Question

Assume the number of hickory nuts that fall on my deck in August is Poisson distributed with a mean of 5.6 per hour. What is the probability of:

a. More than 11 hickory nuts falling on my deck during a given August hour?

b. Only 1 hickory nut falling on my deck during a given hour?

c. Exactly 4 hickory nuts falling on my deck during a given hour?

d. 3 or fewer hickory nuts falling on my deck during a given hour?

e. What is the mean and standard deviation for the above Poisson Distribution?

Answer 1

Suppose X~ Poi ( = 5.6)

a) Here we want to find P( X > 11 ) = 1 - P( X <= 11) .......( 1 )

Let's use excel:

P(X <= 11) = "=POISSON(11,5.6,1)" = 0.9875

Plug this value in equation ( 1 ), we get:

P( X > 11 ) = 1 - P( X <= 11) = 1 - 0.9875 = 0.0125.

b. Only 1 hickory nut falling on my deck during a given hour?

Here we want to find P( X = 1) = "=POISSON(1,5.6,1)" = 0.0207

c. Exactly 4 hickory nuts falling on my deck during a given hour?

Here we want to find P( X = 4) = "=POISSON(4,5.6,0)" = 0.1515

d. 3 or fewer hickory nuts falling on my deck during a given hour?

Here we want to find P(X <= 3) = "=POISSON(3,5.6,1)" = 0.1906

e. What is the mean and standard deviation for the above Poisson Distribution?

mean = = 5.6

standard deviation of Poisson distribution =