1. DISCRETE RANDOM VARIABLES
Section 2
Choose from the following:
Introduction: Car share scheme a success
Example 4.3: A discrete random variable
Example 4.4: Laura’s Milk Bill
End presentation

2. Car share scheme a success

3. Car share scheme a success 1 2 3 4 5 > 5 0.35 0.375 0.205 0.065
Number of people / Outcome r 1 2 3 4 5
> 5
Relative frequency / Probability P(X = r)
0.35
0.375
0.205
0.065
0.005

4. Multiply each r value by P(X = r) to form the
Expectation r
P(X = r)
r P(X = r) 1
0.35 2
0.375 3
0.205 4
0.065 5
0.005
Totals
Multiply each r value by P(X = r) to form the
r P(X = r)
column

5. Multiply each r value by P(X = r) to form the
Expectation r
P(X = r)
r P(X = r) 1
0.35 2
0.375
0.75 3
0.205 4
0.065 5
0.005
Totals
Multiply each r value by P(X = r) to form the
r P(X = r)
column

6. Multiply each r value by P(X = r) to form the
Expectation r
P(X = r)
r P(X = r) 1
0.35 2
0.375
0.75 3
0.205
0.615 4
0.065 5
0.005
Totals
Multiply each r value by P(X = r) to form the
r P(X = r)
column

7. Multiply each r value by P(X = r) to form the
Expectation r
P(X = r)
r P(X = r) 1
0.35 2
0.375
0.75 3
0.205
0.615 4
0.065
0.26 5
0.005
Totals
Multiply each r value by P(X = r) to form the
r P(X = r)
column

8. Multiply each r value by P(X = r) to form the
Expectation r
P(X = r)
r P(X = r) 1
0.35 2
0.375
0.75 3
0.205
0.615 4
0.065
0.26 5
0.005
0.025
Totals
Multiply each r value by P(X = r) to form the
r P(X = r)
column

9. Now find the total of the
Expectation r
P(X = r)
r P(X = r) 1
0.35 2
0.375
0.75 3
0.205
0.615 4
0.065
0.26 5
0.005
0.025
Totals
Now find the total of the
r P(X = r)
column

10. Expectation r
P(X = r)
r P(X = r) 1
0.35 2
0.375
0.75 3
0.205
0.615 4
0.065
0.26 5
0.005
0.025
Totals
Expectation
= E(X) or m
E(X) = m = S r P(X = r)
= 1× × × × ×0.005 =
= 2

11. Variance r
P(X = r)
r P(X = r)
r2 P(X = r) 1
0.35 2
0.375
0.75
1.5 3
0.205
0.615
1.845 4
0.065
0.26
1.04 5
0.005
0.025
0.125
Totals
4.86
Var(X) = s 2 = S r 2 P(X = r) – m 2
= 12× × × × × – 22
= – 4
= – 4 = 0.86

12. Example 4.3
The discrete random variable X has the distribution: r 1 2 3
P(X = r)
0.2
0.3
0.4
0.1
(i) Find E(X).
(ii) Find E(X2).
(iii) Find Var(X) using Var(X)= E(X2) – m2

13. Example 4.3 : (i) Expectation E(X)r
P(X = r)
r P(X = r)
0.2 1
0.3 2
0.4
0.8 3
0.1
Totals
1.4
Expectation
= E(X) or m
E(X) = m = S r P(X = r)
= 0× × × ×0.1 =
= 1.4

14. Example 4.3 : (ii) E(X 2) r
P(X = r)
r P(X = r)
r2 P(X = r)
0.2 1
0.3 2
0.4
0.8
1.6 3
0.1
0.9
Totals
1.4
2.8
Expectation
= E(X2)
E(X2) = S r 2 P(X = r)
= 02× × × ×0.1 =
= 2.8

15. Example 4.3 : Variance r
P(X = r)
r P(X = r)
r2 P(X = r)
0.2 1
0.3 2
0.4
0.8
1.6 3
0.1
0.9
Totals
1.4
2.8
Var(X) = s 2 = S r 2 P(X = r) – m 2
= 02× × × ×0.1 – 1.42
= – 1.42
= – = 0.84

16. Example 4.4 : Laura’s Milk Bill

17. Example 4.4 : Laura’s Milk Bill
Laura has one pint of milk on three days out of every four and none on the fourth day. A pint of milk costs 40p.
Let X represent her weekly milk bill.
(i) Find the probability distribution for her weekly milk bill.
(ii) Find the mean (m) and standard deviation (s) of her weekly milk bill.
(iii) Find (a) P(X > m + s ) and (b) P(X < m −s ).

18. Example 4.4 : (i) Probability distribution
Since Laura has milk delivered, it takes four weeks before the delivery pattern starts to repeat. M Tu W Th F Sa Su
No. pints
Milk bill   6
£2.40 5
£2.00 r
2.00
2.40
P(X = r)
0.75
0.25

19. Example 4.4 : (i) Mean μ or expectation E(X)
P(X = r)
r P(X = r)
2.00
0.75
1.50
2.40
0.25
0.60
Totals 1
2.10
Expectation
= E(X) or m
E(X) = m = S r P(X = r)
= × × 0.25 =
= 2.10

20. Example 4.3 : (ii) Standard deviation σ
P(X = r)
r P(X = r)
r2 P(X = r)
2.00
0.75
1.50
3.00
2.40
0.25
0.60
1.44
Totals 1
2.10
4.44
Var(X) = s 2 = S r 2 P(X = r) – m 2
= 22 × × – 2.12
= – 2.12
= – = 0.03
Hence s = √0.03 = (to 2 d.p.)

21. Example 4.4 : (iii) Calculating probabilities
The probability distribution for Laura’s weekly milk bill: r
2.00
2.40
P(X = r)
0.75
0.25
P(X > μ + σ) = P(X > )
= P(X > 2.27)
= 0.25
(b) P(X < μ − σ) = P(X < 2.10 − 0.17)
= P(X < 1.93)
= 0