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4.1B – Probability Distribution MEAN of discrete random variable: µ = ΣxP(x) • EACH x is multiplied by its probability and the products are added. • µ = EXPECTED VALUE of discrete random variables
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4.1B – Probability Distribution. MEAN of discrete random variable: µ = Σ xP (x) EACH x is multiplied by its probability and the products are added. µ = EXPECTED VALUE of discrete random variables. Example: Find the Mean. Example: Find the Mean. Example: Find the Mean. - PowerPoint PPT Presentation

### Transcript of 4.1B – Probability Distribution

4.1B – Probability Distribution

• MEAN of discrete random variable:

µ = ΣxP(x)

• EACH x is multiplied by its probability and the products are added.

• µ = EXPECTED VALUE of discrete random variables

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24

2 33

3 42

4 30

5 21

Σ=

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24

2 33

3 42

4 30

5 21

Σ=150

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16

2 33

3 42

4 30

5 21

Σ=150

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16

2 33 .22

3 42 .28

4 30 .2

5 21 .14

Σ=150

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16

2 33 .22

3 42 .28

4 30 .2

5 21 .14

Σ=150 Σ=1.0

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16 1(.16)=.16

2 33 .22

3 42 .28

4 30 .2

5 21 .14

Σ=150 Σ=1.0

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16 1(.16)=.16

2 33 .22 2(.22)=.44

3 42 .28

4 30 .2

5 21 .14

Σ=150 Σ=1.0

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16 1(.16)=.16

2 33 .22 2(.22)=.44

3 42 .28 3(.28)=.84

4 30 .2

5 21 .14

Σ=150 Σ=1.0

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16 1(.16)=.16

2 33 .22 2(.22)=.44

3 42 .28 3(.28)=.84

4 30 .2 4(.2)=.80

5 21 .14

Σ=150 Σ=1.0

Example: Find the MeanScore,

xFrequency,

fP(x)f/Σf

xP(x)

1 24 24/150=.16 1(.16)=.16

2 33 .22 2(.22)=.44

3 42 .28 3(.28)=.84

4 30 .2 4(.2)=.80

5 21 .14 5(.14)=.70

Σ=150 Σ=1.0

Example: Find the MeanScore,

xFrequency, f P(x)

f/ΣfxP(x)

1 24 24/150=.16 1(.16)=.16

2 33 .22 2(.22)=.44

3 42 .28 3(.28)=.84

4 30 .2 4(.2)=.80

5 21 .14 5(.14)=.70

Σ=150 Σ=1.0 ΣxP(x)=2.94 = µ

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498

\$248

\$148

\$73

\$-2

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500

\$248

\$148

\$73

\$-2

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500

\$248 1/1500

\$148

\$73

\$-2

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500

\$248 1/1500

\$148 1/1500

\$73

\$-2

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500

\$248 1/1500

\$148 1/1500

\$73 1/1500

\$-2

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500

\$248 1/1500

\$148 1/1500

\$73 1/1500

\$-2 1496/1500

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500 498(1/1500)=498/1500

\$248 1/1500

\$148 1/1500

\$73 1/1500

\$-2 1496/1500

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500 498(1/1500)=498/1500

\$248 1/1500 248(1/1500)=248/1500

\$148 1/1500

\$73 1/1500

\$-2 1496/1500

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500 498(1/1500)=498/1500

\$248 1/1500 248(1/1500)=248/1500

\$148 1/1500 148(1/1500)=148/1500

\$73 1/1500

\$-2 1496/1500

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500 498(1/1500)=498/1500

\$248 1/1500 248(1/1500)=248/1500

\$148 1/1500 148(1/1500)=148/1500

\$73 1/1500 73(1/1500)=73/1500

\$-2 1496/1500

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500 498(1/1500)=498/1500

\$248 1/1500 248(1/1500)=248/1500

\$148 1/1500 148(1/1500)=148/1500

\$73 1/1500 73(1/1500)=73/1500

\$-2 1496/1500 -2(1/1500)=-2992/1500

Prize-\$2 EV = µ=ΣxP(x)

Example: Find the EXPECTED VALUE of your gain.

