A certain type of weapon has probability p of working successfully. We test n we

Question

A certain type of weapon has probability p of working successfully. We test n weapons, and the stockpile is replaced if the number of failures, X, is at least one. How large must n be to have P[X>1] = 0.99 when p=0.95? (a) Use exact binomial (b) Use normal approximation (c) Use Poisson approximation (d) Rework (a) through (c) with p=0.90

Explanation / Answer

P[X>1]=1-(P[X=0]+P[X=1])=0.99..so .P[X=0]+P[X=1])=0.01.;............for binomial distribution P[X=r]=(nCr)(p^r)((1-p)^n-r).;.........for normal distribution mean=np,standard deviation=sqrt(np(1-p)).... it is better to use the table(or you can find the values in this link http://www.mathsisfun.com/data/standard-normal-distribution-table.html).;.for poison distribution P[X=x]=e^(-u).*(u^x)/x!..(where u is the mean).........(a)frpm the above formula P[X=0]=(1-p)^n=0.05^n.;..P[X=1]=n*p*(1-p)^(n-1)=n*0.95*0.05^n-1...so P[X=0]+P[X=1]=(0.05^n-1)(0.05+0.95n).....now here trial values of n should be sustituted so that the above value is equal to 0.01....n=3....(b).mean=0.95p,standard deviation=sqrt(n*0.95*0.05)=0.218sqrt(n)..;......P[X>1]=P[(X-0.95n/0.218sqrt(n))>(1-0.95n/0.218sqrt(n))].=0.99...(by converting to standard variable).from the table Z= (1-0.95n)/0.218sqrt(n)=-2.33...So by solving n=2;....(c)u=np=0.95n;.....P[X=0]=e^-(0.95n);....P[X=1]=e^-(0.95n) * 0.95n....P[X=0]+P[X=1]=e^-(0.95n) (1+0.95n)...again by substituting values for n we get n=7 (d) can be done similarly

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