Statistics 1 Discrete Random Variables Section 2

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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: Laura’s Milk Bill

End presentation

Statistics 1 Discrete Random Variables Section 2

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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 )

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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

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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

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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

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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

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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 0.025

Totals 1

Multiply each r value by P(X = r)

to form ther P(X = r)column

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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 0.025

Totals 1 2

Now find the total of the

r P(X = r)column

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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

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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

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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

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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

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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

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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)

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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

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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

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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 ).

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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 1.502.40 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

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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

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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