Answer to Question #340172 in Statistics and Probability for Maneo

Question #340172

Suppose that the probability that a corn seed from a certain batch does not germinate equals 0.02. If we plant 200 of these seeds, what is the probability that

• A) At most 5 seeds will not germinate?

• B) Exactly 3 will not germinate?

• C) At least 3 will not germinate?



1
Expert's answer
2022-06-01T13:08:25-0400

XBin(n=200,p=0.02)X\sim Bin(n=200,p=0.02)

Check ifXX has approximately a normal distribution with μ=np\mu=np and σ=npq.\sigma=\sqrt{npq}. .

In practice, the approximation is adequate provided that both np10np\ge10 and nq10,nq\ge10, since there is then enough symmetry in the underlying binomial

distribution.

Check


np=200(0.02)=4<10np=200(0.02)=4<10

When the value of nn  in a binomial distribution is large and the value of pp  is very small, the binomial distribution can be approximated by a Poisson distribution. If n>20n > 20 and np<5np < 5 or nq<5nq < 5 then the Poisson is a good approximation.

Check


n=200>20,np=200(0.02)=4<10n=200>20, np=200(0.02)=4<10


Then λ=np=4\lambda=np=4

XPo(4)X\sim Po(4)


A)


P(X5)=P(X=0)+P(X=1)P(X\le5)=P(X=0)+P(X=1)

+P(X=2)+P(X=3)+P(X=2)+P(X=3)

+P(X=4)+P(X=5)+P(X=4)+P(X=5)

=e4(4)00!+e4(4)11!+e4(4)22!=\dfrac{e^{-4}(4)^0}{0!}+\dfrac{e^{-4}(4)^1}{1!}+\dfrac{e^{-4}(4)^2}{2!}

+e4(4)33!+e4(4)44!+e4(4)55!+\dfrac{e^{-4}(4)^3}{3!}+\dfrac{e^{-4}(4)^4}{4!}+\dfrac{e^{-4}(4)^5}{5!}

=0.78513=0.78513

B)


P(X=3)=e4(4)33!=0.19537P(X=3)=\dfrac{e^{-4}(4)^3}{3!}=0.19537

C)


P(X3)=1P(X=0)P(X=1)P(X\ge3)=1-P(X=0)-P(X=1)

P(X=2)=1e4(4)00!e4(4)11!-P(X=2)=1-\dfrac{e^{-4}(4)^0}{0!}-\dfrac{e^{-4}(4)^1}{1!}

e4(4)22!=0.76190-\dfrac{e^{-4}(4)^2}{2!}=0.76190


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment