(x – 1)
0
2 Inspect the graph to determine if the
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function is always positive or zero, that is,
f(x) ≥ 0 for all x ∈ [a, b].
3 Calculate the area of the shaded region to
determine if 32(x − 1)dx = 1.
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4 Interpret the results.
(2, 2)
2
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over the domain 1 ≤ x ≤ 2, giving an
x-intercept of 1 and an end point of (2, 2).
Make sure to include the horizontal
lines for y = 0 either side of this graph.
Note: This function is known as a
triangular probability function because
of its shape.
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a f(x) = e
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1
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eXaMpLe
(1, 0)
(2, 0)
x
Yes, f(x) ≥ 0 for all x-values.
Method 1: Using the area of triangles
Area of shaded region = 12 × base × height
= 12 × 1 × 2
=1
Method 2: Using calculus
Area of shaded region = 3 2(x − 1)dx
2
1
2
= 3 (2x − 2)dx
1
2
= 3 x2 − 2x 4 1
= (22 − 2(2)) − (12 − 2(1))
=0−1+2
=1
f(x) ≥ 0 for all values, and the area under the
curve = 1. Therefore, this is a probability density
function.
topic 12 COntInuOus prObabILIty DIstrIbutIOns
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b 1 Sketch the graph of f(x) = 0.5 for
2 ≤ x ≤ 4. This gives a horizontal line,
with end points of (2, 0.5) and (4, 0.5).
Make sure to include the horizontal lines
for y = 0 on either side of this graph.
Note: This function is known as a uniform
or rectangular probability density function
because of its rectangular shape.
b
f(x)
0.5
(2, 0.5)
f(x) = 0.5
(2, 0)
(4, 0.5)
(4, 0)
x
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Yes, f(x) ≥ 0 for all x-values.
2 Inspect the graph to determine if the
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f­ unction is always positive or zero, that
is, f(x) ≥ 0 for all x ∈ [a, b] .
3 Calculate the area of the shaded region
Again, it is not necessary to use calculus to find
the area.
Method 1:
Area of shaded region = length × width
= 2 × 0.5
=1
to determine if 30.5dx = 1.
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Method 2:
Area of shaded region = 30.5dx
4
= 3 0.5x 4 2
= 0.5(4) − 0.5(2)
=2−1
=1
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c 1 Sketch the graph of f(x) = 2e−x for
c
f(x)
(0, 2)
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0 ≤ x ≤ 2. End points will be (0, 2)
and (2, e–2). Make sure to include the
horizontal lines for y = 0 on either side of
this graph.
2
f(x) ≥ 0 for all values, and the area under the
curve = 1. Therefore, this is a probability density
function.
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4 Interpret the results.
4
f(x) = 2e–x
(2, –e2)
2
(0, 0)
458 x
(2, 0)
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Yes, f(x) ≥ 0 for all x-values.
2 Inspect the graph to determine if the
function is always positive or zero, that
is, f(x) ≥ 0 for all x ∈ [a, b] .
−x
−x
32e dx = 2 3 e dx
2
2
to determine if 32e−xdx = 1.
2
0
0
0
2
= 2 3 −e−x 4 0
= 2(−e−2 + e0)
= 2(−e−2 + 1)
= 1.7293
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3 Calculate the area of the shaded region
f(x) ≥ 0 for all values. However, the area under
the curve ≠ 1. Therefore this is not a probability
density function.
Given that the functions below are probability density functions, find the
value of a in each function.
a f(x) = e
0,
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a 1 As the function has already been defined as a
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probability density function, this means that the
area under the graph is definitely 1.
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2 Remove a from the integral, as it is a constant.
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3 Antidifferentiate and substitute in the terminals.
4 Solve for a.
b f(x) = e
elsewhere
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tHinK
a(x − 1) 2, 0 ≤ x ≤ 4
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2
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eXaMpLe
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4 Interpret the results.
ae −4x, x > 0
0,
elsewhere
3f(x)dx = 1
4
a
0
2
3a(x − 1) dx = 1
4
0
a3 (x − 1) 2dx = 1
4
0
a3 (x − 1) 2dx = 1
4
0
ac
ac
(x − 1) 3
3
4
d =1
0
3
33 (−1)
−
d =1
3
3
a a 9 + 13 b = 1
a×
28
3
=1
3
a = 28
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3 f(x)dx = 1
∞
b 1 As the function has already been defined as a
b
p­ robability density function, this means that the
area under the graph is definitely 1.
∞
0
3 ae
−4x
dx = 1
0
∞
a 3 e−4xdx = 1
2 Remove a from the integral, as it is a constant.
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0
k
of the ­terminals, we find the appropriate limit.
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a × lim 3 e−4xdx = 1
k→∞
3 To evaluate an integral containing infinity as one
k
4 Antidifferentiate and substitute in the terminals.
a × lim 3 e−4xdx = 1
k→∞
0
k
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a × lim c − e−4x d = 1
k→∞
4
0
a × lim a−
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k→∞
5 Solve for a. Remember that a number divided
k→∞
a × lim a−
k→∞
e−4k 1
+ b=1
4
4
1
1
+ b=1
4k
4
4e
1
a a0 + b = 1
4
a
=1
4
a=4
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by an ­extremely large number is effectively
1
zero, so lim a 4k b = 0.
k→∞ e
a × lim a−
e−4k 1
+ b=1
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Exercise 12.2 Continuous random variables and ­probability functions
PRactise
1
Work without CAS
Sketch each of the following functions and determine whether each one is a
probability density function.
1 2x
0.25, −2 ≤ x ≤ 2
e , 0 ≤ x ≤ loge 3
a f(x) = • 4
b f(x) = e
0,
elsewhere
WE1
0,
elsewhere
2 Sketch each of the following functions and determine whether each one is a
probability density function.
π
π
1
cos(x), − ≤ x ≤
2
2
a f(x) = • 2
0,
460 elsewhere
1
1
,
≤x≤4
b f(x) = • 2 !x 2
0,
elsewhere
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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3
WE2
Given that the function is a probability density function, find the value of n.
n(x3 − 1), 1 ≤ x ≤ 3
f(x) = e
0,
elsewhere
4 Given that the function is a probability density function, find the value of a.
−ax, −2 ≤ x < 0
f(x) = • 2ax, 0 ≤ x ≤ 3
0,
5 A small car-hire firm keeps note of the age and kilometres covered by each of the
26
2<x≤3
28
3<x≤4
20
4<x≤5
11
5<x≤6
4
6<x≤7
1
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a Determine:
i Pr(X ≤ 2)
b Determine:
i Pr(1 < X ≤ 4)
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1<x≤2
y
30
25
20
15
10
5
0
Age of rental car
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Frequency
10
Frequency
Age
0<x≤1
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cars in their fleet. Generally, cars are no longer used once they have either covered
350 000 kilometres or are more than five years old. The following information
describes the ages of the cars in their current fleet.
Apply the most
appropriate
mathematical
processes and tools
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Consolidate
elsewhere
0
1
2
3
4
5
Age in years
6
7
x
ii Pr(X > 4).
ii Pr(X > 1│X ≤ 4).
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6 The battery life for batteries in television remote controls was investigated
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Hours of life
0 < x ≤ 15
Frequency
15
15 < x ≤ 30
33
30 < x ≤ 45
23
45 < x ≤ 60
26
60 < x ≤ 75
3
Frequency
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in a study.
y
35
30
25
20
15
10
5
0
Remote battery life
0
15 30 45 60
Battery life in hours
75
x
a How many remote control batteries were included in the study?
b What is the probability that a battery will last more than 45 hours?
c What is the probability that a battery will last between 15 and 60 hours?
d A new battery producer is advocating that their batteries have a long life
of 60+ hours. If it is known that this is just advertising hype because these
­batteries are no different from the batteries in the study, what is the probability
that these new batteries will have a life of 60+ hours?
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 461
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7 A number of experienced shot-putters were asked to aim for a line
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10 metres away.
1 < x ≤ 1.5
45
1.5 < x ≤ 2
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Shot‐puts
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Frequency
0.5 < x ≤ 1
y
80
70
60
50
40
30
20
10
0
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0 < x ≤ 0.5
Frequency
75
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Metres
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After each of them put their shot, its distance from the 10-metre line was
measured. All of the shots were on or between the 8- and 10-metre lines. The
results of the measurements are shown, where X is the distance in metres from the
10-metre line.
