Discrete Random Variables 2
•To understand and calculate with cumulative
distribution functions
•To be able to calculate the mean or expected
value of a discrete random variable
•To be able to calculate the variance of a
discrete random variable
Cumulative Distribution Function
The probability that X is less than or equal to x is
written as F(x)
F(x) = P(X ≤ x)
2 coins are thrown. X is the number of heads
Sample space HH, HT, TH, TT
Cumulative distribution function
x
P(X=x)
F(x)
0
¼
¼
1
½
¾
2
¼
1
Problem 1
F(x) = (x + k)
8
x = 1,2,3
a) Find k
b) Write down the distribution table for the
cumulative distribution function
c) Write down F(2.6)
d) Find the probability distribution of X
Problem 1 - Solution
F(x) = (x + k)
8
x = 1,2,3
a) Find k
F(3) = 1 so 3 + k = 1 therefore 3+k = 8 and k=5
8
b) F(2) = 2 + 5 = 7
8
8
F(1) = 1 + 5 = 6
8
8
x
1
F(x)
6/
2
8
7/
3
8
1
Problem 1 - Solution
F(x) = (x + k)
8
x = 1,2,3
c) F(2.6) means P(X≤2.6) = F(2) = 7/8
d) x
F(x)
1
2
3
6/
1/
1/
8
8
8
Mean or Expected value of a DRV
NB p(x) is the same as P(X=x)
Expected value of X =
E(X) = ΣxP(X=x) = Σxp(x)
Statistics experiment
• Collect data
• Frequency distribution
• Mean value Σfx
Σx
Theoretical approach
• Probability distribution
• Expected value Σxp(x)
Problem 1
A survey of 100 houses is conducted about the
number of TV sets
No of sets
Frequency
0
10
1
75
2
10
3
5
a) Calculate the mean number of sets
b) Find the probability distribution where X is the
number of TV sets in a house picked at
random
c) Calculate the expected value of X
Problem 1 - Solution
A survey of 100 houses is conducted about the
number of TV sets
No of sets
0
1
2
3
Frequency
10
75
10
5
a) Calculate the mean number of sets
Mean value Σfx = 0 + 75 + 20 + 15 = 110 = 1.1
Σx
100
100
b)
x
0
1
2
3
p(x)
0.1
0.75
0.10 0.05
c) Expected value of X = Σxp(x)
Σxp(x) = 0 + 0.75 + 0.20 + 0.15 = 1.1
Note that the mean value = E(X) = 1.1
E(X) is sometimes called the mean of X
Problem 2
x
p(x)
1
0.1
2
p
3
0.3
4
q
5
0.2
Given that E(X)=3 write 2 equations involving p
and q and solve the find the value of p and q.
Problem 2 - Solution
x
p(x)
1
0.1
2
p
3
0.3
4
q
5
0.2
Given that E(X)=3 write 2 equations involving p
and q and solve the find the value of p and q.
0.1 + p + 0.3 + q + 0.2 = 1
p + q + 0.6 = 1
p + q = 0.4
0.1 + 2p + 0.9 + 4q + 1 = 3
2p + 4q + 2 = 3
2p + 4q = 1
Problem 2 - Solution
x
p(x)
1
0.1
2
p
3
0.3
4
q
5
0.2
Given that E(X)=3 write 2 equations involving p
and q and solve the find the value of p and q.
p + q = 0.4
and so
p = 0.4 - q
Substitute into 2p + 4q = 1
2(0.4 – q) + 4q = 1
0.8 – 2q + 4q = 1
0.8 + 2q = 1
2q = 0.2 and q = 0.1
p = 0.3
Problem 3 – Expected value of X²
E(X²) = Σx²p(x)
E(Xn) = Σxnp(x)
A discrete random variable X has a probability
distribution
x
1
P(X=x) 12/25
2
6/
25
3
4/
25
4
3/
25
Write down the probability distribution for X²
b) Find E(X²)
a)
Problem 3 – Solution
a)
a)
x
1
x²
1
P(X=x) 12/25
2
4
6/
25
E(X²) = Σx²p(X=x²)
= 12/25 + 24/25 + 36/25 + 48/25
= 120/25
= 4.8
3
9
4/
25
4
16
3/
25
Variance
Var(X) = E(X²) – (E(X))²
Example
2 four sided die numbered 1,2,3,4 are spun and their faces are
added (X).
a) Find the probability distribution of X
b) Find E(M)
c) Find Var(M)
a)
+
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
2
3
x
p(x)
1/
16
2/
16
4
3/
16
5
4/
16
6
3/
16
7
2/
16
8
1/
16
Variance
Var(X) = E(X²) – (E(X))²
b) Find E(M)
x
p(x)
2
1/
16
3
2/
16
4
3/
16
5
4/
16
6
3/
16
7
2/
16
8
1/
16
E(M) = Σxp(x)
= 2/16 + 6/16 + 12/16 + 20/16 + 18/16 + 14/16 + 8/16
= 80/16 = 5
Var(X) = E(X²) – (E(X))²
=(4/16 +18/16 + 48/16 + 100/16 + 108/16 + 98/16 + 64/16)-25
= 440/16 – 25 = 2.5