CA Foundation · Quantitative Aptitude · Theoretical Distributions
A Poisson variable X has P(X=1) = P(X=2). What is the mean of the distribution?
The mean is 2. Equating P(X=1) = e^-m m with P(X=2) = e^-m m^2/2 gives m = m^2/2, which yields m = 2 after rejecting zero. The variance is therefore also 2.
- A1
- B4
- C2Correct
- D0.5
Explanation
P(X=1) = e^-m * m and P(X=2) = e^-m * m^2/2. Equating gives m = m^2/2, so m = 2 (m=0 is rejected). Check: m=1 would give 1 versus 0.5, which are unequal.
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