0
0.5
1
1.5
2
Distance in metres
x
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a How many shot-put throws were measured?
b Calculate:
i Pr(X > 0.5)
ii Pr(1 < X ≤ 2)
c A guest shot-putter is visiting the athletics club where the measurements are
being conducted. His shot-putting ability is equivalent to the abilities of the club
members. Find the probability that he puts the shot within 50 cm of the 10-metre
line if it is known that he put the shot within 1 metre of the 10-metre line.
8 Sketch each of the following functions and determine whether each function is a
probability density function. Note: Use CAS where appropriate.
π
3π
1
cos(x) + 1,
≤x≤
− , −e ≤ x ≤ −1
4
4
a f(x) = • x
b f(x) = •
0,
elsewhere
0, elsewhere
1
sin(x), 0 ≤ x ≤ π
c f(x) = • 2
0,
elsewhere
1
, 1<x≤2
d f(x) = • 2 !x − 1
0,
elsewhere
9 The rectangular function, f , is defined by the rule
f(x) = e
c, 0.25 < x < 1.65
.
0, elsewhere
Find the value of the constant c, given that f is a probability density function.
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10 The graph of a function, f , is shown.
y
(0, z)
(5, 0)
x
(–1, 0) 0
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If f is known to be a probability density function, show that the value of z is 13.
probability density function.
m(6 − 2x), 0 ≤ x ≤ 2
a f(x) = e
0,
elsewhere
me2x, 0 ≤ x ≤ loge 3
0,
me−2x, x ≥ 0
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c f(x) = e
b f(x) = e
O
11 Find the value of the constant m in each of the following if each function is a
elsewhere
0,
elsewhere
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12 Let X be a continuous random variable with the probability density function
x2 + 2kx + 1, 0 ≤ x ≤ 3
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f(x) = e
0,
elsewhere
Show that the value of k is −11
.
9
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13 X is a continuous random variable
such that
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x
1
loge a b, 2 ≤ x ≤ a
2
f(x) = • 2
0,
f(x)
(a, –12 log (a))
e
1 log (a)
–
2 e
elsewhere
and 3f(x)dx = 1. The graph of this
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a
2
0
(2, 0)
x
(a, 0)
function is shown.
Find the value of the constant a.
14 X is a continuous random
variable such that
−x, −1 ≤ x < 0
f(x) = • x, 0 ≤ x ≤ a
0, elsewhere
where a is a constant.
Y is another continuous random variable such that
1
, 1≤y≤e
.
f(y) = • y
0, elsewhere
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 463
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a Sketch the graph of the function for X and find 3 f(x)dx.
a
−1
e
b Sketch the graph of the function for Y and find 3f(y)dy.
1
c Find the value of the constant a if 3 f(x)dx = 3f(y)dy.
e
−1
1
15 X is a continuous random variable such that
f(x) = •
0,
π
12 .
elsewhere
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n sin(3x) cos(3x), 0 < x <
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Master
a
16 A function f is defined by the rule
f(x) = e
a
loge (x), x > 0
0,
elsewhere
.
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If f is known to be a probability density function, find the value of the constant, n.
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a If 3 f(x)dx = 1, find the value of the real constant a.
1
b Does this function define a probability density function?
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12.3
The continuous probability density function
As stated in section 12.2, if X is a continuous random variable, then
Pr(a ≤ X ≤ b) = 3f(x)dx.
EC
b
Units 3 & 4
In other words, by finding the area between
the curve of the continuous probability
function, the x-axis, the line x = a and the
line x = b, providing f(x) ≥ 0, then we are
finding Pr(a ≤ X ≤ b). It is worth noting
that because we are dealing with a
continuous random variable, Pr(X = a) = 0,
and consequently:
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Topic 3
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Concept 2
Calculating
probabilities
Concept summary
Practice questions
a
R
AOS 4
f (x)
0
a
b
x
Pr(a ≤ X ≤ b) = Pr(a < X ≤ b) = Pr(a ≤ X < b) = Pr(a < X < b)
Also,
Pr(a ≤ X ≤ b) = Pr(a ≤ X ≤ c) + Pr(c < X ≤ b), where a < c < b.
This property is particularly helpful when the probability density function is a hybrid
function and the required probability encompasses two functions.
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WORKED
EXAMPLE
3
A continuous random variable, Y, has a probability density function, f ,
defined by
−ay, −3 ≤ y ≤ 0
0<y≤3
f(y) = • ay,
0,
where a is a constant.
a Sketch the graph of f .
c Determine Pr(1 ≤ Y ≤ 3).
b Find the value of the constant, a.
d Determine Pr(Y < 2│Y > −1)
WRITE/DRAW
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THINK
elsewhere
a The hybrid function contains three sections. a f(−3) = 3a and f(3) = 3a
The first graph, f(y) = −ay, is a straight
line with end points of (0, 0) and (–3, 3a).
The second graph is also a straight line
and has end points of (0, 0) and (3, 3a).
Don’t forget to include the f(y) = 0 lines
for x > 3 and x < −3.
(–3, 3a)
b Use the fact that 3 f(y)dy = 1 to solve
for a.
EC
3
PR
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–2
–1
1
2
(3, 0)
3 y
3
b 3 f(y)dy = 1
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−3
(0, 0)
0
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3
(3, 3a)
3a
(–3, 0)
–3
O
f (y)
−3
Using the area of a triangle, we find:
1
2
× 3 × 3a + 12 × 3 × 3a
9a 9a
+
2
2
9a
a
=1
=1
=1
= 19
1
1
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c Pr(1 ≤ Y ≤ 3) = 3 f(y)dy. Identify the part c Pr(1 ≤ Y ≤ 3) = 3 f(y)dy
3
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of the function that the required y-values
sit within: the values 1 ≤ Y ≤ 3 are within
1
the region where f(y) = ay = y.
9
= 3 a 91 y b dy
3
1
=
c
1 2
y d
18
3
1
1
1
= 18
(3) 2 − 18
(1) 2
8
= 18
= 49
Note: The method of finding the area of a
trapezium could also be used.
Topic 12 CONTINUOUS PROBABILITY DISTRIBUTIONS
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Pr(Y < 2 ∩ Y > −1)
Pr(Y > −1)
Pr(−1 < Y < 2)
=
Pr(Y > −1)
d Pr(Y < 2 ∣ Y > −1) =
probability.
2 Find Pr(−1 < Y < 2). As the interval is
across two functions, the interval needs
to be split.
Pr(−1 < Y < 2) = Pr(−1 < Y < 0) + Pr(0 ≤ Y < 2)
0
3 To find the probabilities we need to find
the areas under the curve.
=
1
3 −9 ydy
−1
2
+ 319 ydy
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d 1 State the rule for the conditional
0
2
­substituting the terminals.
1 2
= − c 18
y d
=
=
5 Find Pr(Y > −1). As the interval is
=
1
3 −9 ydy
−1
R
7 Antidifferentiate and evaluate after
U
N
C
O
substituting the terminals.
8 Now substitute into the formula to find
Pr(−1 < Y < 2)
Pr(Y < 2 ∣ Y > −1) =
.
Pr(Y > −1)
3
+ 319 ydy
0
0
= − 3 19 ydy + 12
−1
R
EC
find the areas under the curve. As
Pr(0 ≤ Y ≤ 3) covers exactly half the
area under the curve, Pr(0 ≤ Y ≤ 3) = 12.
(The entire area ­under the curve is always
1 for a probability ­density function.)
1
1
+ 18
(2) 2 − 18
(0) 2
Pr(Y > −1) = Pr(−1 < Y < 0) + Pr(0 ≤ Y ≤ 3)
0
6 To find the probabilities we need to
1 2 2
y d
18
0
−1
1
1
− a 18 (0) 2 − 18
(−1) 2 b
1
4
+ 18
18
5
18
TE
D
across two functions, the interval needs
to be split.
c
PA
G
=
+
E
4 Antidifferentiate and evaluate after
0
0
PR
O
−1
O
= − 3 19 ydy + 319 ydy
0
1 2
= − c 18
y d
0
−1
+ 12
1
1
= − a18
(0) 2 − 18
(−1) 2 b + 12
1
9
= 18
+ 18
= 10
18
= 59
Pr(Y < 2 ∣ Y > −1) =
Pr(−1 < Y < 2)
Pr(Y > −1)
5
= 18 ÷
5
= 18
×
5
9
9
5
= 12
466 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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Exercise 12.3 The continuous probability density function
PRactise
1
WE3
The continuous random variable Z has a probability density function given by
−z + 1,
Work without CAS
f(z) = • z − 1,
0,
0≤z<1
1≤z≤2
elsewhere.
a Sketch the graph of f .
b Find Pr(Z < 0.75).
c Find Pr(Z > 0.5).
