Suppose that on average 4.5 red ants are caught in an ant trap in one hour. On a

Question

Suppose that on average 4.5 red ants are caught in an ant trap in one hour. On average 7.1 black ants are caught in the ant trap in one hour. The ants are trapped randomly in time so the number of ants trapped can be considered to have a Poisson distribution. Let X-the number of red ants caught in an hour. Let Y-the number of black ants caught in an hour. X and Y are independent. So, X has a Poisson distribution with parameter 2, . Y has a Poisson distribution with parameter a,what is the value of ? b. What os the value of ? c. What is the probability that 2 red ants are caught in an hour? d. What is the probability that 3 black ants are caught in an hour? e. What is the probability that 2 red ants and 3 black ants are caught in an hour? f. What is the standard deviation of xY

Explanation / Answer

POSSION DISTRIBUTION
pmf of P.D is = f ( k ) = e- x / x!
where   
= parameter of the distribution.
x = is the number of independent trials
i.
number of red ant caught in an hour = mean = 4.5
ii.
number of black ant caught in an hour = mean = 7.1
iii.
P( X = 2 ) = e ^-4.5 * 4.5^2 / 2! = 0.11248
iv.
P( X = 3 ) = e ^-7.1 * 7.1^3 / 3! = 0.04922
v.
P( 2 red ants & 3 black ant caught in an hour) = P( X = 2 ) +P( X = 3 ) = 0.11248+0.04922 = 0.1617
vi.
standard deviation x + y = sqrt(4.5 + 7.1) = 3.4058

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