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
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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
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 )
Statistics 1 Discrete Random Variables Section 2
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Expectation
r P(X = r) r P(X = r)1 0.35 0.352 0.3753 0.2054 0.0655 0.005
Totals 1
Multiply each r value by P(X = r)
to form ther P(X = r)column
Statistics 1 Discrete Random Variables Section 2
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Expectation
r P(X = r) r P(X = r)1 0.35 0.352 0.375 0.753 0.2054 0.0655 0.005
Totals 1
Multiply each r value by P(X = r)
to form ther P(X = r)column
Statistics 1 Discrete Random Variables Section 2
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Expectation
r P(X = r) r P(X = r)1 0.35 0.352 0.375 0.753 0.205 0.6154 0.0655 0.005
Totals 1
Multiply each r value by P(X = r)
to form ther P(X = r)column
Statistics 1 Discrete Random Variables Section 2
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Expectation
r P(X = r) r P(X = r)1 0.35 0.352 0.375 0.753 0.205 0.6154 0.065 0.265 0.005
Totals 1
Multiply each r value by P(X = r)
to form ther P(X = r)column
Statistics 1 Discrete Random Variables Section 2
7
Expectation
r P(X = r) r P(X = r)1 0.35 0.352 0.375 0.753 0.205 0.6154 0.065 0.265 0.005 0.025
Totals 1
Multiply each r value by P(X = r)
to form ther P(X = r)column
Statistics 1 Discrete Random Variables Section 2
8
Expectation
r P(X = r) r P(X = r)1 0.35 0.352 0.375 0.753 0.205 0.6154 0.065 0.265 0.005 0.025
Totals 1 2
Now find the total of the
r P(X = r)column
Statistics 1 Discrete Random Variables Section 2
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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 0.352 0.375 0.753 0.205 0.6154 0.065 0.265 0.005 0.025
Totals 1 2
Expectation= E(X) or m
Statistics 1 Discrete Random Variables Section 2
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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 0.352 0.375 0.75 1.5 03 0.205 0.615 1.845 0.2054 0.065 0.26 1.04 0.265 0.005 0.025 0.125 0.045
Totals 1 2 4.86 0.86
Statistics 1 Discrete Random Variables Section 2
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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 0.352 0.375 0.75 1.5 03 0.205 0.615 1.845 0.2054 0.065 0.26 1.04 0.265 0.005 0.025 0.125 0.045
Totals 1 2 4.86 0.86Var(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
Statistics 1 Discrete Random Variables Section 2
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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
Statistics 1 Discrete Random Variables Section 2
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Example 4.3 : (i) Expectation E(X)
r P(X = r) r P(X = r)0 0.2 01 0.3 0.32 0.4 0.83 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
Statistics 1 Discrete Random Variables Section 2
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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 01 0.3 0.3 0.32 0.4 0.8 1.63 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)
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 0.3921 0.3 0.3 0.3 0.0482 0.4 0.8 1.6 0.1443 0.1 0.3 0.9 0.256
Totals 1 1.4 2.8 0.84Var(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
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 0.3921 0.3 0.3 0.3 0.0482 0.4 0.8 1.6 0.1443 0.1 0.3 0.9 0.256
Totals 1 1.4 2.8 0.84Var(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
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 ).
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
Statistics 1 Discrete Random Variables Section 2
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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 1.502.40 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
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 3.002.40 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
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
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