1) The document discusses discrete random variables through examples and calculations of expectation, variance, and probability. 2) It provides the probability distribution for a car share scheme and calculates the expectation of 2. It also gives an example of a discrete random variable and calculates its expectation, variance using two methods. 3) The final example gives Laura's probability distribution for her weekly milk bill and calculates the mean, standard deviation, and probabilities.

1. Statistics 1 Discrete Random Variables Section 2
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. Statistics 1 Discrete Random Variables Section 2
2
Car share scheme a success
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 0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 1 2 3 4 5 6
r
P(X = r )

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

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

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

6. Statistics 1 Discrete Random Variables Section 2
6
Expectation
r P(X = r) r P(X = r)
1 0.35 0.35
2 0.375 0.75
3 0.205 0.615
4 0.065 0.26
5 0.005
Totals 1
Multiply each r
value by P(X = r)
to form the
r P(X = r)
column

7. Statistics 1 Discrete Random Variables Section 2
7
Expectation
r P(X = r) r P(X = r)
1 0.35 0.35
2 0.375 0.75
3 0.205 0.615
4 0.065 0.26
5 0.005 0.025
Totals 1
Multiply each r
value by P(X = r)
to form the
r P(X = r)
column

8. Statistics 1 Discrete Random Variables Section 2
8
Expectation
r P(X = r) r P(X = r)
1 0.35 0.35
2 0.375 0.75
3 0.205 0.615
4 0.065 0.26
5 0.005 0.025
Totals 1 2
Now find the total
of the
r P(X = r)
column

9. Statistics 1 Discrete Random Variables Section 2
9
Expectation
E(X) = m = S r P(X = r)
= 1×0.35 + 2×0.375 + 3×0.205 + 4×0.065 + 5×0.005
= 0.35 + 0.75 + 0.615 + 0.26 + 0.025
= 2
r P(X = r) r P(X = r)
1 0.35 0.35
2 0.375 0.75
3 0.205 0.615
4 0.065 0.26
5 0.005 0.025
Totals 1 2
Expectation
= E(X) or m

10. Statistics 1 Discrete Random Variables Section 2
10
Var(X) = s 2 = S r 2 P(X = r) – m 2
= 12×0.35 + 22×0.375 + 32×0.205 + 42×0.065 + 52×0.005 – 22
= 0.35 + 1.5 + 1.845 + 1.04 + 0.125 – 4
= 4.86 – 4 = 0.86
Variance – Method A
(a) (b)
r P(X = r) r P(X = r) r2 P(X = r) (r – m)2 P(X = r)
1 0.35 0.35 0.35 0.35
2 0.375 0.75 1.5 0
3 0.205 0.615 1.845 0.205
4 0.065 0.26 1.04 0.26
5 0.005 0.025 0.125 0.045
Totals 1 2 4.86 0.86

11. Statistics 1 Discrete Random Variables Section 2
11
Variance – Method B
(a) (b)
r P(X = r) r P(X = r) r2 P(X = r) (r – m)2 P(X = r)
1 0.35 0.35 0.35 0.35
2 0.375 0.75 1.5 0
3 0.205 0.615 1.845 0.205
4 0.065 0.26 1.04 0.26
5 0.005 0.025 0.125 0.045
Totals 1 2 4.86 0.86
Var(X) = s 2 = S (r – m)2 P(X = r)
= (1 – 2)2×0.35 + (2 – 2)2×0.375 + (3 – 2)2×0.205
+ (4 – 2)2×0.065 + (5 – 2)2×0.005
= 0.35 + 0 + 0.205 + 0.26 + 0.045 = 0.86

12. Statistics 1 Discrete Random Variables Section 2
12
Example 4.3
The discrete random variable X has the distribution:
(i) Find E(X).
(ii) Find E(X2).
(iii) Find Var(X) using
(a) E(X2) – m2 and (b) E([X – m]2) .
r 0 1 2 3
P(X = r) 0.2 0.3 0.4 0.1

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

14. Statistics 1 Discrete Random Variables Section 2
14
Example 4.3 : (ii) E(X 2)
r P(X = r) r P(X = r) r2 P(X = r)
0 0.2 0 0
1 0.3 0.3 0.3
2 0.4 0.8 1.6
3 0.1 0.3 0.9
Totals 1 1.4 2.8
E(X2) = S r 2 P(X = r)
= 02×0.2 + 12×0.3 + 22×0.4 + 32×0.1
= 0 + 0.3 + 1.6 + 0.9
= 2.8
Expectation
= E(X2)

15. Statistics 1 Discrete Random Variables Section 2
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Example 4.3 : Variance – Method A
(a) (b)
r P(X = r) r P(X = r) r2 P(X = r) (r – m)2 P(X = r)
0 0.2 0 0 0.392
1 0.3 0.3 0.3 0.048
2 0.4 0.8 1.6 0.144
3 0.1 0.3 0.9 0.256
Totals 1 1.4 2.8 0.84
Var(X) = s 2 = S r 2 P(X = r) – m 2
= 02×0.2 + 12×0.3 + 22×0.4 + 32×0.1 – 1.42
= 0 + 0.3 + 1.6 + 0.9 + 0.125 – 1.42
= 2.8 – 1.96 = 0.84

16. Statistics 1 Discrete Random Variables Section 2
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Example 4.3 : Variance – Method B
(a) (b)
r P(X = r) r P(X = r) r2 P(X = r) (r – m)2 P(X = r)
0 0.2 0 0 0.392
1 0.3 0.3 0.3 0.048
2 0.4 0.8 1.6 0.144
3 0.1 0.3 0.9 0.256
Totals 1 1.4 2.8 0.84
Var(X) = s 2 = S (r – m)2 P(X = r)
= (0 – 1.4)2×0.2 + (1 – 1.4)2×0.3 +
(2 – 1.4)2×0.4 + (3 – 1.4)2×0.3
= 0.392 + 0.048 + 0.144 + 0.256 = 0.84

17. Statistics 1 Discrete Random Variables Section 2
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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. Statistics 1 Discrete Random Variables Section 2
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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
       5 £2.00
       5 £2.00
r 2.00 2.40
P(X = r) 0.75 0.25

19. Statistics 1 Discrete Random Variables Section 2
19
Example 4.4 : (i) Mean μ or expectation E(X)
r 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)
= 2.00 × 0.75 + 2.40 × 0.25
= 1.50 + 0.60
= 2.10

20. Statistics 1 Discrete Random Variables Section 2
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Example 4.3 : (ii) Standard deviation σ
(a)
r 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 × 0.75 + 2.42 × 0.25 – 2.12
= 3.00 + 1.44 – 2.12
= 4.44 – 4.41 = 0.03
Hence s = √0.03 = 0.17 (to 2 d.p.)
Using
Method A

21. Statistics 1 Discrete Random Variables Section 2
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Example 4.4 : (iii) Calculating probabilities
r 2.00 2.40
P(X = r) 0.75 0.25
The probability distribution for Laura’s weekly milk bill:
(a) P(X > μ + σ) = P(X > 2.10 + 0.17)
= P(X > 2.27)
= 0.25
(b) P(X < μ − σ) = P(X < 2.10 − 0.17)
= P(X < 1.93)
= 0