Asssume the mean number of Guinea hen’s sold per day is 3 and it is Poisson dist

Question

Asssume the mean number of Guinea hen’s sold per day is 3 and it is Poisson distributed. Since we assumed the number of Guinea’s hens sold its Poisson distributed it means the waiting time between sales is exponentially distributed with a mean of 10/3 hours (its open7am-5pm). Assuming its 3pm what are the probability the rest of the guinea hen’s get to live to see tomorrow?

What is the expected average number of Guinea hens to be sold per day over 30 days? If we were to repeat this 30 day experiment over and over what is the expected standard deviation of the (30 day total / 30 days)?

Explanation / Answer

Assume that its 3 pm so that means now 2 hour left in work.

So, expected number of guinear hen's to be sold in next two hours = (3/10) * 2 = 0.6

Now, there is no guniea hen's being sold in this period. Its probability

Pr(X = 0) = POISSON (X = 0, 0.6) = e- 0.6  = 0.5488

expected average number of Guinea hens to be sold per day over 30 days = 3 * 30 /30= 3  

Now we repeat this 30 day experiment over and over our standard deviation of the (30 day total / 30 days) would be sqrt(3/30)= 0.3162 guinea' hen

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.