If X~P(m) and its coefficient of variation is 50,what is the probabili...
So, the formula for coefficient of variation is given by -
COV = (SD/Mean) X 100
Now, we know that COV is 50% (it is always assumed that the COV is a percentage unless otherwise mentioned)
so by the above formula -
50 = (√m / m) X 100 (This is because mean =variance = m, in a Poisson distribution)
now when you simplify this further ,
you will get √m= 2 , therefore m=4
Now with m=4… we can solve the probability function in a Poisson distribution given by (e^-m . m^x)/X!
Now with this info,we need to find P(X>0).
however, P(x>o) = 1- P(X=0)
= 1- (e^-4 . 4^0)/ 0!
=1 - (1/e^4)
= 1- 0.01832
= 0.982
If X~P(m) and its coefficient of variation is 50,what is the probabili...
Solution:
Given, X ~ P(m) and coefficient of variation = 50
The coefficient of variation of a Poisson distribution is given by the formula:
CV = sqrt(m)
where m is the mean of the Poisson distribution.
Therefore, sqrt(m)/m = 50/100
Solving for m, we get:
m = 4
Now, we need to find the probability that X assumes only non-zero values.
P(X > 0) = 1 - P(X = 0)
The probability mass function of a Poisson distribution is given by:
P(X = k) = (e^-m * m^k) / k!
Therefore, P(X = 0) = e^-4 = 0.0183 (approx.)
Hence, the probability that X assumes only non-zero values is:
P(X > 0) = 1 - P(X = 0) = 1 - 0.0183 = 0.9817 (approx.)
Therefore, the answer is b) 0.982.
Summary:
- Given X~P(m) and coefficient of variation = 50.
- Using formula, sqrt(m)/m = 50/100, we get m = 4.
- P(X > 0) = 1 - P(X = 0).
- Using probability mass function, we get P(X = 0) = e^-4 = 0.0183 (approx.).
- Therefore, the probability that X assumes only non-zero values is P(X > 0) = 1 - P(X = 0) = 1 - 0.0183 = 0.9817 (approx.).
- Hence, the answer is b) 0.982.
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