Question

An automobile manufacturer finds that 1 in every 2000 automobiles produced has a particular manufacturing defect. (a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 5000 cars (b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p. Repeat (a) using a Poisson distribution and compare the results (a) The probability using the binomial distribution is (Round to five decimal places as needed.) (b) The probability using the Poisson distribution is (Round to five decimal places as needed.) Compare the results. Choose the correct answer below. O A. The probability calculated using the binomial distribution is much larger. O B. The two probabilities are approximately the same. O C. The probability calculated using the Poisson distribution is much larger.

Answer 1

Solution : ( a )

$\mathrm{p}(x) = \mathrm{nC}_{x}*\mathrm{p}^{x} * \mathrm{q}^{(\mathrm{n}-x)}$

$\mathrm{Probability\:of\:finding\:4\:cars\:with\:the\:defect\:in\:a\:random\:sample\:of\:5000\:cars\::}$

$\mathrm{p}(4) = \mathrm{5000C}_{4}*\mathrm{\left ( \frac{1}{2000} \right )}^{4} * \mathrm{\left ( 1-\frac{1}{2000} \right )}^{(\mathrm{5000}-4)}$

$= \frac{5000!}{4!*(5000-4)!}*\mathrm{\left ( \frac{1}{2000} \right )}^{4} * \mathrm{\left ( 1-\frac{1}{2000} \right )}^{(\mathrm{5000}-4)}$

$= \frac{5000!}{4!*4996!}*\mathrm{\left ( \frac{1}{2000} \right )}^{4} * \mathrm{\left ( \frac{1999}{2000} \right )}^{4996}$

$= \frac{5000*4999*4998*4997*4996!}{4!*4996!}*\mathrm{\left ( \frac{1}{2000} \right )}^{4} * \mathrm{\left ( \frac{1999}{2000} \right )}^{4996}$

$= \frac{5000*4999*4998*4997}{4!}*\mathrm{\left ( \frac{1}{2000} \right )}^{4} * \mathrm{\left ( \frac{1999}{2000} \right )}^{4996}$

$= \frac{5000*4999*4998*4997}{4*3*2*1}*\mathrm{\left ( \frac{1}{2000} \right )}^{4} * \mathrm{\left ( \frac{1999}{2000} \right )}^{4996}$

$= \frac{5000*4999*4998*4997}{24}*\mathrm{\left ( \frac{1}{2000} \right )}^{4} * \mathrm{\left ( \frac{1999}{2000} \right )}^{4996}$

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Solution : ( b )

$\mu =\mathrm{n*p=5000*\left ( \frac{1}{2000} \right )=2.5}$

$\mathrm{Using\:a\:Poisson\:distribution}$

$e^{-\mu }\frac{\mu ^{x}}{x!}$

$\mathrm{Here\:\:\mu =2.5\:\:and\:\:}x=4;$

$=e^{-2.5 }*\frac{(2.5) ^{4}}{4!}$

$=e^{-2.5 }*\frac{(2.5) ^{4}}{4*3*2*1}$

$=e^{-2.5 }*\frac{(2.5) ^{4}}{24}$

$=\frac{1}{e^{2.5 }}*\frac{(2.5) ^{4}}{24}$

$=\frac{1}{e^{2.5 }}*\frac{39.0625}{24}$

$=\frac{39.0625}{24e^{2.5}}$

$=\frac{39.0625}{24*12.18249}$

$=\frac{39.0625}{292.37976}$

$=0.13360$

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Solution :

$\mathrm{p=0.78\quad\:and\quad\:n=6}$

( a )

$\mathrm{The\:mean\:of\:the\:binomial\:distribution\:\:(\mu )=\:n*p\:=\:6*0.78=4.68\approx 4.7}$

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( b )

$\mathrm{The\:variance\:of\:the\:binomial\:distribution\:\:(\sigma ^{2})\:=\:n*p*(1-p)\:=\:6*0.78*(1-0.78)=6*0.78*0.22=1.0296\approx 1.0}$

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( c )

$\mathrm{The\:standard-deviation\:of\:the\:binomial\:distribution\:\:(\sigma )\:=\:\sqrt{n*p*(1-p)}\:=\:\sqrt{6*0.78*(1-0.78)}=\sqrt{6*0.78*0.22}=\sqrt{1.0296}=1.01469\approx 1.0}$