CA Foundation · Quantitative Aptitude · Theoretical Distributions
The number of cheques returned unpaid at a Nashik bank branch per day follows a Poisson distribution with mean 2. Using e^-2 = 0.1353 and e^-4 = 0.0183, what is the probability that exactly 2 cheques are returned in one day, and what is the expected count over 2 days with the Poisson mean for the two-day period?
The probability of exactly 2 returns in a day is 0.2706, from e^-2 times 2 squared over 2 factorial. For two days the Poisson mean scales with time to 4, because independent Poisson counts add their means.
- A0.2706 and mean 4Correct
- B0.2706 and mean 2
- C0.1353 and mean 4
- D0.5413 and mean 4
Explanation
For one day, P(2) = e^-2 * 2^2/2! = 0.1353 * 2 = 0.2706. For two days, independent Poisson counts add, so the mean is 2 + 2 = 4. Mean 2 ignores the longer period. 0.1353 forgets the factor 2^2/2!. 0.5413 uses 4 in the numerator by mistake (0.1353 * 4).
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