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

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

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

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)((14^(0))/(0!) +(14^(1))/(1!) + (14^(2))/(2!) + (14^(3))/(3!) +(14^(4))/(4!) + (14^(5))/(5!) +(14^(6))/(6!)+(14^(7))/(7!)+(14^(8))/(8!) +(14^(9))/(9!)+(14^(10))/(10!)+(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)=(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`