This is for my stats class

Question

A wildlife biologist observes four brown bears as they attempt to catch salmon in a river in Alaska. Assume that each bear has a .7 probability of catching a fish and that the bears fishing attempts are independent. Let the random variable X be the number of bears that catch a fish. Create the probability distribution for X (rounding each to THREE decimal places).

What is the mean of the above distribution?

Answer 1

This is an example of a binomial distribution.

n = 4, p = 0.7, q = 1 - p = 0.3

 

General prob dist: P(X = x) = 0.7^x * 0.3^4-x * ⁴C_x =  0.7^x * 0.3^4-x * 4!/(x!(4-x)!)

where P(x = x) is the probability that x bears catch a fish. x could be 0,1,2,3 or 4.

 

P(X = 0) = 0.7⁰ * 0.3⁴ * ⁴C₀ = 0.0081

 

P(X = 1) = 0.7¹ * 0.3³ * ⁴C₁ = 0.7 * 0.027 * 4 = 0.0756

 

P(X = 2) = 0.7² * 0.3² * ⁴C₂ = 0.49 * 0.09 * 6 = 0.2646

 

P(X = 3) = 0.7³ * 0.3¹ * ⁴C₃ = 0.343 * 0.3 * 4 = 0.4116

 

P(X = 4) = 0.7⁴ * 0.3⁰ * ⁴C₄ = 0.2401

 

Mean of a binomial distribution is : np = 4*0.7 = 2.8