Question

In his doctoral thesis, L. A. Beckel (University of Minnesota, 1982) studied the social behavior of river otters during the mating season. An important role in the bonding process of river otters is very short periods of social grooming. After extensive observations, Dr. Beckel found that one group of river otters under study had a frequency of initiating grooming of approximately 1.7 for each 10 minutes. Suppose that you are observing river otters for 40 minutes. Let r = 0, 1, 2, ... be a random variable that represents the number of times (in a 40-minute interval) one otter initiates social grooming of another.

What is λ?
λ =

(b) Find the probabilities that in your 40 minutes of observation, one otter will initiate social grooming four times, five times, and six times. (Round your answers to four decimal places.)
(c) Find the probability that one otter will initiate social grooming four or more times during the 40-minute observation period. (Round your answer to four decimal places.)
(d) Find the probability that one otter will initiate social grooming less than four times during the 40-minute observation period. (Round your answer to four decimal places.)

Answer 1

a) β = 1.7 = 2

b) P(x =4) = 0.09024

P(x=5) = 0.03610

P(x=6) = 0.006016

c) -0.2183

d) 1.2183

Explanation:

a) β = 1.7 = 2

P(x) = β∧x.e∧-β/x!

b) P(x =4) = 2∧4*e∧-2/4! = 16/24*e∧2 = 2/3 X 7.3875 = 0.09024

P(x=5) = 2∧5*e∧-2/5! = 32/120*e∧2 = 4/15 X 7.3875 = 0.03610

P(x=6) = 2∧6*e∧-2/6! = 64/720*e∧2 = 2/45 X7.3875 = 0.006016

c) P(x=0) = 2∧0*e∧-2/0!= 1//7.3875 = 0.1354

P(x=1) = 2∧1*e-2/1! = 2/7.3875 = 0.2707

P(x=2) = 2∧2*e-2/2! = 2/ 7.3875 = 0.2707

P(x=3) = 2∧3*e-2/3! = 4/7.3875 = 0.5415

For probability that one other will initiate four or more times:

1 - (0.1354 + 0.2707 +0.2707 + 0.5415) = 1 - 1.2183 = -0.2183

d) the probability is less than four times = 1.2183