Question

The average number of field mice per acre in a 5​-acre wheat field is estimated to be 14. ​(a) Find the probability that fewer than 12 field mice are found on a given acre. ​(b) Find the probability that fewer than 12 field mice are found on 2 of the next 3 acres inspected.

Answer 1

Answer:

(a) $P(X < 12)=0.26$

(b) $P(X=2)=0.15$

Step-by-step explanation:

Question a

This is a Poisson distribution. The average/mean, μ = 14

So, probability that fewer than 12 field mice are found on a given acre is:

$P(X < 12) = e^{-14}(\frac{14^{0}}{0!} +\frac{14^{1}}{1!} + \frac{14^{2}}{2!} + \frac{14^{3}}{3!} +\frac{14^{4}}{4!} + \frac{14^{5}}{5!} +\frac{14^{6}}{6!}+\frac{14^{7}}{7!}+\frac{14^{8}}{8!} +\frac{14^{9}}{9!}+\frac{14^{10}}{10!}+\frac{14^{11}}{11!})\\ \\P(X < 12) = e^{-14}(1+14+98+457.33+1600.67+4481.87+10457.69+20915.38+36601.91+56936.31+79710.83+101450.15)\\\\P(X < 12) = 8.315*10^{-7}(312725.1248)=0.26 \\\\P(X < 12)=0.26$

Question b

This is a Binomial distribution with:

Probability of success, p = 0.26

n = 3

$P(X=2)= (3C2)p^{2}(1-p)=\frac{3!}{2!(3-2)!}*(0.26^{2})*(1-0.26)\\ \\P(X=2)=3(0.0676)(0.74)=0.15\\\\P(X=2)=0.15$