28% of US 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 they 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:

a

  [tex]P(X = 2) =  0.2547[/tex]  

b

  [tex]P(X > 2) =  0.5623 [/tex]  

c

   [tex]P(2 \le X  \le 5 ) = 0.7817 [/tex]

Step-by-step explanation:

From the question we are told that

   The  proportion of US adults who say that they are likely to make purchase during a sale tax holiday is  

         [tex]p = 0.28[/tex]

  The  sample size is  n  =  10  

Generally the distribution of US adults who say that they are likely to make purchase during a sale tax holiday, follows a binomial distribution  

i.e  

         [tex]X  \~ \ \ \  B(n , p)[/tex]

and the probability distribution function for binomial  distribution is  

      [tex]P(X = x) =  ^{n}C_x *  p^x *  (1- p)^{n-x}[/tex]

Here C stands for combination hence we are going to be making use of the combination function in our calculators  

Generally the probability number of adult is exactly 2 is mathematically represented as

     [tex]P(X = 2) =  ^{10}C_2 *  (0.28)^2 *  (1- 0.28)^{10-2}[/tex]  

=>  [tex]P(X = 2) =  45 *  0.0784 * 0.0722[/tex]  

=>  [tex]P(X = 2) =  0.2547[/tex]  

Generally the probability number of adult is more than 2 is mathematically represented as

    [tex]P(X > 2) =  1- [P(X \le  2) ][/tex]  

    [tex]P(X > 2) =  1- [[P(X = 0)] +[P(X = 1) ] + [P(X = 2)] ][/tex]  

=>  [tex]P(X > 2) =  1- [^{10}C_0 *  (0.28)^0 *  (1- 0.28)^{10-0}] +[^{10}C_1 *  (0.28)^1 *  (1- 0.28)^{10-1}] + [^{10}C_2 *  0.28^2 *  (1- 0.28)^{10-2}] ][/tex]  

=>  [tex]P(X > 2) =  1- [0.0374 + 0.1456 + 0.2547 [/tex]  

=>  [tex]P(X > 2) =  0.5623 [/tex]  

Generally the probability number of adult is between two and​ five, inclusive is mathematically represented as

    [tex]P(2 \le X  \le 5 ) =  [P(X = 2 ) + P(X = 3 ) +P(X = 4) + P(X= 5 )] [/tex]

=>  [tex]P(2 \le X  \le 5 ) =  [[^{10}C_32 *  0.28^2 *  (1- 0.28)^{10-2}] + [^{10}C_3*  0.28^3 *  (1- 0.28)^{10-3}] +[^{10}C_4 *  0.28^4 *  (1- 0.82)^{10-4}] + [^{10}C_5 *  0.28^5 *  (1- 0.28)^{10-5}]] [/tex]

=>  [tex]P(2 \le X  \le 5 ) =  [[0.2547] + [120*  0.02195 *  0.1003] +[210 *  0.00615 *  0.1393] + [252 *  0.0017 *  0.1935]] [/tex]

=>  [tex]P(2 \le X  \le 5 ) =  [[0.2547] + [0.2642] +[0.1799] + [0.0829] [/tex]

=>  [tex]P(2 \le X  \le 5 ) = 0.7817 [/tex]

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