contestada

Of all customers purchasing automatic garage-door openers, 75% purchase a chain-driven model. Let X = the number among the next 15 purchasers who select the chain-driven model. (a) What is the pmf of X? nb(x; 15, 0.75) b(x; 15, 0.25) h(x; 3, 15, 75) b(x; 15, 0.75) h(x; 12, 15, 75) nb(x; 15, 0.25) Correct: Your answer is correct. (b) Compute P(X > 12). (Round your answer to three decimal places.) P(X > 12) = (c) Compute P(9 ≤ X ≤ 12). (Round your answer to three decimal places.) P(9 ≤ X ≤ 12) =

Respuesta :

bayosanuade

Answer: (a) b(x; 15, 0.75)

               (b) 0.197

               (c) 0.559

Step-by-step explanation:

P( X=x)=(n C x)) p^x q^(n-x)

From the question, p = 0.75

 [tex]\\[/tex]q = 0.25

   [tex]\\[/tex] n = 15

[tex]\\[/tex]So, the Pmf of X implies

[tex]\\[/tex]P( X=x)=(15C x)) 〖(0.75)〗^x 〖(0.25)〗^(15-x)

  [tex]\\[/tex] b( x:15,0.75)

[tex]\\[/tex](b) p(x>12)=p(x=13)+p(x=14)+p(x=15)

[tex]\\[/tex]=0.1169302838+0.06681730505+ 0.01336346101

[tex]\\[/tex]=0.19711104986

[tex]\\[/tex]≈0.197

[tex]\\[/tex](c) p(9≤x≤12)=p(x=9)+p(x=10)+p(x=11)+p(x=12)

 [tex]\\[/tex] =0.09174776729+0.16514598113+ 0.22519906517                            +0.22519906517

  [tex]\\[/tex]=0.55866049575

  [tex]\\[/tex]≈0.559

   

PMF is the probability function of discrete distributions. The pmf and the needed probabilities for the given situations are:

  • PMF of X = [tex]P(X = x) = \: ^{15}C_x{15}^x(0.25)^{{15}-x}[/tex]
  • P(X > 12) = 0.236 approx.
  • P(9 ≤ X ≤ 12) = 0.707 approx.

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability [tex]1- p = q[/tex] (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

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

The probability that out of n trials, there'd be x successes is given by

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

For the given case, as each purchaser can independently be purchaser of a chain driven model, so if we take:

X = the number among the next 15 purchasers who select the chain-driven model

Then, [tex]X \sim B(n=15, p = 75\% = 0.75)[/tex]

where we define success as a purchaser being purchaser of chain driven model, and success is that purchaser not being purchaser of chain driven model.

The pmf of X, thus, is given by:

[tex]P(X =x) = \: ^{15}C_x{15}^x(1-0.75)^{{15}-x}\\\\P(X = x) = \: ^{15}C_x{15}^x(0.25)^{{15}-x}[/tex]

Also, we get:

[tex]P(X > 12) = P(X = 13) + P(X = 14) + P(X = 15)\\\\P(X > 12) = \: ^{15}C_{13}{15}^{13}(0.25)^{2} + \: ^{15}C_{14}{15}^{14}(0.25)^{1} + \: ^{15}C_{15}{15}^{15}(0.25)^{0}\\\\P(X > 12) \approx 0.156 + 0.067 + 0.013 = 0.236[/tex]

And, similarly,

[tex]P(9 \leq X \leq 12) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)\\P(9 \leq X \leq 12) = \:^{15}C_915}^9(0.25)^{6} + \: ^{15}C_{10}15}^{10}(0.25)^{5} + \\ \: ^{15}C_{11}15}^{11}(0.25)^{4} + \: ^{15}C_{12}15}^{12}(0.25)^{3} \\P(9 \leq X \leq 12) \approx 0.092 + 0.165 + 0.225 + 0.225\\P(9 \leq X \leq 12) \approx 0.707[/tex]

Thus, the pmf and the needed probabilities for the given situations are:

  • PMF of X = [tex]P(X = x) = \: ^{15}C_x{15}^x(0.25)^{{15}-x}[/tex]
  • P(X > 12) = 0.236 approx.
  • P(9 ≤ X ≤ 12) = 0.707 approx.

Learn more about binomial distributions here:

https://brainly.com/question/13609688

Otras preguntas