Download Slide 1 - WordPress.com

Six fuses, of which two are defective and four are good, are to be
tested one after the other in random order until both defective
fuses are identified. Find the probability that the number of fuses
that will be tested is:
(a) three
(b) four or fewer.
X = finding the two defective fuses. DISCRETE RANDOM VARIABLE
x
2
P(X=x)
1/
P(X≤x)
1/
15
15
3
4
2/
15
3/
15
4/
15
5/
3/
6/
15
10/
15
1
15
(a) P(X=3)= 2/15
5
6
15
(b) P(X≤4)= 6/15
PROBABILITY FUNCTION
CUMULATIVE FUNCTION
The probability distribution for the discrete random variable, X,
representing the score on a biased tetrahedral die is:
x
1
2
3
4
3/
4/
P(X=x) 12/25 6/25
25
25
How many times would you expect each number to occur after
100 throws of this die?
x
Expected Frequency
1
48
2
24
3
16
4
12
What is the mean (expected) score over 100 throws?
[(1x48)+(2x24)+(3x16)+(4x12)]÷100 = 1.92
Consider we do a similar thing using probabilities instead of
expected frequencies:
x
1
2
3
4
3/
4/
P(X=x) 12/25 6/25
25
25
[(1x 12/25 )+(2x 6/25 )+(3x 4/25 )+(4x 3/25 )] ÷ 1 = 1.92
In other words, the number of throws are irrelevant.
In probability, we call the mean of a random variable the EXPECTED
VALUE or E(X).
  E ( X )   xP (X  x )
x
It can also be shown that the VARIANCE of X is:
 2  Var (X )   x 2 P (X  x )   2  E (X 2 )   2
x
Ex8B Qu8
A statistically minded parent is discussing pocket money with a young
child. The parent has £1 and a 50p coin. Both coins are spun and if a
coin lands as a head the child can have it as pocket money, otherwise
the parent keeps it. The child’s parent is generous and, rather than
disappointing the child, in the event of both coins landing tails the
child will receive 50p. How much pocket money can the child expect
to receive?
x
P(X=x)
50
1/
2
£1
1/
4
£1.50
1/
4
E(X) = (50 x ½) + (100 x ¼) + (150 x ¼) = 87.5p
Ex8B Qu13
A random variable R takes the integer value r with probability p(r)
Where:
p(r) = kr3
r = 1, 2, 3, 4
p(r) = 0
otherwise
Find:
(a) The value of k and display the distribution on graph paper
(b) The mean and variance of the distribution.
r
1
2
3
4
0.8
k
P(R=r)
8k
27k 64k
0.6
k = 1/100
k + 8k + 27k + 64k = 1
0.4
r
1
2
3
4
0.2
P(R=r) 1/100 8/100 27/100 64/100
1
2
3
4
r
P(R=r)
1
1/
100
2
8/
100
3
27/
100
4
64/
100
E ( X )   xP ( X  x )
x
E(X) = (1 x 1/100) + (2 x 8/100) + (3 x 27/100) + (4 x 64/100)
E(X) = 3.54
Var ( X )   x 2 P ( X  x )   2
x
Var(X) = (12 x 1/100) + (22 x 8/100) + (32 x 27/100) + (42 x 64/100) – 3.542
Var(X) = 0.4684
Going back to the biased tetrahedral die. We know E(X) = 1.92,
what is Var (X)?
x
1
2
3
4
Var(X) = 1.1136
12
3
6
4
/25
/25
P(X=x)
/25
/25
Imagine we alter the scores using the formula R = 4X – 3.
What would E(R) and Var(R) be? Is there a connection between
these values and E(X) and Var(X), the original values?
E(R) = 4.68
r
1
5
9
13
3/
4/
P(R=r) 12/25 6/25
Var(R) = 17.8176
25
25
What you should notice is:
E(R) = 4E(X) - 3 and
Var(R) = 16Var(X)
In general…
E(aX ± b) = aE(X) ± b
Var(aX ± b) = a2 Var(X)
Does E(X2) = [E(X)]2?
x
x2
1
1
P(X=x)
2
4
12/
25
6/
25
3
9
4/
Remember:
Var ( X )  E ( X 2 )   2
25
4
16
3/
25
E(X) = 1.92
[E(X)]2 = 3.6864
E(X2) = 4.8
So, E(X2) ≠ [E(X)]2
You can get E(X2)
from this.
Ex8C Qu1
The random variable Y has mean 2 and variance 9. Find:
(a) E(3Y + 1)
(b) E(2 – 3Y)
(c) Var(3Y + 1)
(d) Var(2 – 3Y)
(e) E(Y2)
(f) E[(Y-1)(Y+1)
(a) E(3Y + 1) = 3E(Y) + 1
= (3 x 2) + 1 = 7
(b) E(2 – 3Y) = 2 – 3E(Y)
= 2 – (3 x 2) = -4
(c) Var(3Y + 1) = 32Var(Y)
=9x9
= 81
(d) Var(2 – 3Y) = (-3)2Var(Y) = 9 x 9 = 81
(e) E(Y2):
Var(Y) = E(Y2) – [E(Y)]2
So, 9 = E(Y2) - 22 So, E(Y2) = 13
(f) E[(Y-1)(Y+1)] = E(Y2 – 1) = E(Y2) – 1 = 13 – 1 = 12
Ex 8C Qu6
The discrete random variable X has a probability distribution
specified by the following table:
x
-2
0
2
4
P(X=x)
0.3
0.4
0.2
0.1
(a) Find E(X)
(b) Find Var(X)
(c) Find E(½X + 1)
(d) Find Var(½X + 1)
= 0.2
(a) E(X) = (-2 x 0.3) + (0 x 0.4) + (2 x 0.2) + (4 x 0.1)
(b) Var(X) = (-22 x 0.3) + (02 x 0.4) + (22 x 0.2) + (42 x 0.1) – 0.22 = 3.56
(c) E(½X + 1) = ½E(X) + 1 = ½(0.2) + 1 = 1.1
(d) Var(½X + 1) = (½)2 Var(X) = ¼ (3.56) =
0.89
DISCRETE UNIFORM DISTRIBUTION
The conditions for a discrete uniform distribution (drv) are:
 the drv, X, is defined over a set of n distinct values
 each value is equally likely.
It follows from this last condition that if there are n distinct values
and each is equally likely:
1
P ( X  xr ) 
r  1, 2, 3, 4, ... , n
n
It can be shown that for a discrete uniform distribution:
(n  1)(n  1)
n 1
2
  Var ( X ) 
  E( X ) 
12
2
Ex 8D Qu1
A fair die is thrown once and the random variable X represents the
value on the uppermost face.
(a) Find the mean and variance of X.
(b) Calculate the probability that X is within one standard deviation
of the mean.
 X is defined over a distinct number of values (6)
 each value is equally likely, 1/6
 X is a discrete uniform distribution.
(a) Find the mean and variance of X.
(n  1)(n  1)
n 1
2
  Var ( X ) 
  E( X ) 
12
2
61
35
(6  1)(6  1)
E( X ) 

 3.5
Var ( X ) 
2
12
12
(b) Calculate the probability that X is within one standard deviation
of the mean.
35
Standard Deviation = √ Variance 
= 1.708
12
P(X is 1 σ from mean) = P(3.5 – 1.708 < X < 3.5 + 1.708)
= P(1.792< X < 5.208) = P(2 ≤ X ≤ 5)
4 2
 
6 3