Respuesta :
Answer:
(a) P(X=3) = 0.093
(b) P(X≤3) = 0.966
(c) P(X≥4) = 0.034
(d) P(1≤X≤3) = 0.688
(e) The probability that none of the 25 boards is defective is 0.277.
(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.
Step-by-step explanation:
We are given that when circuit boards used in the manufacture of compact disc players are tested, 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 probability distribution for the binomial distribution is given by;
[tex]P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......[/tex]
where, n = number of trials (samples) taken = 25
r = number of success
p = probability of success which in our question is percentage
of defectivs, i.e. 5%
(a) P(X = 3) = [tex]\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}[/tex]
= [tex]2300 \times 0.05^{3}\times 0.95^{22}[/tex]
= 0.093
(b) P(X [tex]\leq[/tex] 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= [tex]\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}[/tex]
= [tex]1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}[/tex]
= 0.966
(c) P(X [tex]\geq[/tex] 4) = 1 - P(X < 4) = 1 - P(X [tex]\leq[/tex] 3)
= 1 - 0.966
= 0.034
(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
= [tex]\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}[/tex]
= [tex]25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}[/tex]
= 0.688
(e) The probability that none of the 25 boards is defective is given by = P(X = 0)
P(X = 0) = [tex]\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}[/tex]
= [tex]1 \times 1\times 0.95^{25}[/tex]
= 0.277
(f) The expected value of X is given by;
E(X) = [tex]n \times p[/tex]
= [tex]25 \times 0.05[/tex] = 1.25
The standard deviation of X is given by;
S.D.(X) = [tex]\sqrt{n \times p \times (1-p)}[/tex]
= [tex]\sqrt{25 \times 0.05 \times (1-0.05)}[/tex]
= 1.089
