A random variable X has Poisson distribution with mean 2. Then P(X 1.5) equals :

Question

A random variable X has Poisson distribution with mean 2. Then P(X > 1.5) equals : (A) (3/e2) (B) 1-(3/e2) (C) 0 (D) (2/e2)

Tardigrade Admin · Accepted Answer

Key Idea: In a poisson distribution, P(X = r) = (e-λλr/r) (λ = mean). ∴ P(X = r > 1.5) = P(2)+P(3) + ...∞ = 1-((P(0)+P(1))) = 1-(e-2+(e-2×2/1)) = 1-(3/e2)

Q. A random variable X has Poisson distribution with mean 2. Then $P (X > 1.5)$ equals :

$\frac{3}{e ^{2}}$

$1 - \frac{3}{e ^{2}}$

$0$

$\frac{2}{e ^{2}}$

Key Idea : In a poisson distribution,
$P (X = r) = \frac{e ^{- λ} λ ^{r}}{r} (λ = m e an) .$
$∴ P (X = r > 1.5) = P (2) + P (3) + ...\infty$
$= 1 - ((P (0) + P (1)))$
$= 1 - (e^{- 2} + \frac{e ^{- 2} \times 2}{1}) = 1 - \frac{3}{e ^{2}}$

Q. A random variable X has Poisson distribution with mean 2. Then P(X>1.5) equals :

e23​

1−e23​

0

e22​

Key Idea : In a poisson distribution, P(X=r)=re−λλr​(λ=mean). ∴P(X=r>1.5)=P(2)+P(3)+...∞ =1−((P(0)+P(1))) =1−(e−2+1e−2×2​)=1−e23​

Q. A random variable X has Poisson distribution with mean 2. Then $P (X > 1.5)$ equals :

$\frac{3}{e ^{2}}$

$1 - \frac{3}{e ^{2}}$

$0$

$\frac{2}{e ^{2}}$

Key Idea : In a poisson distribution,
$P (X = r) = \frac{e ^{- λ} λ ^{r}}{r} (λ = m e an) .$
$∴ P (X = r > 1.5) = P (2) + P (3) + ...\infty$
$= 1 - ((P (0) + P (1)))$
$= 1 - (e^{- 2} + \frac{e ^{- 2} \times 2}{1}) = 1 - \frac{3}{e ^{2}}$