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).
Required:
a. Determine P(X=3).
b. Determine P(X≤3).
c. Determine P(X≥4).
d. Determine P(1≤X≤3)..
e. What is the probability that none of the 25 boards is defective?
f. Calculate the expected value and standard deviation of X.

Answer 1

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;

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

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

                   =  $2300 \times 0.05^{3}\times 0.95^{22}$

                   =  0.093

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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

=  $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

=  0.966

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

                    =  1 - 0.966

                    =  0.034

(d) P(1 ≤ X ≤ 3) =  P(X = 1) + P(X = 2) + P(X = 3)

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

=  $25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

=  0.688

(e) The probability that none of the 25 boards is defective is given by = P(X = 0)

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

                   =  $1 \times 1\times 0.95^{25}$

                   =  0.277

(f) The expected value of X is given by;

       E(X)  =  $n \times p$

                =  $25 \times 0.05$  = 1.25

The standard deviation of X is given by;

        S.D.(X)  =  $\sqrt{n \times p \times (1-p)}$

                     =  $\sqrt{25 \times 0.05 \times (1-0.05)}$

                     =  1.089