Download S2 Continuous Random Variables

1.
If y =
5x2
– 2x + 1, find
𝑑𝑦
.
𝑑𝑥
𝑑𝑦
= 10𝑥 − 2
𝑑𝑥
2.
Find the turning point of the function given in question 1.
10𝑥 − 2 = 0
1
𝑥=
5
2
4
1
1
𝑦=5
−2
+1 =
5
5
5
1 4
𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡: ,
5 5
3.
Solve 3x2 – 8x + 1 = 0 giving your answers to 2 decimal places.
𝑎 = 3,
𝑥=
𝑏 = −8,
8±
−8
2
𝑐=1
− 4(3)(1)
6
8 ± 52
𝑥=
6
𝑥 = 2.54, 0.13
4.
Calculate
2
−1
𝑥 2 − 2𝑥 + 5 𝑑𝑥
2
x

   x 2  5x 
3
 1
3
2
 2
  (1)

2
2
   2  5(2)   
 (1)  5(1) 
  3
 1
 3
26
19
 15
 
3
3
3
3
Suppose the heights of a very large number of 17 year olds was
surveyed
The total area under the graph is 1 and particular areas represent
probabilities
Suppose the class widths were refined to 0.1m
Assuming that enough data was collected the class widths
could be further reduced until the ‘relative frequency
polygon’ approximates to a curve. When the function ‘f(x)’
is found for the curve, it is called the probability density
function, or pdf
To be a probability density function (pdf), f(x) must satisfy these
basic properties:
 f(x) ≥ 0 for all x, so that no probabilities are negative

∞
𝑓
−∞
𝑥 𝑑𝑥 = 1 ; often f(x) is only defined over a small
range, in which case the integral over that range will be 1.
You can find the probability that a
random variable lies between
x = a and x = b from the area
under the curve represented by
f(x) between these two points.
Example 1
(a)
Show that f(x) is a probability density function where
1
𝑓 𝑥 = 2 𝑥 − 3 𝑓𝑜𝑟 3 ≤ 𝑥 ≤ 5
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
 f(x) ≥ 0 for all x, so that no probabilities are negative
f(x) is positive for all values of x

∞
𝑓
−∞
𝑥 𝑑𝑥 = 1 ; often f(x) is only defined over a small
range, in which case the integral over that range will be 1.
5
2
2
1
x



1
5
3




1


3
x
 2 (x  3)dx 2  2  3  2  2  3(5)    2  3(3) 
5
2
3
1 5
9
1
  


2 2
2
Example 1
(a)
Show that f(x) is a probability density function where
1
𝑓 𝑥 = 2 𝑥 − 3 𝑓𝑜𝑟 3 ≤ 𝑥 ≤ 5
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
(b)
Find P(X < 4)
4
1
P( X  4)   (x  3)dx
2
3
4
2
1 x

P( X  4)    3x 
2 2
3
1  4 2
  32

P( X  4)    3(4)     3(3) 
2  2
 2

1
9
P(X  4)   4   
2 
2 
1
P( X  4) 
4
𝒃
𝒇
𝒂
𝑷 𝒂<𝑿<𝒃 =
If f(x) is a pdf, then
𝒙 𝒅𝒙
Example 2
𝑓 𝑥 =
𝑥
2−𝑥
0
𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 1
𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 2 gives a pdf.
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Draw the pdf and find P(0.5 < x < 1.3)
1.3
1
P(0.5 < x < 1.3)   xdx   (2  x)dx
f(x)
0.5
1
1
1.3
x 
x  
    2 x  
2 1
 2  0.5 
1
2
2
 0.63
0
1
2
x
Example 3
2
𝑘
9
−
𝑥
𝑓𝑜𝑟 − 3 ≤ 𝑥 ≤ 3
𝑓 𝑥 =
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
(a)
Find the value of k and calculate P(-1 < X < 2)
(b)
Find
P(X = 2)
∞
(a) 
𝑓 𝑥 𝑑𝑥 = 1 ; often f(x) is only defined over a small
−∞
range, in which case the integral over that range will be 1.
3
3
3
3 3