• 1500 tickets are sold for \$2 each for 4 prizes of \$500, \$250, \$150, and \$75. You buy 1 ticket. What is the expected value of your gain?Gain, x P(x) xP(x)

\$498 1/1500 498(1/1500)=498/1500

\$248 1/1500 248(1/1500)=248/1500

\$148 1/1500 148(1/1500)=148/1500

\$73 1/1500 73(1/1500)=73/1500

\$-2 1496/1500 -2(1496/1500)=-2992/1500

Prize-\$2 EV = µ=ΣxP(x)=-2025/1500 = -\$1.35

Standard Deviation

• VARIANCE of discrete random variableσ² = Σ(x-µ)²P(x) ORσ² = [Σx²P(x)] - µ²

• STANDARD DEVIATION of discrete random variable

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16

2 33 .22 2(.22)=.44

3 42 .28 .84

4 30 .20 .80

5 21 .14 .70

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

2 33 .22 2(.22)=.44

3 42 .28 .84

4 30 .20 .80

5 21 .14 .70

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

2 33 .22 2(.22)=.44 2-2.94=-.94

3 42 .28 .84

4 30 .20 .80

5 21 .14 .70

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

2 33 .22 2(.22)=.44 2-2.94=-.94

3 42 .28 .84 .06

4 30 .20 .80

5 21 .14 .70

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

2 33 .22 2(.22)=.44 2-2.94=-.94

3 42 .28 .84 .06

4 30 .20 .80 1.06

5 21 .14 .70

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

2 33 .22 2(.22)=.44 2-2.94=-.94

3 42 .28 .84 .06

4 30 .20 .80 1.06

5 21 .14 .70 2.06

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

2 33 .22 2(.22)=.44 2-2.94=-.94

3 42 .28 .84 .06

4 30 .20 .80 1.06

5 21 .14 .70 2.06

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

3 42 .28 .84 .06

4 30 .20 .80 1.06

5 21 .14 .70 2.06

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

3 42 .28 .84 .06 .004

4 30 .20 .80 1.06

5 21 .14 .70 2.06

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

3 42 .28 .84 .06 .004

4 30 .20 .80 1.06 1.124

5 21 .14 .70 2.06

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

3 42 .28 .84 .06 .004

4 30 .20 .80 1.06 1.124

5 21 .14 .70 2.06 4.244

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

.16(3.764)=.602

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

3 42 .28 .84 .06 .004

4 30 .20 .80 1.06 1.124

5 21 .14 .70 2.06 4.244

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

.16(3.764)=.602

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

.22(.884)=.194

3 42 .28 .84 .06 .004

4 30 .20 .80 1.06 1.124

5 21 .14 .70 2.06 4.244

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

.16(3.764)=.602

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

.22(.884)=.194

3 42 .28 .84 .06 .004 .001

4 30 .20 .80 1.06 1.124

5 21 .14 .70 2.06 4.244

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

.16(3.764)=.602

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

.22(.884)=.194

3 42 .28 .84 .06 .004 .001

4 30 .20 .80 1.06 1.124 .225

5 21 .14 .70 2.06 4.244

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

.16(3.764)=.602

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

.22(.884)=.194

3 42 .28 .84 .06 .004 .001

4 30 .20 .80 1.06 1.124 .225

5 21 .14 .70 2.06 4.244 .594

Σ=150 Σ=1.0 Σ=2.94=µ

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

.16(3.764)=.602

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

.22(.884)=.194

3 42 .28 .84 .06 .004 .001

4 30 .20 .80 1.06 1.124 .225

5 21 .14 .70 2.06 4.244 .594

Σ=150 Σ=1.0 Σ=2.94=µ

Σ=1.616

σ = √σ²

Example: Find Variance & Standard Deviation

Score, x

Freq.f

P(x)f/Σf

xP(x) x-µx-ΣxP(x)

(x-µ)² P(x)(x-µ)²

1 24 24/150=.16 1(.16)=.16 1-2.94=-1.94

(-1.94)²=3.764

.16(3.764)=.602

2 33 .22 2(.22)=.44 2-2.94=-.94

(-.94)²=.884

.22(.884)=.194

3 42 .28 .84 .06 .004 .001

4 30 .20 .80 1.06 1.124 .225

5 21 .14 .70 2.06 4.244 .594

Σ=150 Σ=1.0 Σ=2.94=µ

Σ=1.616

σ = √σ² = √1.616 = 1.27