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 = <em>the number of defective boards in a random sample of size, n = 25</em>
So, X ∼ Bin(25,0.05)
The probability distribution for the binomial distribution is given by;
![P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......](https://tex.z-dn.net/?f=P%28X%3Dr%29%3D%20%5Cbinom%7Bn%7D%7Br%7D%20%5Ctimes%20p%5E%7Br%7D%5Ctimes%20%281-p%29%5E%7Bn-r%7D%20%20%3B%20x%20%3D%200%2C1%2C2%2C......)
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) = ![\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}](https://tex.z-dn.net/?f=%5Cbinom%7B25%7D%7B3%7D%20%5Ctimes%200.05%5E%7B3%7D%5Ctimes%20%281-0.05%29%5E%7B25-3%7D)
= ![2300 \times 0.05^{3}\times 0.95^{22}](https://tex.z-dn.net/?f=2300%20%5Ctimes%200.05%5E%7B3%7D%5Ctimes%200.95%5E%7B22%7D)
= <u>0.093</u>
(b) P(X
3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= ![\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}](https://tex.z-dn.net/?f=%5Cbinom%7B25%7D%7B0%7D%20%5Ctimes%200.05%5E%7B0%7D%5Ctimes%20%281-0.05%29%5E%7B25-0%7D%2B%5Cbinom%7B25%7D%7B1%7D%20%5Ctimes%200.05%5E%7B1%7D%5Ctimes%20%281-0.05%29%5E%7B25-1%7D%2B%5Cbinom%7B25%7D%7B2%7D%20%5Ctimes%200.05%5E%7B2%7D%5Ctimes%20%281-0.05%29%5E%7B25-2%7D%2B%5Cbinom%7B25%7D%7B3%7D%20%5Ctimes%200.05%5E%7B3%7D%5Ctimes%20%281-0.05%29%5E%7B25-3%7D)
= ![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}](https://tex.z-dn.net/?f=1%20%5Ctimes%201%20%5Ctimes%200.95%5E%7B25%7D%2B25%20%5Ctimes%200.05%5E%7B1%7D%5Ctimes%200.95%5E%7B24%7D%2B300%20%5Ctimes%200.05%5E%7B2%7D%5Ctimes%200.95%5E%7B23%7D%2B2300%20%5Ctimes%200.05%5E%7B3%7D%5Ctimes%200.95%5E%7B22%7D)
= <u>0.966</u>
(c) P(X
4) = 1 - P(X < 4) = 1 - P(X
3)
= 1 - 0.966
= <u>0.034</u>
<u></u>
(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
= ![\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}](https://tex.z-dn.net/?f=%5Cbinom%7B25%7D%7B1%7D%20%5Ctimes%200.05%5E%7B1%7D%5Ctimes%20%281-0.05%29%5E%7B25-1%7D%2B%5Cbinom%7B25%7D%7B2%7D%20%5Ctimes%200.05%5E%7B2%7D%5Ctimes%20%281-0.05%29%5E%7B25-2%7D%2B%5Cbinom%7B25%7D%7B3%7D%20%5Ctimes%200.05%5E%7B3%7D%5Ctimes%20%281-0.05%29%5E%7B25-3%7D)
= ![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}](https://tex.z-dn.net/?f=25%20%5Ctimes%200.05%5E%7B1%7D%5Ctimes%200.95%5E%7B24%7D%2B300%20%5Ctimes%200.05%5E%7B2%7D%5Ctimes%200.95%5E%7B23%7D%2B2300%20%5Ctimes%200.05%5E%7B3%7D%5Ctimes%200.95%5E%7B22%7D)
= <u>0.688</u>
(e) The probability that none of the 25 boards is defective is given by = P(X = 0)
P(X = 0) = ![\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}](https://tex.z-dn.net/?f=%5Cbinom%7B25%7D%7B0%7D%20%5Ctimes%200.05%5E%7B0%7D%5Ctimes%20%281-0.05%29%5E%7B25-0%7D)
= ![1 \times 1\times 0.95^{25}](https://tex.z-dn.net/?f=1%20%5Ctimes%201%5Ctimes%200.95%5E%7B25%7D)
= <u>0.277</u>
(f) The expected value of X is given by;
E(X) = ![n \times p](https://tex.z-dn.net/?f=n%20%5Ctimes%20p)
=
= 1.25
The standard deviation of X is given by;
S.D.(X) =
= ![\sqrt{25 \times 0.05 \times (1-0.05)}](https://tex.z-dn.net/?f=%5Csqrt%7B25%20%5Ctimes%200.05%20%5Ctimes%20%281-0.05%29%7D)
= <u>1.089</u>