3
(

3
)




x


2
k
(
9

x
)
dx

1
k
9
(
3
)


9
(

3
)

k 9 x    1

 
  1

3
3 
3 
3  3


1
k(18  18)  1 36k  1 k 
36
2
3
3
3

1
2
(

1
)




1
x  
 9(2)     9(1) 

P(1  X  2)  9 x  

3 
3 
36 
3  1 36 
1  46
26  2
    
36  3
3 3
(b) For any CONTINUOUS distribution, P(X = a) = 0
P(X = 2) = 0
In S1, the cumulative distribution
function (cdf) F(x0) was defined as
P(X ≤ x0) for discrete random variables.
For continuous random variables, you
can find the cdf by integrating the pdf.
Similarly, you can find the pdf from the
cdf by differentiating.
Probability Density
Function (pdf)
f(x)
Cumulative Distribution
Function (cdf)
F(x)
Example 4
𝑓 𝑥 =
1
2
𝑥− 3
𝑓𝑜𝑟 3 ≤ 𝑥 ≤ 5
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
is a pdf.
Find the cumulative distribution function.
𝐹 𝑥 =
0
𝑥<3
x 2 3x 9
 
3 x 5
4 2 4
1
𝑥>5
2
2
2

1
x
3




1
1 x

3 2 (x  3)dx  2  2  3x   2  2  3x    2  3(3) 
3
x
x
1  x2
9  x 2 3x 9
   3x      
2 2
2
4 2 4
Example 5
𝐼𝑓 𝐹 𝑥 =
0
1 2
𝑥
9
1
𝑥≤0
0 ≤𝑥 ≤3
𝑥>3
dF (x)
f (x) 
dx
𝑓 𝑥 =
2
 x
9
2
x 0 x 3
9
0
otherwise
find f(x)
The mean or expected value of a discrete probability distribution is
defined as:
μ = E(X) = Σpx
For a continuous random variable:
𝝁=𝑬 𝑿 =
∞
𝒙𝒇
−∞
𝒙 𝒅𝒙
where in practice, the limits will be the interval over which f(x) is
defined.
Example 6
1
2
𝑓 𝑥 =
𝑥− 3
0
3≤𝑥 ≤5
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find E(X)
5
𝐸 𝑋 =
5
𝑥𝑓 𝑥 𝑑𝑥
=
3
3
5
1 2
𝑥 − 3𝑥 𝑑𝑥
2
1  53 3(5)2   33 3(3)2 
1  x 3x 
  
 

  

2  3
2 
2 3
2  3 2  3
3
2
1 25 27  13



3
2  6
6 
For a discrete random variable, variance, Var(X) = σ2 = E(X2) – μ2
For a continuous random variable,𝑬 𝑿𝟐 =
∞ 𝟐
𝒙 𝒇
−∞
𝒙 𝒅𝒙
Example 7
Find the standard deviation for
𝑓 𝑥 =
𝐸
𝑋2
1
2
0
𝑥− 3
5
𝑥2𝑓
=
3
3≤𝑥 ≤5
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
5
𝑥 𝑑𝑥 =
3
1 3
𝑥 − 3𝑥 2 𝑑𝑥
2
1  54 3   34 3 
   5     3   19
2  4
 4

2
2
13


2
2
Var(X) = E(X ) – μ  19    
9
3
2
2
Standard Deviation 

9
3
5
1 x
3
  x 
2 4
3
4
*
Example 8
The continuous random variable, X, has probability density function
𝑘𝑥 0 ≤ 𝑥 ≤ 5
𝑓 𝑥 =
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
(a)
Find the value of k
(b)
Find the mean and variance of X.
(c)
Calculate
i
P(X > μ)
ii
P(X > μ + σ)
5
10
2
(
c
)
i
P
(
X


)

