Respuesta :
Answer:
(a) The value of P (X ≤ 2) is 0.873.
(b) The value of P (X ≥ 5) is 0.007.
(c) The value of P (1 ≤ X ≤ 4) is 0.716.
(d) The probability that none of the 25 boards is defective is 0.277.
Step-by-step explanation:
The random variable X is defined as the number of defective circuit boards.
It is provided that X follows a Binomial distribution with parameter n = 25 and p = 0.05.
The probability mass function of Binomial distribution is:
[tex]P(X=x) = {n\choose x}p^{x}(1-)^{n-x};\ x=0,1,2,3...[/tex]
(a)
Compute the value of P (X ≤ 2) as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
[tex]= {25\choose 0}(0.05)^{0}(1-0.05)^{25-0}+{25\choose 1}(0.05)^{1}(1-0.05)^{25-1}\\+{25\choose 2}(0.05)^{2}(1-0.05)^{25-2}\\=0.2774+0.3650+0.2305\\=0.8729[/tex]
Thus, the value of P (X ≤ 2) is 0.873.
(b)
Compute the value of P (X ≥ 5) as follows:
P (X ≥ 5) = 1 - P (X < 5)
= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)+- P (X = 4)
[tex]=1- {25\choose 0}(0.05)^{0}(1-0.05)^{25-0}-{25\choose 1}(0.05)^{1}(1-0.05)^{25-1}\\-{25\choose 2}(0.05)^{2}(1-0.05)^{25-2}-{25\choose 3}(0.05)^{3}(1-0.05)^{25-3}\\-{25\choose 4}(0.05)^{4}(1-0.05)^{25-4}\\=1-0.2774-0.3650-0.2305-0.0930-0.027\\=0.0071[/tex]
Thus, the value of P (X ≥ 5) is 0.007.
(c)
Compute the value of P (1 ≤ X ≤ 4) as follows:
P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
[tex]={25\choose 1}(0.05)^{1}(1-0.05)^{25-1}+{25\choose 2}(0.05)^{2}(1-0.05)^{25-2}\\+{25\choose 3}(0.05)^{3}(1-0.05)^{25-3}+{25\choose 4}(0.05)^{4}(1-0.05)^{25-4}\\=0.3650+0.2305+0.0930+0.027\\=0.7155[/tex]
Thus, the value of P (1 ≤ X ≤ 4) is 0.716.
(d)
Compute the probability of X = 0 as follows:
[tex]P(X=0) = {25\choose 0}(0.05)^{0}(1-0.05)^{25-0}\\=1\times1\times0.2773896\\=0.2773896\\\approx 0.277[/tex]
Thus, the probability that none of the 25 boards is defective is 0.277.
Answer:
(a) P(X < 2) = 0.6424
(b) P(X > 5) = 0.99879
(c) P(1 < X < 4) = 0.3235
(d) P(X = 0) = 0.2774.
Step-by-step explanation:
We are given that the long-run percentage of defectives is 5%. Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.05).
The binomial probability is given by;
[tex]P(X=r) =\binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....[/tex]
(a) P(X < 2) = P(X = 0) + P(X = 1)
= [tex]\binom{25}{0}0.05^{0}(1-0.05)^{25-0} + \binom{25}{1}0.05^{1}(1-0.05)^{25-1}[/tex]
= [tex]1*1*0.95^{25} + 25*0.05*0.95^{24}[/tex] = 0.6424
(b) P(X > 5) = 1 - P(X <= 5)
P(X <= 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) +P(X =5)
= [tex]\binom{25}{0}0.05^{0}(1-0.05)^{25-0} + \binom{25}{1}0.05^{1}(1-0.05)^{25-1} + \binom{25}{2}0.05^{2}(1-0.05)^{25-2} + \binom{25}{3}0.05^{3}(1-0.05)^{25-3}+\binom{25}{4}0.05^{4}(1-0.05)^{25-4} + \binom{25}{5}0.05^{5}(1-0.05)^{25-5}[/tex]
= [tex]1*1*0.95^{25} + 25*0.05*0.95^{24} + 300*0.05^{2} *0.95^{23} + 2300*0.05^{3} *0.95^{22} + 12650*0.05^{4} *0.95^{21} + 53130*0.05^{5} *0.95^{20}[/tex]
= 0.99879
(c) P(1 < X < 4) = P(X = 2) + P(X = 3)
= [tex]\binom{25}{2}0.05^{2}(1-0.05)^{25-2} + \binom{25}{3}0.05^{3}(1-0.05)^{25-3}[/tex]
= [tex]300*0.05^{2} *0.95^{23} + 2300*0.05^{3} *0.95^{22}[/tex] = 0.3235
(d) Probability that none of the 25 boards is defective = P(X = 0)
P(X = 0) = [tex]\binom{25}{0}0.05^{0}(1-0.05)^{25-0}[/tex]
= [tex]1*1*0.95^{25}[/tex] = 0.2774.
