Question

24% of U.S. adults say they are more likely to make purchases during a sales tax holiday. you randomly select 10 adults. find the probability that the number of adults who say theu are more likely to make purchases during a sales tax holiday is (a) exactly two, (b) more than two, and (c) between two and five inclusive​

Answer 1

Answer:

a) $p(X= 2) = 0.261$

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 $p(X= 2) = e^{-\alpha } \frac{\alpha^r }{r! }$

$p(X= 2) = e^{2.4 } \frac{\(2.4)^2 }{2! }$

$p(X= 2) = e^{2.4 } \frac{\(2.4)^2 }{2! }= 0.261$

b) The probability of getting more than '2'

$P(X>2) = 1- {p(x=0)+p(x=1)+p(x=2)}$

                $= e^{-2.4 } \frac{(2.4)\^0 }{0! }+ e^{-2.4 } \frac{(2.4)\^1 }{1!}+e^{-2.4 } \frac{(2.4)\^2 }{2!}$

              = 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) =$= e^{-2.4 } \frac{(2.4)\^3 }{3! }+ e^{-2.4 } \frac{(2.4)\^4 }{4!}+$

P(2<x<5) = 0.2090 + 0.125 = 0.334