This requires the Poisson distribution, where
area = 5-acres
and mean number of field mice = 12 (in 5-acres of field)
therefore
lambda=12 (mean, given)
and the probability of k mice in the 5-acre field is given by the Poisson distribution as
P(X=k)=lambda^k * e^(-lambda) / k! ..............(1)
To find the probability of having LESS than 7 field mice, we add the probabilities of 0 to 6, which is
P(X<7)=P(X=0)+P(X=1)+...+P(X=6)
evaluating with equation (1) for X=0 to 6, we get:
0 0.0000061 0.0000742 0.0004423 0.0017704 0.0053095 0.0127416 0.025481Total = 0.045822
Answer: The probability that fewer than 7 field mice are found in the 5-acre field is 0.0458.
The rate of change for both functions is 2.
C) Its members are not required to disclose their last names ; <span>is true about the recovery program Alcoholics Anonymous.</span>
Answer:
7 4 and 15 because term means numbers
Answer:
Step-by-step explanation:
If each day equal chance then p = Prob that a person is borne on a particular day = 1/365
Each person is independent of the other and there are two outcomes either borne in July or not
p = prob for one person not borne in July = (365-31)/365 = 334/365
a)Hence prob that no one from n people borne in July = 
b) p = prob of any one borne in July or Aug = 
=0.1698
X- no of people borne in July or Aug
n =15
P(X>=2) =
=0.7505