0,
elsewhere
where a is a constant.
a Find the value of the constant a.
b Sketch the graph of f .
c Find Pr(0.5 ≤ X ≤ 1).
3 Let X be a continuous random variable with a probability density function
defined by
Apply the most
appropriate
mathematical
processes and tools
1
sin(x),
2
0≤x≤π
0,
elsewhere
.
TE
D
f(x) = e
PA
G
Consolidate
O
0≤x≤a
PR
O
4x3,
E
f(x) = e
FS
2 The continuous random variable X has a probability density function given by
a Sketch the graph of f .
π
4
EC
b Find Pra < X <
π
4
3π
b.
4
c Find PraX > │X <
3π
b.
4
U
N
C
O
R
R
4 A probability density function is defined by the rule
k(2 + x),
f(x) = • k(2 − x),
0,
−2 ≤ x < 0
0≤x≤2
elsewhere
where X is a continuous random variable and k is a constant.
a Sketch the graph of f .
1
b Show that the value of k is 4.
c Find Pr(−1 ≤ X ≤ 1).
d Find Pr(X ≥ −1│X ≤ 1).
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 467
467
23/08/15 6:59 PM
5 The amount of petrol sold daily
0,
a Sketch the graph of f .
1≤x≤6
PA
G
f(x) =
1
,
u5
E
PR
O
O
FS
Frequency
by a busy service station is a
uniformly distributed probability
(18, k)
(30, k)
density function. A minimum
k
of 18 000 litres and a maximum
of 30 000 litres are sold on any
given day. The graph of the
function is shown.
a Find the value of the constant k.
b Find the probability that
between 20 000 and
25 000 litres of petrol are sold
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32
on a given day.
Petrol sold (thousands of litres)
c Find the probability that as
much as 26 000 litres of petrol
were sold on a particular day,
given that it was known that at
least 22 000 litres were sold.
6 The continuous random variable X has a uniform rectangular probability density
function defined by
elsewhere
.
b Determine Pr(2 ≤ X ≤ 5).
7 The continuous random variable Z has a probability density function defined by
TE
D
1
,
f(z) = • 2z
0,
1 ≤ z ≤ e2
.
elsewhere
e2
EC
a Sketch the graph of f and shade the area that represents 3 f(z)dz.
e2
b Find 3 f(z)dz. Explain your result.
R
R
1
1
U
N
C
O
The continuous random variable U has a probability function defined by
e4u, u ≥ 0
f(u) = e
.
0,
elsewhere
c Sketch the graph of f and shade the area that represents 3 f(u)du, where a is
a
a constant.
e2
0
d Find the exact value of the constant a if 3 f(z)dz is equal to 3 f(u)du.
1
a
0
8 The continuous random variable Z has a probability density function defined by
π
π
1
cos(z), − ≤ z ≤
2
2 .
f(z) = • 2
0,
elsewhere
468 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 468
23/08/15 6:59 PM
a Sketch the graph of f and verify that y = f(z) is a probability density function.
π
π
6
4
9 The continuous random variable U has a probability density function defined by
1
1 − (2u − 3u2), 0 ≤ u ≤ a
4
f(u) = •
0,
elsewhere
where a is a constant. Find:
a the value of the constant a
c Pr(0.1 < U < 0.5)
b Pr(U < 0.75)
d Pr(U = 0.8).
FS
b Find Pra− ≤ Z ≤ b.
PR
O
3 2
x, 0≤z≤2
f (x) = • 8
.
0,
elsewhere
O
10 The continuous random variable X has a probability density function defined by
PA
G
E
Find:
a P(X > 1.2)
b P(X > 1│X > 0.5), correct to 4 decimal places
c the value of n such that P(X ≤ n) = 0.75.
11 The continuous random variable Z has a probability density function defined by
z
0≤z≤a
0,
elsewhere
TE
D
f(z) = c
e−3,
where a is a constant. Find:
a
a the value of the constant a such that 3 f(z)dz = 1
EC
0
R
b Pr(0 < Z < 0.7), correct to 4 decimal places
c Pr(Z < 0.7│Z > 0.2), correct to 4 decimal places
d the value of α, correct to 2 decimal places, such that Pr(Z ≤ α) = 0.54.
U
N
C
O
R
12 The continuous random variable X has a probability density function given as
Master
f(x) = e
3e−3x,
x≥0
0,
elsewhere
.
a Sketch the graph of f .
b Find Pr(0 ≤ X ≤ 1), correct to 4 decimal places.
c Find Pr(X > 2), correct to 4 decimal places.
13 The continuous random variable X has a probability density function defined by
f(x) = e
loge (x2),
x≥1
0,
elsewhere
.
Find, correct to 4 decimal places:
a the value of the constant a if 3 f(x)dx = 1
1
b Pr(1.25 ≤ X ≤ 2).
a
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 469
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14 The graph of the probability function
f(z) =
is shown.
1
π(z + 1)
2
f (z)
(0, 1–π )
–1
0
1
2
3 z
O
–2
PR
O
–3
FS
1
f (z) = —
π(z2 + 1)
a Find, correct to 4 decimal places, Pr(−0.25 < Z < 0.25).
Suppose another probability density function is defined as
1
,
+1
E
−a ≤ x ≤ a
0,
.
elsewhere
PA
G
f(x) = •
x2
b Find the value of the constant a.
The commonly used measures of central tendency and spread in statistics are the
mean, median, variance, standard deviation and range. These same measurements are
appropriate for continuous probability functions.
EC
Units 3 & 4
TE
D
12.4
Measures of centre and spread
Measures of central tendency
AOS 4
The mean
Remember that for a discrete random variable,
R
Topic 3
R
Concept 3
O
Mean and median
Concept summary
Practice questions
E(X) = μ = a xnPr(X = xn).
x=n
x=1
U
N
C
This definition can also be applied to a continuous random variable.
Interactivity
Mean
int-6435
∞
We define E(X) = μ = 3 xf(x)dx.
−∞
If f(x) = 0 everywhere except for x ∈ [a, b], where the
function is defined, then
b
E(X) = μ = 3xf(x)dx.
a
470 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 470
23/08/15 6:59 PM
Consider the continuous random variable, X, which has a probability density function
defined by
x2, 0 ≤ x ≤ 1
f(x) = e
0, elsewhere
For this function,
1
E(X) = μ = 3xf(x)dx
0
= 3x(x2)dx
FS
1
= 3x3dx
PR
O
0
O
0
1
1
PA
G
E
x4
= c d
4 0
4
1
−0
=
4
1
=
4
Similarly, if the continuous random variable X has a probability density function of
TE
D
f(x) = u
elsewhere,
0,
∞
E(X) = μ = 3 xf(x)dx
EC
then
7e−7x, x ≥ 0
0
R
= lim 37xe−7xdx
k→∞
k
R
0
U
N
C
O
= 0.1429
where CAS technology is required to determine the integral.
The mean of a function of X is similarly found.
The function of X, g(x), has a mean defined by:
∞
E(g(x)) = μ = 3 g(x)f(x)dx.
−∞
So if we again consider
f(x) = e
x2,
0≤x≤1
0,
elsewhere
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 471
471
23/08/15 6:59 PM
then
1
E(X2) = 3x2 f(x)dx
0
1
= 3x4dx
0
1
FS
x5
= c d
5 0
15 0
−
5
1
=
5
This definition is important when we investigate the variance of a continuous
random variable.
PR
O
O
=
Median and percentiles
The median is also known as the 50th percentile, Q2, the halfway mark or the middle
value of the distribution.
PA
G
E
Interactivity
Median and
percentiles
int-6436
For a continuous random variable, X, defined by the probability
m
function f, the median can be found by solving 3 f (x)dx = 0.5.
