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Law Incorporation [45]
2 years ago
15

Assume that the Poisson distribution applies and that the mean number of hurricanes in a certain area is 6.7 per year. a. Find t

he probability​ that, in a​ year, there will be 4 hurricanes. b. In a 35​-year ​period, how many years are expected to have 4 ​hurricanes? c. How does the result from part​ (b) compare to a recent period of 35 years in which 3 years had 4 ​hurricanes? Does the Poisson distribution work well​ here?
Mathematics
1 answer:
igor_vitrenko [27]2 years ago
6 0

Answer:

a) 10.34% probability​ that, in a​ year, there will be 4 hurricanes.

b) 3.62 years are expected to have 4 ​hurricanes

c) Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

6.7 per year.

This means that \mu = 6.7

a. Find the probability​ that, in a​ year, there will be 4 hurricanes.

This is P(X = 4).

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 4) = \frac{e^{-6.7}*(6.7)^{4}}{(4)!} = 0.1034

10.34% probability​ that, in a​ year, there will be 4 hurricanes.

b. In a 35​-year ​period, how many years are expected to have 4 ​hurricanes?

Each year, 0.1034 probability of 10 hurricanes.

In 35 years

35*0.1034 = 3.62

3.62 years are expected to have 4 ​hurricanes

c. How does the result from part​ (b) compare to a recent period of 35 years in which 3 years had 4 ​hurricanes? Does the Poisson distribution work well​ here?

Either 3 or 4 hurricanes(discrete number) are close to the mean of 3.62, which means that the Poisson distribution works well in this case.

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A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L
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Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

nCx=\frac{n!}{x!(n-x)!}

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448

P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224

P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192

P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

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2 years ago
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