CA Foundation 2026 · Paper 3 · Quantitative AptitudeQ94 · 1 mark↻ Appears in 6 of 6 yearsOfficial key verified
If the standard deviation of a Poisson distribution is 3, then P(X=0)P(X = 0) is ________.
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Poisson: λ=σ2=9\lambda=\sigma^2=9, so P(X=0)=e−λ=e−9P(X=0)=e^{-\lambda}=e^{-9}.

Step 1 — Get the parameter λ\lambda

For a Poisson variable the mean and variance are both λ\lambda, hence the standard deviation is λ\sqrt{\lambda}:

σ=λ=3  ⇒  λ=9\sigma=\sqrt{\lambda}=3 \;\Rightarrow\; \lambda=9

Step 2 — Apply the Poisson probability formula

P(X=x)=e−λλxx!P(X=x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}

Step 3 — Put x=0x=0

P(X=0)=e−9 900!=e−9P(X=0)=\dfrac{e^{-9}\,9^{0}}{0!}=e^{-9} …

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