Question

As black cats age, they develop some white hairs on their bodies. A particular black car has an average of 1.52 white hairs per square inch on its back.

A) what is the probability that this car will only have three white hairs on a randomly selected one square inch section of its back?

B) what is the probability this cat will have eight white hairs over a randomly selected five square inch section of its back?

C) if C is a random variable representing the number of white hairs this far has on a given one square inch section of its back, what is E[X^2]?

Answer 1

Solution

Let X = Number of white hairs per square inch on the back of the particular cat.

We assume X ~ Poisson (λ), where

λ = average number of white hairs per square inch on its back = 1.52 [given] ………………….. (A)

Back-up Theory

If a random variable X ~ Poisson(λ), i.e., X has Poisson Distribution with mean λ then

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) …………………..………..(1)

where x = 0, 1, 2, ……. , ∞

Values of p(x) for various values of λ and x can be obtained by using Excel Function,

POISSON(x,Mean,Cumulative) ……………………………………………………………………………………………………………… (1a)

Mean = E(X) = λ ......................................................................……………………………………… (2)

Variance = V(X) = λ ……………................................................................………………………… (3)

For any random variable, V(X) = E(X²) – {E(X)}^{2.............}…….……………………………………… (4)

Now to work out the solution,

Part (a)

Probability that this cat will only have three white hairs on a randomly selected one square inch section of its back

= P(X = 3)

= e ^{– 1.52}.(1.52)³/(3!) [vide (1)]

= 0.1280 [vide (1a)] ANSWER

Part (b)

Probability that this cat will have eight white hairs on a randomly selected one square inch section of its back

= P(X = 8)

= e ^{– 1.52}.(1.52)⁸/(8!) [vide (1)]

= 0.000155 [vide (1a)] ANSWER

Part (c)

Vide (4),

E(X²) = V(X) + {E(X)}²

= 1.52 + 1.5² [vide (A), (2) and (3)]

= 3.8304 ANSWER

DONE