How do I Find P ( 2 < X < 7 )? (Stats)

I Got the part A, but don't get how to do part B.enter image source here

1 Answer
May 16, 2018

P(2 < X < 7) = 54.43%P(2<X<7)=54.43%.

Explanation:

The picture is a bit blurry, but it appears as though XX is the number of calls over a 1.5 minute interval. Thus, X" ~ Poisson"(lambda)X ~ Poisson(λ) where lambda = 1.5 xx 4 = 6.λ=1.5×4=6.

Thus, P(X=x)" "=(e^-6 xx6^x)/(x!)P(X=x) =e6×6xx!

Since XX is discrete, we know

P(2 < X < 7) = sum_(x=3)^6 P(X = x)P(2<X<7)=6x=3P(X=x)

= P(X"=3")+P(X"=4")+P(X"=5")+P(X"=6")=P(X=3)+P(X=4)+P(X=5)+P(X=6)

We just need to find these 4 discrete probabilities.

P(X=3) = (e^-6 xx6^3)/(3!) = 0.0892P(X=3)=e6×633!=0.0892

P(X=4) = (e^-6 xx6^4)/(4!) = 0.1339P(X=4)=e6×644!=0.1339

P(X=5) = (e^-6 xx6^5)/(5!) = 0.1606P(X=5)=e6×655!=0.1606

P(X=6) = (e^-6 xx6^6)/(6!) = 0.1606P(X=6)=e6×666!=0.1606

Then

P(2 < X < 7)P(2<X<7)

= P(X"=3")+P(X"=4")+P(X"=5")+P(X"=6")=P(X=3)+P(X=4)+P(X=5)+P(X=6)

= 0.0892+0.1339+0.1606+0.1606=0.0892+0.1339+0.1606+0.1606

= 0.5443=0.5443
=54.43%=54.43%