Answer to Question #340172 in Statistics and Probability for Maneo

Question

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?

Answer 1

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

Check if$X$ has approximately a normal distribution with $\mu=np$ and $\sigma=\sqrt{npq}.$ .

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

distribution.

Check

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

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

Check

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

Then $\lambda=np=4$

$X\sim Po(4)$

A)

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

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

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

$$=\dfrac{e^{-4}(4)^0}{0!}+\dfrac{e^{-4}(4)^1}{1!}+\dfrac{e^{-4}(4)^2}{2!}$$

$$+\dfrac{e^{-4}(4)^3}{3!}+\dfrac{e^{-4}(4)^4}{4!}+\dfrac{e^{-4}(4)^5}{5!}$$

$$=0.78513$$

B)

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

C)

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

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

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