TE
D
−∞
EC
Other percentiles, which are frequently calculated, are the 25th percentile or lower
quartile, Q1, and the 75th percentile or upper quartile, Q3.
R
The interquartile range is calculated as:
IQR = Q3 − Q1
O
R
Consider a continuous random variable, X, that has a probability density function of
2
f(x) = e
0.21e2x−x ,
−3 ≤ x ≤ 5
U
N
C
0,
elsewhere
To find the median, m, we solve for m as follows:
3
m
2
0.21e2x−x dx
.
f(x)
= 0.5
−3
The area under the curve is equated to 0.5,
giving half of the total area and hence the
50th percentile. Solving via CAS, the result
is that m = 0.9897 ≃ 1.
This can be seen on a graph as follows.
f (x) = 0.21e2x – x
2
0.5
–3
472 0
x=1
5
x
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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Consider the continuous random variable X, which has a probability density
function of
x3
f(x) = 4 , 0 ≤ x ≤ 2
•
0,
elsewhere.
The median is given by Pr(0 ≤ x ≤ m) = 0.5:
m
x3
3 4 dx = 0.5
0
m
FS
x4
1
d =
16 0
2
O
m4
1
−0=
16
2
m4 = 8
4
m = ±"
8
PR
O
c
PA
G
E
m = 1.6818 (0 ≤ m ≤ 2)
To find the lower quartile, we make the area under the curve equal to 0.25. Thus the
lower quartile is given by Pr(0 ≤ x ≤ a) = 0.25:
a
x3
3 dx = 0.25
4
0
a
TE
D
x4
1
d =
4
16 0
a4
1
−0=
4
16
a4 = 4
R
EC
c
R
4
a = ±"
4
U
N
C
O
a = Q1 = 1.4142 (0 ≤ a ≤ m)
Similarly, to find the upper quartile, we make the area under the curve equal to 0.75.
Thus the upper quartile is given by Pr(0 ≤ x ≤ n) = 0.75:
n
x3
3 4 dx = 0.75
0
n
3
x4
c d =
16 0 4
3
n
−0=
16
4
n = 12
4
n = ±"
12
n = Q3 = 1.8612 (m ≤ x ≤ 2)
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 473
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23/08/15 7:00 PM
So the interquartile range is given by Q3 − Q1 = 1.8612 − 1.4142
= 0.4470.
These values are shown on the following graph.
f (x)
(2, 2)
Upper quartile
x = 1.8612
Median
x = 1.6818
Lower quartile
x = 1.4142
(0, 0)
PR
O
x
A continuous random variable, Y, has a probability density function, f ,
defined by
ky, 0 ≤ y ≤ 1
f( y) = e
0, elsewhere
E
4
where k is a constant.
TE
D
a Sketch the graph of f .
PA
G
WOrKeD
eXaMpLe
(2, 0)
FS
2
x3
—
4
O
f (x) =
b Find the value of the constant k.
c Find:
EC
i the mean of Y
ii the median of Y.
R
d Find the interquartile range of Y.
R
tHinK
O
a The graph f(y) = ky is a straight line with end points
a
f (y)
k
(1, k)
U
N
C
at (0, 0) and (1, k). Remember to include the lines
f(y) = 0 for y > 1 and y < 0.
WritE/draW
(0, 0)
474
(1, 0) y
Maths Quest 12 MatheMatICaL MethODs VCe units 3 and 4
c12ContinuousProbabilityDistributions.indd 474
23/08/15 7:00 PM
1
b Solve 3 ky dy = 1 to find the value of k.
3 ky dy = 1
1
b
0
0
1
k3 y dy = 1
0
k(1) 2
2
y2
2
1
d =1
0
−0=1
FS
kc
i 1 State the rule for the mean.
c
μ = 3 y(2y)dy
1
i
0
1
= 3 2y2dy
TE
D
c
PA
G
E
PR
O
O
k
=1
2
k=2
Using the area of a triangle also enables
you to find the value of k.
1
×1×k=1
2
k
=1
2
k=2
0
1
2
= c y3 d
3 0
EC
2 Antidifferentiate and simplify.
2(1) 3
−0
3
2
=
3
C
O
R
R
=
U
N
ii 1 State the rule for the median.
m
ii 3f(y)dy = 0.5
0
m
32ydy = 0.5
0
2 Antidifferentiate and solve for m. Note that
m must be a value within the domain of the
function, so within 0 ≤ y ≤ 1.
3 y2 4 m
0 = 0.5
2
m − 0 = 0.5
m =± 1
Å2
1
m =
(0 < m < 1)
"2
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 475
475
23/08/15 7:00 PM
Median =
3 Write the answer.
a
d
i 1 State the rule for the lower quartile, Q1.
d
1
"2
3 f(y)dy = 0.25
0
3 2ydy = 0.25
a
0
a
3 y2 4 0 = 0.25
2 Antidifferentiate and solve for Q1.
O
a = ±!0.25
FS
a2 − 0 = 0.25
PR
O
a = Q1 = 0.5 a0 < Q1 <
n
3 f(y)dy = 0.75
3 State the rule for the upper quartile, Q3.
0
1
b
!2
n
PA
G
E
3 2ydy = 0.75
0
n
3 y2 4 = 0.75
4 Antidifferentiate and solve for Q3.
0
TE
D
n2 − 0 = 0.75
EC
5 State the rule for the interquartile range.
n = ±!0.75
n = Q3 = 0.8660 a0 < Q3 <
= 0.8660 − 0.5
= 0.3660
R
R
6 Substitute the appropriate values and simplify.
IQR = Q3 − Q1
1
b
!2
U
N
AOS 4
Variance, standard deviation and range
The variance and standard deviation are important measures of spread in statistics.
From previous calculations for discrete probability functions, we know that
C
Units 3 & 4
O
Measures of spread
Topic 3
Concept 4
Variance and
standard deviation
Concept summary
Practice questions
Var(X ) = E(X2) − [E(X )] 2 and SD(X ) = !Var(X )
For continuous probability functions,
∞
Var(X) = 3 (x − μ)2f(x) dx
−∞
Interactivity
Variance, standard
deviation and range
int-6437
476 ∞
= 3 (x2 − 2xμ + μ2)f(x)dx
−∞
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 476
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∞
∞
∞
= 3 x2f(x)dx − 3 2xf(x)μdx + 3 μ2f(x)dx
−∞
−∞
−∞
∞
∞
−∞
−∞
= E(X ) − 2μ 3 xf(x)dx + μ2 3 1f(x)dx
2
= E(X2) − 2μ × E(X) + μ2
= E(X2) − 2μ2 + μ2
= E(X2) − [E(X)] 2
FS
= E(X2) − μ2
∞
∞
O
Two important facts were used in this proof: 3 f(x)dx = 1 and 3 xf(x)dx = μ = E(X).
PR
O
−∞
Substituting this result into SD(X) = !Var(X) gives us
−∞
SD(X) = "E(X2) − 3 E(X) 4 2 .
PA
G
E
The range is calculated as the highest value minus the lowest value, so for the
1
, 1≤x≤6
probability density function given by f(x) = • 5
, the highest possible
0,
elsewhere
For a continuous random variable, X, with a probability density function, f ,
defined by
1
x + 2, −4 ≤ x ≤ −2
f(x) = 2
•
0,
elsewhere
find:
EC
5
R
R
WOrKeD
eXaMpLe
TE
D
x-value is 6 and the lowest is 1. Therefore, the range for this function = 6 – 1
= 5.
b the median
c the variance
d the standard deviation, correct to 4 decimal places.
U
N
tHinK
C
O
a the mean
a 1 State the rule for the mean and simplify.
WritE
−2
a μ = 3 xf(x)dx
−4
−2
= 3 xa12x + 2bdx
−4
−2
= 3 a12x2 + 2xbdx
−4
topic 12 COntInuOus prObabILIty DIstrIbutIOns
c12ContinuousProbabilityDistributions.indd 477
477
23/08/15 7:00 PM
=
2 Antidifferentiate and evaluate.
=
=
c
1 3
x
6
+ x2 d
−2
−4
3
a 1 (−2) +
6
4
+ 4 + 32
3
3
(−2) 2 b −
− 16
a
+ (−4) 2 b
1 (−4) 3
6
= −223
3 f(x)dx = 0.5
m
b
−4
3 a 2x + 2 b dx = 0.5
m
FS
b 1 State the rule for the median.
