Question

1)

Population (parametric) mean= 53.501

Standard deviation = 1.79208

Imagine that 5 individuals are sampled at random from this population. Calculate the probability that the average calculated will be less than the value: (population - (st.dev/3))

2) The average number of squirrels you come across on campus is 4.1. What is the probability of observing at most three squirrels? Assume that squirrels are randomly distributed and that each squirrel is an independent observation.

3) 15% of a specific population of rainbow trout carry intestinal parasites. Assume 9 random samples are obtained from this population.

a) calculate the probability that 1 carry intestinal parasites.

b) calculate the probability that at least 2 individuals carry intestinal parasites.

Please show/explain equations used.

Answer 1

(1)

Population (parametric) mean= 53.501

Standard deviation = 1.79208

sample size n = 5

the value = mean - standard deviation/3 = 53.501 - 1.79208/3 = 52.9036

Here standard error of sample mean = 1.79208/sqrt(5) = 0.80144

Pr( < 52.9036 ; 53.501 ; 0.80144)

Z = (52.9036 - 53.501)/0.80144 = -0.7454

Pr( < 52.9036 ; 53.501 ; 0.80144) = Pr(Z < -.07454) = 0.2280

Question 2

Here the distributionis poission distribution with parameter = 4.1

Pr(x <= 3) = POISSON(x <= 3 ; 4.1) = 0.4142

Question 3

Here the distribution is binomial one where n = 9 and p = 0.15

(a) If x is the number of rainbow trout that carry intestinal parasites

Pr(x = 1) = ⁹C₁ (0.15)¹ (0.85)⁸ = 0.3679

(b) Pr(x >= 2) = 1 - Pr(x = 0) - Pr(x = 1) = 1 - ⁹C₀ (0.15)⁰ (0.85)⁹ - ⁹C₁ (0.15)¹ (0.85)⁸

= 1 - 0.2316 - 0.3679 = 0.4005