DISCRETE%20RANDOM%20VARIABLES

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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 Lauras Milk Bill End
presentation
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Car share scheme a success
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Car share scheme a success
Number of people / Outcome r 1 2 3 4 5 gt 5
Relative frequency / Probability P(X r) 0.35 0.375 0.205 0.065 0.005 0
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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
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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
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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
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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
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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
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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
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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
Expectation E(X) or m
E(X) m S r P(X r) 10.35 20.375
30.205 40.065 50.005 0.35 0.75
0.615 0.26 0.025 2
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Variance
r P(X r) r P(X r) r2 P(X r)
1 0.35 0.35 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 1 2 4.86
Var(X) s 2 S r 2 P(X r) m 2
120.35 220.375 320.205 420.065
520.005 22 0.35 1.5 1.845 1.04
0.125 4 4.86 4 0.86
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Example 4.3
The discrete random variable X has the
distribution
r 0 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
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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) 00.2 10.3
20.4 30.1 0 0.3 0.8 0.3 1.4
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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
Expectation E(X2)
E(X2) S r 2 P(X r) 020.2 120.3
220.4 320.1 0 0.3 1.6 0.9 2.8

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Example 4.3 Variance
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
Var(X) s 2 S r 2 P(X r) m 2
020.2 120.3 220.4 320.1 1.42 0
0.3 1.6 0.9 1.42 2.8 1.96
0.84
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Example 4.4 Lauras Milk Bill
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Example 4.4 Lauras 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 gt m s ) and (b) P(X lt
m -s ).
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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
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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
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Example 4.3 (ii) Standard deviation s
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
v0.03 0.17 (to 2 d.p.)
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Example 4.4 (iii) Calculating probabilities
The probability distribution for Lauras weekly
milk bill
r 2.00 2.40
P(X r) 0.75 0.25

• P(X gt µ s) P(X gt 2.10 0.17)
• P(X gt 2.27)
• 0.25

(b) P(X lt µ - s) P(X lt 2.10 - 0.17) P(X
lt 1.93) 0