Idea Transcript
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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.
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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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y 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
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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.
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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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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
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Probability density functions
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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
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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 WritE/draW
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, 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. 4
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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)
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x (2, 0)
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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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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WOrKeD 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
topic 12 COntInuOus prObabILIty DIstrIbutIOns
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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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1 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 4 4
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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,
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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 −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
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7 Antidifferentiate and evaluate after
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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
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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
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across two functions, the interval needs to be split.
c
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=
+
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4 Antidifferentiate and evaluate after
0
0
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= − 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 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
.
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f(x) = e
PA G
Consolidate
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0≤x≤a
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4x3,
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f(x) = e
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2 The continuous random variable X has a probability density function given by
a Sketch the graph of f .
π 4
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b Find Pra < X <
π 4
3π b. 4
c Find PraX > │X <
3π b. 4
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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
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5 The amount of petrol sold daily
0,
a Sketch the graph of f .
1≤x≤6
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f(x) =
1 , u5
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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
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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.
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1
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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
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Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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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).
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b Find Pra− ≤ Z ≤ b.
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3 2 x, 0≤z≤2 f (x) = • 8 . 0, elsewhere
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10 The continuous random variable X has a probability density function defined by
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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
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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
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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.
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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
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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
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–2
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–3
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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
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−a ≤ x ≤ a
0,
.
elsewhere
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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.
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Units 3 & 4
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12.4
Measures of centre and spread
Measures of central tendency
AOS 4
The mean Remember that for a discrete random variable,
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Topic 3
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Concept 3
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Mean and median Concept summary Practice questions
E(X) = μ = a xnPr(X = xn). x=n x=1
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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
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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
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1
= 3x3dx
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0 1
1
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x4 = c d 4 0 4 1 −0 = 4 1 = 4
Similarly, if the continuous random variable X has a probability density function of
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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
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= 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
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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.
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=
Median and percentiles The median is also known as the 50th percentile, Q2, the halfway mark or the middle value of the distribution.
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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.
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−∞
EC
Other percentiles, which are frequently calculated, are the 25th percentile or lower quartile, Q1, and the 75th percentile or upper quartile, Q3.
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The interquartile range is calculated as: IQR = Q3 − Q1
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Consider a continuous random variable, X, that has a probability density function of 2
f(x) = e
0.21e2x−x ,
−3 ≤ x ≤ 5
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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
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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
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x4 1 d = 16 0 2
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m4 1 −0= 16 2 m4 = 8 4 m = ±" 8
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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
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x4 1 d = 4 16 0
a4 1 −0= 4 16 a4 = 4
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c
R
4 a = ±" 4
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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
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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)
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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.
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tHinK
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a The graph f(y) = ky is a straight line with end points
a
f (y) k
(1, k)
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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
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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
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c
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E
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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
=
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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
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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
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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
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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
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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
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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
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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).
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−∞
Substituting this result into SD(X) = !Var(X) gives us
−∞
SD(X) = "E(X2) − 3 E(X) 4 2 .
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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
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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.
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tHinK
C
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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
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=
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
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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
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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.
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So m =
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3 Write the answer.
R
first.
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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
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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)
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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: i the mean of Z ii the 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
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C
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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
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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
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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
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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 i Var(X) 0.25 0.5 0.75 1 ii SD(X). d Find the median value of f correct to 4 decimal places.
480
Maths Quest 12 MATHEMATICAL METHODS VCE Units 3 and 4
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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
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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