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Normal Distribution - Find the Area Property in Normal Curve

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Normal Distribution - Find the Area Property in Normal Curve

1. 1. Normal Distribution –Find the area property SUNDARA B. N. Assistant Professor
2. 2. Problem If the random variable X is normally distributed with mean 80 and standard deviation 5, then find a) P(X>95) b) P(X<72) c) P(60.5<X<90) d) P(85<X<97) and e) P(64<X<76)
3. 3. Given Information Here we are given that μ = 80 σ = 5  X~N(80,25). We know that is X~N(μ, σ²),then the S.N.V is given by Z = X – μ σ
4. 4. a) P(X>95) Consider X = 95  P(X>95)=P(Z>3) = 0.5-P(0<Z<3) = 0.5- 0.4987(TV) =0.0013 X=9 5 Z=3 X=8 0 Z=0
5. 5. b) P(X<72) Consider X = 72  P(X<72)=P(Z<-1.6)  = P(Z<1.6) = 0.5- P(0<Z<1.6) = 0.5- 0.4452(TV) =0.0548 X=8 0 Z=0 X=7 2 Z=- 1.6
6. 6. c) P(60.5<X<90) Consider X = 60.5 and X = 90  P(60.5<X<90)=P(- 3.9<X<2) = P(-3.9<X<0)+P(0<Z<2) = P(0<X<3.9)+P(0<Z<2) = 0.5000+0.4772 (TV) =0.9772 X=8 0 Z=0 X=9 0 Z=2 X=60.5 Z=- 3.9
7. 7. d) P(85<X<97 Consider X = 85 and X = 97  P(85<X<97)=P(1<X<3. 4) = P(0<Z<3.4)-P(0<Z<1) = 0.4997-0.34139(TV) =0.1584 X=8 0 Z=0 X=8 5 Z=1 X=9 7 Z=3. 4
8. 8. e) P(64<X<76) Consider X = 64 and X = 76  P(64<X<76)=P(-3.2<Z<- 0.8) = P(0.8<Z<3.2) = P(0<Z<3.2)-P(0<Z<0.8) = 0.4993-0.2881(TV) =0.2112 X=8 0 Z=0 X=7 6 Z=- 0.8 Z=- 3.2 X=6 4