−4
2 Antidifferentiate and solve for m.
a
1 2
m
4
+ 2m b −
a
1 2
x
4
+ 2x d
m
−4
= 0.5
PR
O
The quadratic formula is needed
as the quadratic equation
formed cannot be factorised.
Alternatively, use CAS to solve for m.
c
O
1
(−4) 2
4
+ 2(−4) b = 0.5
1 2
m
4
2
+ 2m + 4 = 0.5
m + 8m + 16 = 2
PA
G
E
m2 + 8m + 14 = 0
−8 ± "(8) 2 − 4(1)(14)
2(1)
−8 ±!8
m=
2
= −4 ± "2
∴ m = −4 + "2 as m ∈ 3 −4, 2 4
The median is −4 + "2.
TE
D
So m =
EC
3 Write the answer.
R
first.
U
N
C
O
2 Find
E(X2)
R
c 1 Write the rule for variance.
The median is −4 + "2.
c Var (X) = E (X2) − [E (X)]2
= 3x2f(x)dx
b
E(X2)
a
−2
= 3 x2 a 12 x + 2 b dx
−4
−2
= 3 a 12 x3 + 2x2 b dx
−4
=
=
1
c 8 x4
+ 23 x3 d
1
4
a (−2)
8
+
−2
−4
2
(−2) 3 b
3
= 2 − 16
− 32 + 128
3
3
−
a
1
(−4) 4
8
+ 23 (−4) 3 b
= −30 + 112
3
= 22
3
478 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 478
23/08/15 7:00 PM
Var(X) = E(X2) − 3 E(X) 4 2
3 Substitute E(X) and E(X2) into the rule
for variance.
= 22
− a −83 b
3
2
= 22
− 64
3
9
= 66
− 64
9
9
= 29
2
Ä9
= 0.4714
and evaluate.
Exercise 12.4 Measures of centre and spread
1
WE4
Work without CAS
Question 1
The continuous random variable Z has a probability density function of
1
, 1≤z≤a
!z
f(z) = •
elsewhere
PA
G
0,
E
PRactise
FS
=
O
2 Substitute the variance into the rule
d SD(X) = !Var(X)
PR
O
d 1 Write the rule for standard deviation.
EC
TE
D
where a is a constant.
a Find the value of the constant a.
b Find:
ithe mean of Z
iithe median of Z.
2 The continuous random variable, Y, has a probability density function of
f(y) = e
!y,
0≤y≤a
3e−3x,
x≥0
0,
elsewhere
0,
elsewhere
U
N
C
O
R
R
where a is a constant. Find, correct to 4 decimal places:
a the value of the constant a
b E(Y)
c the median value of Y.
3 WE5 For the continuous random variable Z, the probability density function is
e
2 loge (2z), 1 ≤ z ≤
2
f(z) = •
elsewhere.
0,
Find the mean, median, variance and standard deviation correct to
4 decimal places.
4 The function
f(x) = e
defines the probability density function for the continuous random variable, X.
Find the mean, median, variance and standard deviation of X.
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 479
479
23/08/15 7:00 PM
Consolidate
1
, 0≤x≤1
.
f(x) = • 2 !x
0,
elsewhere
a Prove that f is a probability density function.
b Find E(X).
c Find the median value of f .
6 The time in minutes that an individual must wait in line to be served at the local
bank branch is defined by
f(t) = 2e−2t, t ≥ 0
where T is a continuous random variable.
a What is the mean waiting time for a customer in the queue, correct to
1 decimal place?
b Calculate the standard deviation for the waiting time in the queue, correct to
1 decimal place.
c Determine the median waiting time in the queue, correct to 2 decimal places.
7 The continuous random variable Y has a probability density function defined by
y2
3
, 0≤y≤"
9
.
f(y) = • 3
0, elsewhere
Find, correct to 4 decimal places:
a the expected value of Y
b the median value of Y
c the lower and upper quartiles of Y
d the inter-quartile range of Y.
8 The continuous random variable Z has a probability density function defined by
a
, 1≤z≤8
f(y) = • z
EC
TE
D
PA
G
E
PR
O
O
FS
Apply the most
appropriate
mathematical
processes and tools
5 Let X be a continuous random variable with a probability density function of
0,
elsewhere
U
N
C
O
R
R
where a is a constant.
a Find the value, correct to four decimal places, of the constant a.
b Find E(Z) correct to 4 decimal places.
c Find Var(Z) and SD(Z).
d Determine the interquartile range for Z.
e Determine the range for Z.
f (x)
9 X is a continuous random variable. The graph of the
probability density function
1
2
f (x) = –
–
π (sin(2x) + 1)
1
π
f(x) = (sin(2x) + 1) for 0 ≤ x ≤ π
π
is shown.
(π, –π1)
(0, –π1 )
a Show that f(x) is a probability density function.
b Calculate E(X) correct to 4 decimal places.
c Calculate, correct to 4 decimal places:
x
0
iVar(X)
0.25 0.5 0.75 1
iiSD(X).
d Find the median value of f correct to 4 decimal places.
480 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 480
23/08/15 7:00 PM
10 The continuous random variable X has a probability density function defined by
f(x) = e
ax − bx2,
0≤x≤2
0,
elsewhere.
Find the values of the constants a and b if E(X) = 1.
11 The continuous random variable, Z, has a probability density function of
3
,
f(z) = • z2
0,
1≤z≤a
elsewhere
E
PR
O
O
FS
where a is a constant.
3
a Show that the value of a is 2.
b Find the mean value and variance of f correct to 4 decimal places.
c Find the median and interquartile range of f .
12 a Find the derivative of "4 − x2.
b Hence, find the mean value of the probability density function defined by
3
, 0 ≤ x ≤ !3
f(x) = • π"4 − x2
.
0,
elsewhere
PA
G
13 Consider the continuous random variable X with a probability density function of
h(2 − x),
0≤x≤2
f(x) = • h(x − 2),
2<x≤4
0,
elsewhere
R
EC
TE
D
where h is a constant.
a Find the value of the constant h.
b Find E(X).
c Find Var(X).
14 Consider the continuous random variable X with a probability density function of
k, a ≤ x ≤ b
f(x) = e
0, elsewhere
U
N
C
O
R
where a, b and k are positive constants.
a Sketch the graph of the function f .
1
b Show that k =
.
b−a
c Find E(X) in terms of a and b.
d Find Var(X) in terms of a and b.
15 The continuous random variable Y has a probability density function
Master
y
0.2 loge a b, 2 ≤ y ≤ 7.9344
.
f( y) = •
2
0,
elsewhere
a Verify that f is a probability density function.
b Find E(Y) correct to 4 decimal places.
c Find Var(Y) and SD(Y) correct to 4 decimal places.
d Find the median value of Y correct to 4 decimal places.
e State the range.
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 481
481
23/08/15 7:00 PM
16 The continuous random variable Z has a probability density function
f(z) = e
!z − 1,
0,
1≤z≤a
elsewhere
where a is a constant.
a Find the value of the constant a correct to 4 decimal places.
b Determine, correct to 4 decimal places:
iE(Z)
iiE(Z2)
iiiVar(Z)
ivSD(Z).
FS
O
Sometimes it is necessary to apply transformations to a continuous random variable.
A transformation is a change that is applied to the random variable. The change may
consist of one or more operations that may involve adding or subtracting a constant or
multiplying or dividing the variable by a constant.
Suppose a linear transformation is applied to the continuous random variable X to
create a new continuous random variable, Y. For instance
PR
O
12.5
Linear transformations
E
Y = aX + b
PA
G
It can be shown that E(Y) = E(aX + b) = aE(X) + b
and Var(Y) = Var(aX + b) = a2Var(X).
First let us show that E(Y) = E(aX + b) = aE(X) + b.
∞
−∞
∞
TE
D
Since E(X) = 3 xf(x) dx,
then E(aX + b) = 3 (ax + b)f(x) dx.