(b) E ( X ) 
(a) k 
9
3
25
25
2
ii P( X     )  0.1857
E(X ) 
2
25
Var ( X ) 
18
Example 9
The continuous random variable Y has probability density function
𝑦
𝑓𝑜𝑟 0 ≤ 𝑦 ≤ 1
𝑓 𝑦 = 2 − 𝑦 𝑓𝑜𝑟 1 ≤ 𝑦 ≤ 2
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find the mean and variance of Y.
1
7
2
Var (Y ) 
E (Y ) 
E (Y )  1
6
6
Example 10
The continuous random variable X has probability density function
𝑓 𝑥 =
(a)
(b)
1
36
9 − 𝑥2
−3≤𝑥 ≤3
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find the mean and variance of X.
Calculate
i
P(X > 2)
(a) E ( X )  0
E(X 2 ) 
(b) i P( X  2) 
2
27
9
5
Var ( X ) 
ii
P(|X| > σ)
9
5
ii P(| X |  )  P( X   )  P( X   )  0.374
Example 11
The weekly petrol consumption, in hundreds of litres, of a sales
representative may be modelled by the random variable X with
probability density function
2 (𝑏 − 𝑥)
𝑎𝑥
0≤𝑥 ≤2
𝑓 𝑥 =
0
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
(a)
Find the values of a and b if the mean consumption is 144 litres.
(b)
Find the standard deviation of the weekly petrol consumption.
3
(a) a 
b4
20
(b)   0.408  40.8litres
The mode of a continuous random variable is the value of x for
which f(x) is a maximum over the interval in which f(x) exists.
This is either a stationary point, at which f ’(x) = 0, or the end value
of the interval over which f(x) is defined.
There may not be a mode if no single value occurs more often that
any other. (Bimodal distributions do occur commonly in real life).
Example 12
𝑓 𝑥 =
1
36
0
9 − 𝑥2
−3≤𝑥 ≤3
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
x
d 1
2 
0

9x   

18
dx  36
This lies in the range - 3  x  3
So the mode  0
x0
Find the mode.
Example 13
𝑓 𝑥 =
1
2
𝑥− 3
0
f (3)  0
3≤𝑥 ≤5
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
f (5)  1
Maximum value of f(x) is when x  5
Mode  5
Find the mode
You can use the cumulative distribution function to find the median
and quartiles for a continuous random variable:
If m is the median, then F(m) = 0.5
If Q1 is the lower quartile then F(Q1) = 0.25
If Q3 is the upper quartile then F(Q3) = 0.75
You can set up an appropriate algebraic equation and solve it to find
the required quartile, decile or percentile.
Example 14
𝑓 𝑥 =
Find
1
2
0
𝑥− 3
3≤𝑥 ≤5
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
(a)
(b)
the median
the lower and upper quartiles for X.
x
x
2
1
1  x 2
  32

1 x

(a) F ( X )   (x  3)dx    3x     3x     3(3) 
2
2 2
 2

 3 2  2
3
1  x2
9  x 2 3x 9
   3x      
2 2
2
4 2 4
Median is when F(X)=0.5
x 2 3x 9 1
2
x
 6x  7  0
  
4 2 4 2
x  3  2  4.41  median
Example 14
𝑓 𝑥 =
1
2
𝑥− 3
0
3≤𝑥 ≤5
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find
(a)
the median
(b)
the lower and upper quartiles for X.
Q3 is when F(X) = 0.75
(b) Q1 is when F(X) = 0.25
x 2 3x 9 1
  
4 2 4 4
x2  6x  8  0
(x  2)(x  4)  0
Q1  4
x 2 3x 9 3
  
4 2 4 4
x2  6x  6  0
Q3  3  3  4.73
If you were asked for the SHAPE of the distribution or the SKEW…
Q1  4
Q2  4.41
Q2 – Q1 = 0.41
Q3 – Q2 = 0.32
Q2 – Q1 > Q3 – Q2
Negative Skew
Q3  4.73