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:
absolutley no solutions
Step-by-step explanation:
Answer:
Step-by-step explanation:
In the past, mean of age of employees
i.e. 
Recently sample was taken
n = sample size = 60
Mean of sample = 45
Std dev of sample s = 16
(Right tailed test)
Since only population std deviation is known we can use t test only
Std error = 
Mean difference = 45-40 =5
Test statistic t=
df = 60
p value =0.008739
Since p < 0.05 we reject null hypothesis
The mean age has increased.
Answer:
3, and -3
Step-by-step explanation:
Because absolute value is the distance away from zero, |3| and |-3| would both be 3 units away from 0. Be sure to have the negative on the inside if you're trying to make something positive, otherwise -|3| means that the absolute value of 3 would then become negative since the negative sign isn't a part of the absolute. hope this helps!
