Question

A recent study of robberies for a certain geographic region showed an average of 1 robbery per 20,000 people. In a city of 80,000 people, find the probability of 2 robberies.

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Concept:
When many alternative outcomes exist, the Poisson distribution formula is used. If a discrete random variable X has a Poisson distribution and  λ  is its average value, then X has a chance of

$P(x; \lambda) = \Biggl\{e^{-\lambda}.\dfrac{\lambda^X}{X!} \Bigg| ~X=0,1,2,3,...,\lambda > 0\Biggl\}$

Given: The average number of robberies is given to be 1 robbery per 20,000 people.

Find: To determine the probability of 2 robberies in a city of 80,000 people.

Solution: To obtain the final result, we must first apply the Poisson Distribution formula to substitute the provided probability values, then simplify.
Step 1: Because there is a specified average number of robberies per given number of persons, the number of robberies is a Poisson random variable with
$p=\dfrac{1}{20,000}= 0.00005$ and
n = 80,000.

Recall the formula used to determine the likelihood that X number of robberies would occur in a city with n inhabitants.

$P(X)=e^{-\lambda}.\dfrac{\lambda^X}{X!}$

Where, $X=0,1,2,3,...,\lambda=np > 0$
So, λ = np = (1/20,000)×80,000 = 4

Step 2: Probability of 2 robbery i.e. X = 2 is
$P(X)=e^{-4}.\dfrac{\lambda^2}{2!}$

= 0.146525

Hence, the probability of 2 robberies given the scenario is 0.146525

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