EC
−∞
O
∞
−∞
∞
−∞
∞
= a 3 xf(x) dx + b 3 f(x) dx
−∞
∞
But E(X) = 3 xf(x) dx, so
C
U
N
∞
E(aX + b) = 3 axf(x) dx + 3 bf(x) dx
R
R
Using the distributive law, it can be shown that this is equal to
−∞
−∞
∞
E(aX + b) = aE(X) + b 3 f(x) dx.
∞
Also, 3 f(x) dx = 1, so
−∞
−∞
E(aX + b) = aE(X) + b.
Also note that E(aX) = aE(X) and E(b) = b.
Now let us show that Var(Y) = Var(aX + b) = a2Var(X).
482 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 482
23/08/15 7:00 PM
Since Var(X) = E(X2) − 3 E(X) 4 2,
then
Var(aX + b) = E 1 (aX + b) 2 2 − 3 E(aX + b) 4 2
∞
= 3 (ax + b) 2f(x) dx − 1 aE(X) + b 2 2
−∞
∞
= 3 (a2x2 + 2abx + b2)f(x) dx − 3 a2 3 E(X) 4 2 + 2abE(X) + b2 4
−∞
Using the distributive law to separate the first integral, we have
∞
−∞
−∞
dx + 3 2abxf(x) dx + 3 b2f(x) dx − a2 3 E(X) 4 2
− 2abE(X) − b2
∞
∞
FS
−∞
∞
O
Var(aX + b) = 3
a2x2f(x)
PR
O
∞
∞
= a 3 x f(x) dx + 2ab 3 xf(x) dx + b 3 f(x) dx − a2 3 E(X) 4 2
2
2
2
−∞
− 2abE(X) −
∞
−∞
E
∞
−∞
b2
∞
−∞
PA
G
But E(X) = 3 xf(x) dx, E(X2) = 3 x2f(x) dx and 3 f(x) dx = 1 for a probability
−∞
density function. Thus,
−∞
TE
D
Var(aX + b) = a2E(X2) + 2abE(X) + b2 − a2 3 E(X) 4 2 − 2abE(X) − b2
= a2E(X2) − a2 3 E(X) 4 2
= a2 (E(X2) − 3 E(X) 4 2)
E(aX + b) = aE(X) + b
and
Var(aX + b) = a2Var(X).
C
O
R
R
Thus,
EC
= a2Var(X)
U
N
WOrKeD
eXaMpLe
6
A continuous random variable, X, has a mean of 3 and a variance of 2. Find:
a E(2X + 1)
b Var(2X + 1)
d E(3X2)
e E(X2 − 5).
tHinK
c E(X2)
WritE
E(2X + 1) = 2E(X) + 1
= 2(3) + 1
=7
a Use E(aX + b) = aE(X) + b to find E(2X + 1).
a
b Use Var(aX + b) = a2Var(X) to find Var(2X + 1).
b Var(2X + 1) = 22Var(X)
=4×2
=8
topic 12 COntInuOus prObabILIty DIstrIbutIOns
c12ContinuousProbabilityDistributions.indd 483
483
23/08/15 7:00 PM
c Use Var(X) = E(X2) − 3 E(X) 4 2 to find E(X2).
c
Var(X)
2
2
2
E(X )
d Use E(aX2) = aE(X2) to find E(3X2).
d
E(3X2) = 3E(X2)
= 3 × 11
= 33
e Use E(aX2 + b) = aE(X2) + b to find E(X2 − 5).
e E(X2 − 5) = E(X5) − 5
O
= 11 − 5
=6
FS
= E(X2) − 3 E(X) 4 2
= E(X2) − 32
= E(X2) − 9
= 11
The graph of the probability density function for the continuous random
variable X is shown. The rule for the probability density function is given by
f(x) = e
3kx,
0≤x≤1
0,
elsewhere
where k is a constant.
E
7
f (x)
3k
PA
G
WOrKeD
eXaMpLe
PR
O
It may also be necessary to find the expected value and variance before using the facts
that E(aX + b) = aE(X) + b and Var(aX + b) = a2Var(X).
(1, 3k)
a Find the value of the constant k.
TE
D
b Calculate E(X) and Var(X).
c Find E(3X − 1) and Var(3X − 1).
(0, 0)
d Find E(2X2 + 3).
EC
tHinK
(1, 0) x
WritE
a Solve 3kx dx = 1 to find k, or alternatively use the a Method 1:
R
1
R
0
U
N
C
O
formula for the area of a triangle to find k.
1
33kxdx = 1
0
c
1
3kx2
d =1
2 0
3k(1) 2
−0=1
2
k=
2
3
Method 2:
1
× 1 × 3k = 1
2
3k
2
=1
3k = 2
k=
484
2
3
Maths Quest 12 MatheMatICaL MethODs VCe units 3 and 4
c12ContinuousProbabilityDistributions.indd 484
23/08/15 7:00 PM
1
b 1 Write the rule for the mean.
b
E(X) = 3 xf(x)dx
0
1
= 3 (x × 2x)dx
0
1
2
= 3 (2x )dx
0
2 3 1
x d
3
0
= 23 (1) 3 − 0
PR
O
= 23
FS
c
O
=
2 Antidifferentiate and evaluate.
Var(X) = E(X2) − [E(X)] 2
3 Write the rule for the variance.
1
E(X2) = 3 x2f(x)dx
PA
G
E
4 Find E(X2).
TE
D
EC
2
2
a b
3
1
4
−
2
9
9
8
−
18
18
1
18
= 12 −
R
=
R
O
C
+ b) =
+ 3).
aE(X2)
1 4 1
x d
2
0
Var(X) = E(X2) − [E(X)] 2
c
E(3X − 1) = 3E(X) − 1
= 3 a 23 b − 1
U
N
E(2X2
a2Var(X)
c
= 12
=2−1
=1
Var(3X − 1) = 32Var(X)
2 Use the property
E(aX2
=
=
E(aX + b) = aE(X) + b to work out
E(3X − 1).
d Use the property
0
=
c 1 Use the property
Var(aX + b) =
Var(3X − 1).
= 3 2x3dx
1
= 12 (1)4 − 0
5 Substitute the appropriate values into the
variance formula.
0
to calculate
1
= 9 a 18
b
= 12
+ b to calculate
d
E(2X2 + 3) = 2E(X2) + 3
= 2 a 12 b + 3
=4
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 485
485
23/08/15 7:00 PM
Exercise 12.5 Linear transformations
WE6
f(x) = • kx,
0,
2k
0<x≤2
FS
Work without CAS
Questions 1–3
If the continuous random variable Y has a mean of 4 and a variance of 3, find:
a E(2Y − 3)
b Var(2Y − 3)
c E(Y 2)
d E(Y(Y − 1))
2 Two continuous random variables, X and Y, are related such that Y = aX + 5
where a is a positive integer and E(aX + 5) = Var(aX + 5). The mean of X is 9
and the variance of X is 2.
a Find the value of the constant a.
b Find E(Y) and Var(Y).
f (x)
3 WE7 The continuous random variable X has a
probability density function defined by
−kx, −2 ≤ x ≤ 0
(–2, 2k)
(2, 2k)
1
O
PRactise
elsewhere
TE
D
PA
G
E
PR
O
where k is a constant. The graph of the
x
0
(–2, 0)
(2, 0)
function is shown.
a Find the value of the constant k.
b Determine E(X) and Var(X).
c Find E(5X + 3) and Var(5X + 3).
d Find E((3X − 2) 2).
4 The continuous random variable X has a probability density function defined by
π
−cos (x),
≤x≤π
2
.
f(x) = •
0,
elsewhere
Consolidate
EC
a Sketch the graph of f and verify that it is a probability density function.
b Calculate E(X) and Var(X).
c Calculate E(3X + 1) and Var(3X + 1).
d Calculate E((2X − 1)(3X − 2)).
5 For a continuous random variable Z, where E(Z) = 5 and Var(Z) = 2, find:
a E(3Z − 2)
b Var(3Z − 2)
c E(Z2)
d E a 3Z 2 − 1 b .
R
Apply the most
appropriate
mathematical
processes and tools
R
1
U
N
C
O
6 The mean of the continuous random variable Y is known to be 3.5, and its
486 standard deviation is 1.2. Find:
Y
a E(2 − Y)
b Ea b
2
c Var(Y)
7 The length of time it takes for an electric
d Var(2 − Y)
Y
2
e Vara b.
kettle to come to the boil is a continuous
random variable with a mean of 1.5 minutes
and a standard deviation of 1.1 minutes.
