Question

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). (Round your probabilities to three decimal places.)
(a) Determine P(X <_ 2).(b) Determine P(X >_ 5).(c) Determine P(1 <_ X <_ 4).(d) What is the probability that none of the 25 boards is defective?

Answer 1

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:

$P(X=x) = {n\choose x}p^{x}(1-)^{n-x};\ x=0,1,2,3...$

(a)

Compute the value of P (X ≤ 2) as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              $= {25\choose 0}(0.05)^{0}(1-0.05)^{25-0}+{25\choose 1}(0.05)^{1}(1-0.05)^{25-1}\\+{25\choose 2}(0.05)^{2}(1-0.05)^{25-2}\\=0.2774+0.3650+0.2305\\=0.8729$

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)

              $=1- {25\choose 0}(0.05)^{0}(1-0.05)^{25-0}-{25\choose 1}(0.05)^{1}(1-0.05)^{25-1}\\-{25\choose 2}(0.05)^{2}(1-0.05)^{25-2}-{25\choose 3}(0.05)^{3}(1-0.05)^{25-3}\\-{25\choose 4}(0.05)^{4}(1-0.05)^{25-4}\\=1-0.2774-0.3650-0.2305-0.0930-0.027\\=0.0071$

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)

                   $={25\choose 1}(0.05)^{1}(1-0.05)^{25-1}+{25\choose 2}(0.05)^{2}(1-0.05)^{25-2}\\+{25\choose 3}(0.05)^{3}(1-0.05)^{25-3}+{25\choose 4}(0.05)^{4}(1-0.05)^{25-4}\\=0.3650+0.2305+0.0930+0.027\\=0.7155$

Thus, the value of P (1 ≤ X ≤ 4) is 0.716.

(d)

Compute the probability of X = 0 as follows:

$P(X=0) = {25\choose 0}(0.05)^{0}(1-0.05)^{25-0}\\=1\times1\times0.2773896\\=0.2773896\\\approx 0.277$

Thus, the probability that none of the 25 boards is defective is 0.277.

Answer 2

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;

$P(X=r) =\binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....$

(a) P(X < 2) = P(X = 0) + P(X = 1)

                   = $\binom{25}{0}0.05^{0}(1-0.05)^{25-0} + \binom{25}{1}0.05^{1}(1-0.05)^{25-1}$

                   = $1*1*0.95^{25} + 25*0.05*0.95^{24}$ = 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)  

= $\binom{25}{0}0.05^{0}(1-0.05)^{25-0} + \binom{25}{1}0.05^{1}(1-0.05)^{25-1} + \binom{25}{2}0.05^{2}(1-0.05)^{25-2} + \binom{25}{3}0.05^{3}(1-0.05)^{25-3}+\binom{25}{4}0.05^{4}(1-0.05)^{25-4} + \binom{25}{5}0.05^{5}(1-0.05)^{25-5}$

= $1*1*0.95^{25} + 25*0.05*0.95^{24} + 300*0.05^{2} *0.95^{23} + 2300*0.05^{3} *0.95^{22} + 12650*0.05^{4} *0.95^{21} + 53130*0.05^{5} *0.95^{20}$

= 0.99879

(c) P(1 < X < 4) = P(X = 2) + P(X = 3)

                         = $\binom{25}{2}0.05^{2}(1-0.05)^{25-2} + \binom{25}{3}0.05^{3}(1-0.05)^{25-3}$

                         = $300*0.05^{2} *0.95^{23} + 2300*0.05^{3} *0.95^{22}$ = 0.3235

(d) Probability that none of the 25 boards is defective = P(X = 0)

     P(X = 0) = $\binom{25}{0}0.05^{0}(1-0.05)^{25-0}$

                   = $1*1*0.95^{25}$ = 0.2774.