Independent solution

How to solve this Exponential Distribution question

Setup

Setup

For an exponential variable, divide the threshold by the mean in the survival function. Here that ratio is one.

Pr⁡(X>20000)=exp⁡ ⁣(−2000020000)=e−1\Pr(X>20000)=\exp\!\left(-\frac{20000}{20000}\right)=e^{-1}

Model

Model

Let N count observations above the threshold in the independent sample.

N∼Bin⁡ ⁣(10,e−1)N\sim\operatorname{Bin}\!\left(10,e^{-1}\right)
Pr⁡(N≥9)=Pr⁡(N=9)+Pr⁡(N=10)\Pr(N\ge9)=\Pr(N=9)+\Pr(N=10)

Compute

Compute

Evaluate the two possible binomial outcomes and combine them.

Pr⁡(N≥9)=(109)e−9(1−e−1)+e−10\Pr(N\ge9)=\binom{10}{9}e^{-9}(1-e^{-1})+e^{-10}
Pr⁡(N≥9)=10e−9e10=0.0008254987…\Pr(N\ge9)=\frac{10e-9}{e^{10}}=0.0008254987\ldots

Answer

Answer

The required binomial tail is represented by choice B.

Pr⁡(N≥9)=10e−9e10(B)\boxed{\Pr(N\ge9)=\frac{10e-9}{e^{10}}\quad\text{(B)}}