This exercise looks at how well the Poisson distribution approximates the binomial distribution. Assume that n=363n=363 and p=0.003p=0.003 for a binomial distribution. Find the exact (actual) probability (using the binomial probability formula) of one success in 363 trials. (Report answer accurate to 6 decimal places.) P(x=1)= Find the approximate probability of one success in (an interval spanning) 363 trials. (Report answer accurate to 6 decimal places.) P(x=1)= Find the relative difference between the two values. (Use all of the decimal places shown in your calculator from the previous two answers) Recall, the formula approximation–actual actual ×100% approximation-actual actual ×100% (Report answer as a percent accurate to 4 decimal places; you need not type the “%” symbol.) rel diff = %

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This exercise looks at how well the Poisson distribution approximates the binomial distribution. Assume that n=363n=363 and p=0.003p=0.003 for a binomial distribution.   Find the exact (actual) probability (using the binomial probability formula) of one success in 363 trials.  (Report answer accurate to 6 decimal places.)  P(x=1)=    Find the approximate probability of one success in (an interval spanning) 363 trials.  (Report answer accurate to 6 decimal places.)  P(x=1)=    Find the relative difference between the two values.  (Use all of the decimal places shown in your calculator from the previous two answers)   Recall, the formula approximation–actual actual ×100% approximation-actual actual ×100% (Report answer as a percent accurate to 4 decimal places; you need not type the “%” symbol.)   rel diff = %

curwyrm · Accepted Answer

Step 1   Given that n=363 and p=0.003   The probability mass function of Binomial distribution with n=363 and p=0.003 is,  P(X=x)=nCx(p)x(1−p)n−x  P(X=x)=363Cx(0.003)x(1−0.003)363−x .....(I)  a) Compute the exact (actual) probability (using the binomial probability formula) of one success in 363 trials.   That is, P(x=1) .  P(X=x)=nCx(p)x(1−p)n−x  P(X=x)=363Cx(0.003)x(1−0.003)363−x  P(X=1)=363C1(0.003)1(0.997)363−1  (since nCr=n!r!(n−r)!)  P(X=1)=363!(363−1)!1!(0.003)(0.997)362  P(X=1)=363×362!362!1!(0.003)(0.33701351)  P(X=1)=363×0.001011041  P(X=1)=0.36700771  P(X=1)=0.367008 (rounded to six decimals)  Step 2   (b) Approximate the Binomial distribution using Poisson distribution.    Here, λ=n⋅p=363⋅0.003=1.089   The probability mass function of Poisson distribution is,  P(X=x)=e−λλxx!  P(X=x)=e−1.089(1.089)xx!  Step 3   Compute the approximation probability of one success in (an interval spanning) 363 trials.  That is, P(x=1).  P(X=x)=e−λλxx!  P(X=x)=e−1.089(1.089)xx!

This exercise looks at how well the Poisson distribution approximates the binomi

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