Answer :
Answer:
Part A : [tex]0.3770[/tex]
Part B : [tex]0.2065[/tex]
Part C : [tex]0.0663[/tex]
Step-by-step explanation:
Let's start defining the random variable.
If X : ''Time between arrivals of customers at a store'' is the exponential random variable, its probability distribution function will be :
[tex]f(x)=g.e^{-gx}[/tex]
Where [tex]g=[/tex] λ is the parameter of the exponential distribution.
Also λ = 1 / μ where μ is the mean of the distribution.
Using the data of the exercise :
[tex]g=[/tex] λ = [tex]\frac{1}{6.34}[/tex]
The cumulative distribution function of the exponential random variable is :
[tex]P(X\leq x)=F(x)=1-e^{-gx}[/tex]
For this exercise :
[tex]F(x)=1-e^{-\frac{x}{6.34}}[/tex] (I)
We are going to use the equation (I) to calculate all the probabilities.
Part A : [tex]P(X<3)[/tex]
[tex]P(X<3)=P(X\leq 3)[/tex] because the exponential distribution is a continuous random variable ⇒
[tex]P(X\leq 3)=F(3)=1-e^{-\frac{3}{6.34}}=1-e^{-\frac{150}{317}}[/tex] ≅ [tex]0.3770[/tex]
Part B : [tex]P(X>10)[/tex] ⇒
[tex]P(X>10)=1-P(X\leq 10)=1-F(10)[/tex] ⇒
[tex]P(X>10)=1-(1-e^{-\frac{10}{6.34}})=e^{-\frac{10}{6.34}}[/tex] ≅ [tex]0.2065[/tex]
Part C : [tex]P(5\leq X\leq 6)[/tex] ⇒
[tex]P(5\leq X\leq 6)=F(6)-F(5)=(1-e^{-\frac{6}{6.34}})-(1-e^{-\frac{5}{6.34}})[/tex] ⇒
[tex]P(5\leq X\leq 6)=-e^{-\frac{6}{6.34}}+e^{-\frac{5}{6.34}}[/tex] ≅ [tex]0.0663[/tex]