Sixty percent of consumers prefer to purchase electronics online. You randomly select 8 consumers. Find the probability that the number of consumers who prefer to purchase electronics online is​ (a) exactly​ five, (b) more than​ five, and​ (c) at most five.

Answer :

opudodennis

Answer:

[tex]a. \ P(X=5)=0.2787\\\\b.\ \ P(X>5)=0.3154\\\\c. \ P(X\leq 5)=0.6846[/tex]

Step-by-step explanation:

a. This a binomial probability distribution problem expressed as:

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

#Given the probability of success, p=0.60 and the number of trials, n=8, the probability of exactly 5 purchases is calculated as:

[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\\\P(X=5)={8\choose 5}0.6^5(1-0.6)^3\\\\=0.2787[/tex]

Hence, the probability of exactly 5 purchases is 0.2787

b. Given n=8 , p=0.60, the probability of more than 5 purchases is calculated as:

[tex]P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\\\P(X>5)=P(X=6)+P(X=7)+P(X=8)\\\\={8\choose 6}0.6^6(1-0.6)^2+{8\choose 7}0.6^7(1-0.6)^1+{8\choose 8}0.6^8(1-0.6)^0\\\\=0.2090+0.0896+0.0168\\\\=0.3154[/tex]

Hence, the probability of more than 5 purchases is 0.3154

c. The probability of at most 5 purchases is equivalent to 1-P(x>5) and is calculated as:

#From b above, P(X>5)=0.3154;

[tex]P(X\leq 5)=1-P(X>5)\\\\=1-0.3154\\\\=0.6846[/tex]

Hence, the probability of at most 5 purchases is 0.6846

(a) The required value is [tex]P(X=5)=0.2787[/tex]

(b) The required value is [tex]P(x > 5)=0.3154[/tex]

(c) The required value is [tex]P(X\le5)=0.6846[/tex]

Probability mass function:

If X be a discrete random variable of a function, then the probability mass function of a random variable X is given by,

[tex]P (x) = P (X = x),[/tex]

Given that,

[tex]\eta=8[/tex]

The pmf of X is,

[tex]P(X=x)=\left\{\begin{matrix}\binom{8(0.60)^x(0.40)^{p-x}}{x}\\0,\ otherwise\end{matrix}\right.[/tex]

Part(a):

Calculating P(Xis exactly five) then,

[tex]P(X=5)=\binom{8}{5}(0.60)^5(0.40)^3\\ =0.2787[/tex]

Part(b): Calculating [tex]P(x > 5)[/tex] then,

[tex]P(x > 5)=P(x\geq 6)\\=P(x=6+P(x=7)+P(x=8)\\=0.3154[/tex]

Part(c): Calculating [tex]P(x \leq 5)[/tex] then,

[tex]P(X\le5)=1-P(x > 5)\\=1-0.3154\\=0.6846[/tex]

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