Question

Hi!

I've been stuck with this question this whole day, not being able to get find out how to solve it.

The question follows "In a zoo lives a hyena who every day eats a certain number of packs of hyena food. The probability that the hyena eats one pack of hyena food in one day is 0.4 and the probability that the hyena eats two packs of hyena food in one day is 0.6. What is the probability that the hyena eats more than 153 packets of hyena food in 100 days?"

What I've figured out so far:

Since I have a known number of trials (n=100), I can use a binomial distribution to calculate the problem, P(X>153).

P(1)=0,4
P(2)=0,6

The lower limit is 1 package of food * 100 days = 100 packages of food
The upper limit is 2 packages of food * 200 days = 200 packages of food

But that's about how far I've come, I can not figure out what I should do with the p-values.

Thank you in advance.

Answer 1

Solution

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, ……. , ∞

This probability can also be obtained by using Excel Function, Statistical, POISSON …............................ (1a)

If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(λ), then Y ~ Poisson (kλ) ………....................................................…….. (2)

Now to work out the solution,

Let X = number of food packs eaten by the hyena in a day.

Then, we will assume X ~ Poisson (1.6) ........................................................................................................ (3)

Where = 1.6 = average number of food packs eaten by the hyena in a day

= (1 x 0.4) + (2 x 0.6)

If Y = number of food packs eaten by the hyena in a day, then vide (2) and (3),

Y ~ Poisson (160) ............................................................................................................................................ (4)

The required probability

= P(Y > 153)

= 0.6929 [vide (1a)] Answer

DONE