Answer to Question #173481 in Statistics and Probability for ANJU JAYACHANDRAN

Question #173481

9. (a) Let X be a binomial variate with n =100, p = .1.0 Find the approximate value of

P 10( ≤ X ≤12) using: (5) 

 (i) normal distribution 

 (ii)poisson distribution 

 [You may like to use the following values. 

 P(Z ≤ 67.0 ) = .0 7486, P(Z ≤ 33.0 ) = .0 6293, P(Z ≤ )0 = ]5.0

 


1
Expert's answer
2021-03-29T15:43:01-0400

i)

μ=np=1000.1=10\mu=np=100\cdot0.1=10

σ=np(1p)=1000.1(10.1)=3\sigma=\sqrt{np(1-p)}=\sqrt{100\cdot0.1(1-0.1)}=3

P(10X12)P(9.5Y12.5)=P(10 \leq X \leq12)\approx P(9.5 \leq Y \leq12.5)=

=P(9.5103Z12.5103)=P(0.17Z0.83)==P(\frac{9.5-10}{3} \leq Z\leq\frac{12.5-10}{3})=P(-0.17\leq Z\leq0.83)=

=0.79670.4325=0.3642=0.7967-0.4325=0.3642


ii)

λ=np=10\lambda=np=10

P(X=k)=λkeλk!P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}

P(10X12)=P(X=10)+P(X=11)+P(X=12)P(10 \leq X \leq12)=P(X=10)+P(X=11)+P(X=12)

P(X=10)=1010e1010!=0.1251P(X=10)=\frac{10^{10}e^{-10}}{10!}=0.1251

P(X=11)=1011e1011!=0.1137P(X=11)=\frac{10^{11}e^{-10}}{11!}=0.1137

P(X=12)=1012e1012!=0.0948P(X=12)=\frac{10^{12}e^{-10}}{12!}=0.0948

P(10X12)=0.1251+0.1137+0.0948=0.3336P(10 \leq X \leq12)=0.1251+0.1137+0.0948=0.3336


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