4.1B – Probability Distribution

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Presentation transcript:

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 Mean Score, x Frequency, f P(x) f/Σf xP(x) 1 24 2 33 42 4 30 5 21 Σ=

Example: Find the Mean Score, x Frequency, f P(x) f/Σf xP(x) 1 24 2 33 42 4 30 5 21 Σ=150

Example: Find the Mean Score, x Frequency, f P(x) f/Σf xP(x) 1 24 24/150=.16 2 33 3 42 4 30 5 21 Σ=150

Example: Find the Mean Score, x Frequency, f P(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 Mean Score, x Frequency, f P(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 Mean Score, x Frequency, f P(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 Mean Score, x Frequency, f P(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 Mean Score, x Frequency, f P(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 Mean Score, x Frequency, f P(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 Mean Score, x Frequency, f P(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 Mean Score, x Frequency, f P(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 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 $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 $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 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 $148 $73 $-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 $248 248(1/1500)=248/1500 $148 $73 $-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 $248 248(1/1500)=248/1500 $148 148(1/1500)=148/1500 $73 $-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 $248 248(1/1500)=248/1500 $148 148(1/1500)=148/1500 $73 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 $248 248(1/1500)=248/1500 $148 148(1/1500)=148/1500 $73 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 $248 248(1/1500)=248/1500 $148 148(1/1500)=148/1500 $73 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 variable σ² = Σ(x-µ)²P(x) OR σ² = [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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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-µ x-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