Hello I would like some help with my statistics studywork I

Hello. I would like some help with my statistics studywork. I have a horrible stat textbook. It goes into such little detail, and merely throws out letters and numbers with no specified instructions for calculating. So if you answer, please don\'t just put a formula, etc. with no explanation. Thanks!

1. The average number of injuries per week in an industry are recorded as 2.1

a. Find the probability that in a given week there can be no injuries

b. Find the probability that in a given week, there will be less than 3 injuries

c. Find the probability that in a given week, there will be at least 3 injuries

d. Find the probability that in a given week, there will be at least 2 but at most 4 injuries

e. What is the mean and standard deviation of this problem?

Solution

a. Find the probability that in a given week there can be no injuries

Given X follows Poisson distribution with mean=2.1

P(X=x)=(2.1^x)*exp(-2.1)/x! for x=0,1,2,...

P(X=0)=(2.1^0)*exp(-2.1)/1=0.1224564

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b. Find the probability that in a given week, there will be less than 3 injuries

P(X<3) = P(X=0)+P(X=1)+P(X=2)

=(2.1^0)*exp(-2.1)/1+(2.1^1)*exp(-2.1)/1+(2.1^2)*exp(-2.1)/2

=0.6496314

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c. Find the probability that in a given week, there will be at least 3 injuries

P(X>=3)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)

=1-(2.1^0)*exp(-2.1)/1-(2.1^1)*exp(-2.1)/1-(2.1^2)*exp(-2.1)/2-(2.1^3)*exp(-2.1)/6

=0.1613572

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d. Find the probability that in a given week, there will be at least 2 but at most 4 injuries

P(2<=X<=4) = P(X=2)+P(X=3)+P(X=4)

=(2.1^2)*exp(-2.1)/2 + (2.1^3)*exp(-2.1)/6 +(2.1^4)*exp(-2.1)/24

=0.558259

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e. What is the mean and standard deviation of this problem?

mean=2.1

standard deviation= sqrt(2.1) =1.44913