Answer:
a) [tex]p(X= 2) = 0.261[/tex]
b) P(x>2) = 0.566
c) P(2<x<5) = 0.334
Step-by-step explanation:
Given 24% of U.S. adults say they are more likely to make purchases during a sales tax holiday
Probability 0f U.S. adults say they are more likely to make purchases during a sales tax holiday (p) = 0.24
n = 10
By using Poisson distribution
mean number of make purchases during a sales tax holiday
λ = np = 10 X 0.24 = 2.4
a)
The probability of getting exactly '2'
The probability [tex]p(X= 2) = e^{-\alpha } \frac{\alpha^r }{r! }[/tex]
[tex]p(X= 2) = e^{2.4 } \frac{\(2.4)^2 }{2! }[/tex]
[tex]p(X= 2) = e^{2.4 } \frac{\(2.4)^2 }{2! }= 0.261[/tex]
b) The probability of getting more than '2'
[tex]P(X>2) = 1- {p(x=0)+p(x=1)+p(x=2)}[/tex]
[tex]= e^{-2.4 } \frac{(2.4)\^0 }{0! }+ e^{-2.4 } \frac{(2.4)\^1 }{1!}+e^{-2.4 } \frac{(2.4)\^2 }{2!}[/tex]
= 0.090 + 0.2177+0.261 = 0.566
P(x>2) = 0.566
c) The probability of getting between two and five
P( 2<x<5) = P(x=3)+p(x=4) =[tex]= e^{-2.4 } \frac{(2.4)\^3 }{3! }+ e^{-2.4 } \frac{(2.4)\^4 }{4!}+[/tex]
P(2<x<5) = 0.2090 + 0.125 = 0.334