Math 1710 Class 11 Normal Approximation Accurate Prop CLT Dr. - PowerPoint PPT Presentation

Math 1710 Class 11 V1cu Normal Approximation Making Math 1710 Class 11 Normal Approximation Accurate Prop CLT Dr. Allen Back Failure Picture A Random Variable Problem Sep. 19, 2016 CLT

Suppose 70% approve the President . . . Math 1710 Class 11 V1cu Normal Approximation Making Normal You poll 100 people. Approximation Accurate What is the probability that between 60 and 65 report Prop CLT approval? Failure Picture (including 60 and 65.) A Random Variable Problem CLT

Suppose 70% approve the President . . . Math 1710 Class 11 V1cu Normal Approximation Making You poll 100 people. Normal Approximation What is the probability that between 60 and 65 report Accurate approval? Prop CLT Failure Picture (including 60 and 65.) A Random Solution: Let X = Binomial (100 , . 7) Variable Problem CLT

Suppose 70% approve the President . . . Math 1710 Class 11 V1cu You poll 100 people. Normal What is the probability that between 60 and 65 report Approximation approval? Making Normal (including 60 and 65.) Approximation Accurate Solution: Let X = Binomial (100 , . 7) Prop CLT Failure Picture X has mean A Random Variable µ = (100)( . 7) = 70 Problem CLT and √ � σ = (100)( . 7)( . 3) = 21 = 4 . 58 .

Suppose 70% approve the President . . . Math 1710 Class 11 You poll 100 people. V1cu What is the probability that between 60 and 65 report Normal approval? Approximation (including 60 and 65.) Making Normal Solution: Let X = Binomial (100 , . 7) Approximation Accurate Prop CLT X has mean Failure Picture µ = (100)( . 7) = 70 A Random Variable Problem and √ CLT � σ = (100)( . 7)( . 3) = 21 = 4 . 58 . X will then be approximated by a normal distribution Y with the same mean and standard deviation.

Suppose 70% approve the President . . . Math 1710 You poll 100 people. Class 11 What is the probability that between 60 and 65 report V1cu approval? Normal Approximation (including 60 and 65.) Making Solution: Let X = Binomial (100 , . 7) Normal Approximation Accurate X has mean Prop CLT µ = (100)( . 7) = 70 Failure Picture A Random and Variable √ Problem � σ = (100)( . 7)( . 3) = 21 = 4 . 58 . CLT X will then be approximated by a normal distribution Y with the same mean and standard deviation. Y = N (70 , 4 . 58) .

Suppose 70% approve the President . . . You poll 100 people. Math 1710 Class 11 What is the probability that between 60 and 65 report V1cu approval? Normal (including 60 and 65.) Approximation Solution: Let X = Binomial (100 , . 7) Making Normal Approximation X has mean Accurate Prop CLT µ = (100)( . 7) = 70 Failure Picture A Random and √ Variable � Problem σ = (100)( . 7)( . 3) = 21 = 4 . 58 . CLT X will then be approximated by a normal distribution Y with the same mean and standard deviation. Y = N (70 , 4 . 58) . We can approximate P (60 ≤ X ≤ 65) by P (60 ≤ Y ≤ 65).

Suppose 70% approve the President . . . Math 1710 Class 11 You poll 100 people. V1cu What is the probability that between 60 and 65 report Normal approval? Approximation (including 60 and 65.) Making Normal Approximation Accurate X will then be approximated by a normal distribution Y with Prop CLT the same mean and standard deviation. Failure Picture A Random Y = N (70 , 4 . 58) . Variable Problem We can approximate P (60 ≤ X ≤ 65) by P (60 ≤ Y ≤ 65). CLT The Z score of 60 is − 10 4 . 58 = − 2 . 18 . − 5 The Z score of 65 is 4 . 58 = − 1 . 09 . So P (60 ≤ Y ≤ 65) = P ( Z < − 1 . 09) − P ( Z < − 2 . 18) = . 1379 − . 0146 = . 1233 .

Suppose 70% approve the President . . . Math 1710 Class 11 V1cu You poll 100 people. Normal What is the probability that between 60 and 65 report Approximation approval? Making Normal (including 60 and 65.) Approximation Accurate We can approximate P (60 ≤ X ≤ 65) by P (60 ≤ Y ≤ 65). Prop CLT The Z score of 60 is − 10 4 . 58 = − 2 . 18 . Failure Picture − 5 The Z score of 65 is 4 . 58 = − 1 . 09 . A Random Variable So P (60 ≤ Y ≤ 65) = P ( Z < − 1 . 09) − P ( Z < − 2 . 18) = Problem . 1379 − . 0146 = . 1233 . CLT Actually P (60 ≤ X ≤ 65) = . 15036 . So . 1233 is not that good an approximation.

