Question

If x is a passion random variable with parameter value 4.5, find p(x=1) and P (X = 7).

Answer 1

SOLUTION

GIVEN

X is a poisson random variable with parameter value 4.5

TO DETERMINE

P(X = 1) and P (X = 7)

CONCEPT TO BE IMPLEMENTED

X is a poisson random variable with parameter μ then

$\displaystyle \sf{P(X = r) = {e}^{ - \mu}. \frac{ { \mu}^{r} }{r \: !} }$

Where μ > 0 and r = 1 , 2 , 3 , ....

EVALUATION

We know if X is a poisson random variable with parameter μ then

$\displaystyle \sf{P(X = r) = {e}^{ - \mu}. \frac{ { \mu}^{r} }{r \: !} }$

Where μ > 0 and r = 1 , 2 , 3 , ....

Here it is given that X is a poisson random variable with parameter value 4.5

∴ μ = 4.5

Now

$\displaystyle \sf{P(X = 1) = {e}^{ - \mu}. \frac{ { \mu}^{1} }{1 \: !} }$

$\displaystyle \sf{ \implies \: P(X = 1) = {e}^{ -4.5}. \frac{ {(4.5)}^{1} }{1 \: !} }$

$\displaystyle \sf{ \implies \: P(X = 1) = 0.0111 \times 4.5 }$

$\displaystyle \sf{ \implies \: P(X = 1) = 0.04995}$

Again

$\displaystyle \sf{ \: P(X = 7) = {e}^{ -4.5}. \frac{ {(4.5)}^{7} }{7 \: !} }$

$\displaystyle \sf{ \implies \: P(X = 7) = 0.0111 \times \frac{37366.94}{5040} }$

$\displaystyle \sf{ \implies \: P(X = 7) =0.0823}$

FINAL ANSWER

$\boxed{ \: \: \displaystyle \sf{P(X = 1) = 0.04995 \: \: }}$

$\boxed{ \: \: \displaystyle \sf{ P(X = 7) =0.0823} \: \: }$

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