If each time the kettle is brought to the boil
is an independent event and the kettle is
boiled five times a day, find the mean and
standard deviation of the total time taken for
the kettle to boil during a day.
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 486
23/08/15 7:00 PM
8 The probability density function for the continuous random variable X is
f(x) = e
mx(2 − x),
0≤x≤2
0,
elsewhere
where m is a constant. Find:
a the value of the constant m
b E(X) and Var(X)
c E(5 − 2X) and Var(5 − 2X).
9 The continuous random variable Z has a probability density function given by
0≤z≤a
FS
2
,
f(z) = • z + 1
0,
elsewhere
PA
G
E
PR
O
O
where a is a constant. Calculate, correct to 4 decimal places:
a the value of the constant a
b the mean and variance of Z
c iE(3Z + 1)
iiVar(3Z + 1)
iiiE(Z 2 + 2).
10 The continuous random variable X is transformed so that Y = aX + 3 where a is
a positive integer. If E(X) = 5 and Var(X) = 2, find the value of the constant a,
given that E(Y) = Var(Y). Then calculate both E(Y) and Var(Y) to verify this
statement.
11 The continuous random variable Y is transformed so that Z = aY − 3 where a is
TE
D
a positive integer. If E(Y) = 4 and Var(Y) = 1, find the value(s) of the constant
a, given that E(Z) = Var(Z). Then calculate both E(Z) and Var(Z) to verify this
statement.
EC
12 The continuous random variable Z has a probability density function given by
3
, 1≤z≤a
f(z) = • !z
0,
elsewhere
U
N
C
O
R
R
where a is a constant.
a Find the value of the constant a.
b Calculate the mean and variance of Z correct to 4 decimal places.
c Find, correct to 4 decimal places:
iE(4 − 3Z)
iiVar(4 − 3Z).
13 The daily rainfall, X mm, in a
particular Australian town has a
probability density function
defined by
x
x
sina b, 0 ≤ x ≤ 3π
3
f(x) = • kπ
0,
elsewhere
where k is a constant.
a Find the value of the constant k.
b What is the expected daily rainfall, correct to 2 decimal places?
c During the winter the daily rainfall is better approximated by W = 2X − 1.
What is the expected daily rainfall during winter, correct to 2 decimal places?
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 487
487
23/08/15 7:00 PM
14 The mass, Y kilograms, of flour sold in bags labelled as 1 kilogram is known to
have a probability density function given by
f(y) = e
k(2y + 1),
0.9 ≤ y ≤ 1.25
0,
elsewhere
f(z) = •
5 loge (z)
!z
,
1≤z≤a
PR
O
Master
O
FS
where k is a constant.
a Find the value of the constant k.
b Find the expected mass of a bag of flour, correct to 3 decimal places.
c On a particular day, the machinery packaging the bags of flour needed to be
recalibrated and produced a batch which had a mass of Z kilograms, where
the probability density function for Z was given by Z = 0.75Y + 0.45. What
was the expected mass of a bag of flour for this particular batch, correct to
3 decimal places?
15 The continuous random variable Z has a probability density function defined by
U
N
C
O
R
R
EC
TE
D
PA
G
E
0,
elsewhere
where a is a constant. Determine, correct to 4 decimal places:
a the value of the constant a
b E(Z) and Var(Z)
c E(3 − 2Z) and Var(3 − 2Z).
16 A continuous random variable, X, is transformed so that Y = aX + 1, where a is
a positive constant. If E(X) = 2 and Var(X) = 7, find the value of the constant
a, given E(Y) = Var(Y). Then calculate both E(Y) and Var(Y) to verify this
statement. Give your answers correct to 4 decimal places.
488 Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 488
23/08/15 7:00 PM
12.6 Review
a summary of the key points covered in this topic is
also available as a digital document.
REVIEW QUESTIONS
Download the Review questions document from
the links found in the Resources section of your
eBookPLUS.
Activities
to access eBookPlUs activities, log on to
TE
D
www.jacplus.com.au
PA
G
ONLINE ONLY
E
PR
O
the review contains:
• short-answer questions — providing you with the
opportunity to demonstrate the skills you have
developed to efficiently answer questions without the
use of CAS technology
• Multiple-choice questions — providing you with the
opportunity to practise answering questions using
CAS technology
• Extended-response questions — providing you with
the opportunity to practise exam-style questions.
FS
the Maths Quest review is available in a customisable
format for you to demonstrate your knowledge of this
topic.
www.jacplus.com.au
O
ONLINE ONLY
Interactivities
Units 3 & 4
Continuous probability
distributions
Sit topic test
U
N
C
O
R
R
EC
A comprehensive set of relevant interactivities
to bring difficult mathematical concepts to life
can be found in the Resources section of your
eBookPLUS.
studyON is an interactive and highly visual online
tool that helps you to clearly identify strengths
and weaknesses prior to your exams. You can
then confidently target areas of greatest need,
enabling you to achieve your best results.
topic 12 COntInuOus prObabILIty DIstrIbutIOns
c12ContinuousProbabilityDistributions.indd 489
489
23/08/15 7:01 PM
12 Answers
9
5a i 25
Exercise 12.2
4
f (x)
ii 25
37
f (x) = –14 e2x
loge 3, –94
(
9
–
4
b i 50
)
37
ii 42
29
6a100
( )
0, –1
4
7a200
3
5
(loge 3, 0) x
(0, 0)
41
b 100 c 50 d 100
FS
1a
b i 8
c
21
46
PR
O
b
O
31
ii 100
This is a probability density function as the
area is 1 unit2.
8a
f (x)
f (x)
(–1, 1)
f (x) = – –1x
f (x) = 0.25
(–e, –1e )
E
0.5
(2, 0.25)
(–e, 0)
(2, 0) x
0
(–2, 0)
2a
TE
D
This is a probability density function as the
area is 1 unit2.
(– –π2, 0)
2
– –π
0
4
π
–
4
R
2
π
–
2
3π
–
2
1
)
1.71
(
0
C
3𝜋 ,
––
4
0.29
( 0)
( 0)
𝜋
–
4,
3𝜋
––
,
4
)
𝜋 x
This is not a probability density function as the
π
area is units2.
2
c
f (x)
U
N
f(x)
𝜋
–
4,
f (x) = cos (x) + 1
x
This is a probability density function as the
area is 1 unit2.
b
f (x)
(
(–π2 , 0)
O
– –π
–
– 3π
b
R
f (x) = 0.5 cos(x)
This is a probability density function as the
area is 1 unit2.
EC
f (x)
x
(–1, 0) 0
PA
G
(–2, 0.25)
1
1
f(x) = –
2x
f (x) =
(0.5, 0.71)
1
–
2
sin(x)
(4, 0.25)
0 (0.5, 0)
(4, 0)x
(0, 0)
This is not a probability density function as the
area is 1.2929 units2.
490 3 n =
1
18
4 a =
1
11
𝜋
–
2
(𝜋, 0) x
This is a probability density function as the
area is 1 unit2.
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 490
23/08/15 7:01 PM
d
f(x)
b
1
f (x) = –––––
2 x–1
f (y)
(1, 1)
f (y) = –1y
(e, )
1
–
e
(2, 0.5)
(2, 0) x
(1, 0)
0
This is a probability density function as the area
is 1 unit2.
5
9 c = 7
e
1
3 y dy = 1
1
c a = 1
5
10 3 f(z)dz = 1
15n = 12
1
2
density function.