Suppose 70% approve the President . . . Math 1710 Class 11 You poll 100 people. V1cu What is the probability that between 60 and 65 report Normal Approximation approval? Making (including 60 and 65.) Normal Approximation We can approximate P (60 ≤ X ≤ 65) by P (60 ≤ Y ≤ 65). Accurate The Z score of 60 is − 10 4 . 58 = − 2 . 18 . Prop CLT Failure Picture − 5 The Z score of 65 is 4 . 58 = − 1 . 09 . A Random So P (60 ≤ Y ≤ 65) = P ( Z < − 1 . 09) − P ( Z < − 2 . 18) = Variable Problem . 1379 − . 0146 = . 1233 . CLT Actually P (60 ≤ X ≤ 65) = . 15036 . So . 1233 is not that good an approximation. P (60 < X < 65) = . 09509 . would also be approximated by . 1233!

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation P (60 ≤ X ≤ 65) = . 15036 Accurate P (60 < X < 65) = . 09509 Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu Normal Approximation Making P (60 ≤ X ≤ 65) = . 15036 Normal Approximation P (60 < X < 65) = . 09509 Accurate Prop CLT Failure Picture P (60 ≤ Y ≤ 65) = . 1233 A Random is not that good an approximation to either. Variable Problem CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu Normal P (60 ≤ X ≤ 65) = . 15036 Approximation P (60 < X < 65) = . 09509 Making Normal Approximation Accurate P (60 ≤ Y ≤ 65) = . 1233 Prop CLT is not that good an approximation to either. Failure Picture A Random The problem is that for a continuous model like Y , Variable Problem P ( Y = 65) = 0 . CLT But P ( X = 65) = . 04678 .

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu Normal Approximation Making P (60 ≤ X ≤ 65) = . 15036 Normal Approximation P (60 < X < 65) = . 09509 Accurate Prop CLT Failure Picture P (60 ≤ Y ≤ 65) = . 1233 A Random So Y doesn’t care about < vs. ≤ . But X does! Variable Problem CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 P (60 ≤ X ≤ 65) = . 15036 Class 11 P (60 < X < 65) = . 09509 V1cu P (60 ≤ Y ≤ 65) = . 1233 Normal Approximation Looking closely at the picture provides the key: Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 P (60 ≤ X ≤ 65) = . 15036 V1cu P (60 < X < 65) = . 09509 Normal P (60 ≤ Y ≤ 65) = . 1233 Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . P (60 ≤ X ≤ 65) = . 15036 Math 1710 Class 11 P (60 < X < 65) = . 09509 V1cu P (60 ≤ Y ≤ 65) = . 1233 Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT It is really the area under the normal curve between 64 . 5 and 65 . 5 which approximates P(X=65).

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu Normal Approximation Making Normal P (60 ≤ X ≤ 65) = . 15036 Approximation Accurate P (60 < X < 65) = . 09509 Prop CLT Failure Picture P (60 ≤ Y ≤ 65) = . 1233 A Random Variable Problem CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu Normal Approximation Making P (60 ≤ X ≤ 65) = . 15036 Normal Approximation P (60 < X < 65) = . 09509 Accurate P (60 ≤ Y ≤ 65) = . 1233 Prop CLT Failure Picture So P (60 ≤ X ≤ 65) = . 15036 should be approximated by A Random P (59 . 5 ≤ Y ≤ 65 . 5) . Variable Problem CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu Normal Approximation P (60 ≤ X ≤ 65) = . 15036 Making P (60 < X < 65) = . 09509 Normal Approximation P (60 ≤ Y ≤ 65) = . 1233 Accurate So P (60 ≤ X ≤ 65) = . 15036 should be approximated by Prop CLT Failure Picture P (59 . 5 ≤ Y ≤ 65 . 5) . A Random The z-scores of 59 . 5 and 65 . 5 are Variable Problem − 10 . 5 4 . 58 = − 2 . 29 and − 4 . 5 4 . 58 = − . 98 resp. CLT

X = Binomial (100 , . 7) and Y = N (70 , 4 . 58) . Math 1710 Class 11 V1cu P (60 ≤ X ≤ 65) = . 15036 Normal Approximation P (60 < X < 65) = . 09509 Making P (60 ≤ Y ≤ 65) = . 1233 Normal Approximation So P (60 ≤ X ≤ 65) = . 15036 should be approximated by Accurate P (59 . 5 ≤ Y ≤ 65 . 5) . Prop CLT Failure Picture The z-scores of 59 . 5 and 65 . 5 are A Random − 10 . 5 4 . 58 = − 2 . 29 and − 4 . 5 4 . 58 = − . 98 resp. Variable Problem So P (59 . 5 ≤ Y ≤ 65 . 5) = P ( − 2 . 29 ≤ Z ≤ − . 98) = CLT . 1635 − . 0110 = . 1525 . Much closer to P (60 ≤ X ≤ 65) = . 15036!