# Hi! I’ve been stuck with this

Hi!

I’ve been stuck with this question this whole day, not beingable to get find out how to solve it.

The question follows “In a zoo lives a hyena who every day eatsa certain number of packs of hyena food. The probability that thehyena eats one pack of hyena food in one day is 0.4 and theprobability that the hyena eats two packs of hyena food in one dayis 0.6. What is the probability that the hyena eats more than 153packets 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 abinomial distribution to calculate the problem, P(X>153).

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

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

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

Thank you in advance.

Answer:

Solution

Back-upTheory

If a random variable X ~ Poisson (λ), i.e., X has PoissonDistribution 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 thesolution,

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 ina day

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

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

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

The required probability

= P(Y > 153)

= **0.6929 [**vide (1a)**]Answer**

**DONE**