=1
×6×z=1
3z = 1
1a
1
8
(0, 1)
b m = 2
c m =
3
0
3
+ kx2 + x d = 1
0
+ k(3) 2 + 3 b − 0 = 1
9 + 9k + 3 = 1
9k + 12 = 1
9k = −11
k = −11
9
C
f(x) = –x
15
32
5
8
2a a = 1
b f (x)
(1, 4)
4
f (x) = 4x3
(0, 0)
(a, a)
c
U
N
f (x) = x
(0, 0) (a, 0)
a
3 −xdx + 3xdx =
0
a2 + 1
2
(1, 0) x
15
16
3a
(–1, 0)
−1
c
(1, 0)
(2, 0) z
f (x)
1
O
(–1, 1)
0
b
(0, 0)
R
13a = 2e
14a
EC
1
(3) 3
3
(2, 1)
R
a
TE
D
12 3 (x2 + 2kx + 1)dx = 1
1 3
x
3
f (z) = –z + 1
f (z) = z – 1
1
4
c
f (z)
PA
G
11a m =
1
Exercise 12.3
1
3
E
z=
PR
O
Atriangle = 1
e
O
16a = e. As f(x) ≥ 0 and 3 f(x)dx = 1, this is a probability
−1
1
bh
2
(e, 0) 𝜋 y
(1, 0)
FS
0
y
x
1 sin(x)
y =–
2
1
–
2
(0, 0)
(𝜋, 0)
x
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 491
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23/08/15 7:01 PM
b
c
"2
2
f (u)
f (u) = e4u
c 2"2 − 2
4a
f (x)
(a, e4a)
(0, 2k)
f (x) = k(2 + x)
f (x) = k(2 – x)
0
(–2, 0)
(0, 1)
x
(2, 0)
(0, 0)
d 3 e4udu = 4 e4a −
a
1
1 = 4k
k =
b
5
12
c
1
2
6a
(1, 0.2)
1
f (x) = –
5
(6, 0.2)
3
5
π
−2
b 3
e2
1
O
1
(e , –
2e )
C
2
(1, 0)
b
2
(e2, 0)
1
dz = 1. As f(z) ≥ 0 and 3 f(z)dz = 1, this is a
2z
e2
1
probability density function.
c
1
2
π
sin(z) d 2 π
−2
π
π
= 12 sina b − 12 sina− b
2
2
z
"2 + 1
4
183
256
d 0
9 a a = 1
b
c 0.371
10a
98
125
b
8
9
1
c 63
3
11a a = 3 loge a 2 b
c 0.5342
492 z
=1
This is a probability density function as the area
­under the curve is 1 and f(z) ≥ 0 for all values of z.
1
f (z) = –
2z
U
N
0
(1, 0.5)
(–π2 , 0)
= 12 + 12
R
7a f (z)
0.5
0
1
3 2 cos(z)dz =
(6, 0)x
(1, 0)
f (z) = –12 cos(z)
π
2
R
b
(0, –12 )
(– –π2 , 0)
f (x)
0
and a = 14 loge 5
f (z)
E
1
12
1
4
PA
G
5a
8a
TE
D
d
3
4
6
7
0
1
4
EC
c
O
1
× 4 × 2k
2
PR
O
1 =
u
(a, 0)
FS
1
b A = bh
2
b 0.6243
d 0.60
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 492
23/08/15 7:01 PM
12a
8a a = 0.4809
f (x)
b 3.3663
(0, 3)
c VAR(Z ) = 3.8195, SD(Z ) = 1.9571
d 3.0751
e 7
π
π
9a 3 (sin(2x) + 1)dx = 3 (sin(2x) + 1)dx
π
π
1
1
0
f (x) = 3e–3x
x
FS
0
b 0.9502
13a a = 2.1555
b 0.7147
14a0.1560
b a = tan a b ≈ 0.5463
9
4
19
ii 1.5625
12
or 25
16
E
b i 2a 1.3104
3
10a = 2, b =
b 0.7863
3 E(Z ) = 0.7305, m = 1.3010, Var(z) = 0.3424,
TE
D
SD(Z ) = 0.5851
1
1
1
4 E(X ) = 3 , m = 0.2310, Var(X ) = 9, SD(X ) = 3
1
1
5a 32 !x dx = 32 x dx
1
1
1
2
0
1
=
EC
0
−
1
x 2dx
23
1
R
=
1
R
0
1
2 1
c 2x d
2
0
O
= 12 (2!1 − 2!0)
= 12 × 2
U
N
C
=1
As f(x) ≥ 0 for all x-values, and the area under the
curve is 1, f(x) is a probability density function.
1
3
c m = 0.25
6a 0.5 min
b 0.5 min
c m = 0.35 min
7a 1.5601
b m = 1.6510
3
4
a
c Median = 0.8255
b
PR
O
1
2
PA
G
1a a =
O
c 0.0025
Exercise 12.4
0
π
1
= c −12 cos(2x) + x d
π
0
1
= a a −12 cos(2π) + π b − a −12 cos(0) + 0 b b
π
1
= a −12 + π + 12 b
π
=1
As f(x) ≥ 0 for all x-values, and the area under the
curve is 1, f(x) is a probability density function.
b 1.0708
c i 0.5725
ii 0.7566
d m = 0.9291
c Q1 = 1.3104, Q3 = 1.8899
11a 3
1
3
dz = 1
z2
a
−2
33z dz = 1
1
3 −3z−1 4 a1 = 1
a
3
c− d = 1
z 1
3 3
− + =1
a 1
3
− +3=1
a
3
− = −2
a
3 = 2a
3
a=
2
b E(Z ) = 1.2164, Var(Z ) = 0.0204
6
c m = 5 , interquartile range =
12a −
x
"4 −
13a h =
b 2
x2
b
8
33
3
π
1
4
c 2
d 0.5795
Topic 12 Continuous probability distributions c12ContinuousProbabilityDistributions.indd 493
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π
3 (−cos(x))dx = c −sin(x) d π
14a f (x)
π
2
π
2
(b, k)
f (x) = k
(a, 0)
x
(b, 0)
3 k dx = 1
b
d E((2X − 1)(3X − 2)) = 24.5079
a
5a 13
3 kx 4 ba = 1
c 27
kb − ka = 1
k(b − a) = 1
1
k=
b−a
8a m =
3
4
b E(X ) = 1, Var(X ) = 0.2
c E(5 − 2X ) = 3, Var(5 − 2X ) = 0.8
9a a = 0.6487
7.9344
y
3 f(y)dy = 3 0.2 loge a2 b dy = 1
2
b E(Z ) = 0.2974, Var(Z ) = 0.0349
c i 1.8922
2
b 5.7278
d m = 3.9816
e 5.9344
ii 3.3085
iii 0.1176
iv 0.3430
EC
b i 1.7863
R
Exercise 12.5
b 12
d 15
R
1a 5
c 19
O
2a a = 5
iii 2.1234
11a = 1 or 3, E(Y ) = 1 or 9, Var(Y ) = 1 or 9
12a a =
16a 2.3104
ii 0.3141
10a = 3, E(Y ) = 18, Var(Y ) = 18
TE
D
c Var(Y ) = 2.1600, SD(Y ) = 1.4697
49
36
b E(Z ) = 1.1759, Var(Z ) = 0.0109
c i 0.4722
13a k = 9
14a k = 0.9070
ii 0.0978
b 5.61 mm
c 10.21 mm
b 1.081 kg
c 1.261 kg
400
15a a =
441
b E(Z ) = 1.4921, Var(Z ) = 0.0361
c E(3 − 2Z ) = 0.0158, Var(3 − 2Z ) = 0.1444
16a = 0.5469, E(Y ) = 2.0938, Var(Y ) = 2.0938
C
b E(Y ) = 50, Var(Y ) = 50
1
4
d 1.44
7 E(5T ) = 7.5 minutes, SD(5T ) = 5.5 minutes
12
3a k =
b 1.75
c 1.44
e 0.36
b+a
2
(a − b) 2
7.9344
15a
6a –1.5
E
d
d 8
PA
G
c
b 18
O
b
PR
O
0
As f(x) ≥ 0 for all x-values and the area under the
curve is 1, f(x) is a probability density function.
b E(X ) = 2.5708, Var(X ) = 0.1416
c E(3X + 1) = 8.7124, Var(3X + 1) = 1.2743
FS
(a, k)
k
π
= −sin(π) + sina b
2
=0+1
=1
U
N
b E(X ) = 0, Var(X ) = 2
c E(5X + 3) = 3, Var(5X + 3) = 50
d E((3X − 2) 2) = 22
4a f (x)
(π, 1)
1
f (x) = –cos(x)
(––π2, 0)
0
494 (π, 0)
π
x
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
c12ContinuousProbabilityDistributions.indd 494
23/08/15 7:01 PM
FS
O
PR
O
E
PA
G
TE
D
EC
R
R
O
C
U
N
c12ContinuousProbabilityDistributions.indd 495
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