I drew this by hand. sorry I it does not help.
Answer:
a) 3.6
b) 1.897
c)0.0273
d) 0.9727
Step-by-step explanation:
Rabies has a rare occurrence and we can assume that events are independent. So, X the count of rabies cases reported in a given week is a Poisson random variable with μ=3.6.
a)
The mean of a Poisson random variable X is μ.
mean=E(X)=μ=3.6.
b)
The standard deviation of a Poisson random variable X is √μ.
standard deviation=S.D(X)=√μ=√3.6=1.897.
c)
The probability for Poisson random variable X can be calculated as
P(X=x)=(e^-μ)(μ^x)/x!
where x=0,1,2,3,...
So,
P(no case of rabies)=P(X=0)=e^-3.6(3.6^0)/0!
P(no case of rabies)=P(X=0)=0.0273.
d)
P(at least one case of rabies)=P(X≥1)=1-P(X<1)=1-P(X=0)
P(at least one case of rabies)=1-0.0273=0.9727
Answer:
SAS congruence postulate. sweetheart.
Step-by-step explanation:
A..................................................
