Question

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let of defective boards in a random sample of size , so . a. Determine . b. Determine . c. Determine . d. What is the probability that none of the 25 boards is defective? e. Calculate the expected value and standard deviation of X

Answer 1

Answer:

(a) The value of P (X ≤ 2) is 0.8729.

(b) The value of P (X ≥ 5) is 0.0072.

(c) The value of P (1 ≤ X ≤ 4) is 0.7154.

(d) The probability that none of the 25 boards is defective is 0.2774.

(e) The expected value and standard deviation of X are 1.25 and 1.09 respectively.

Step-by-step explanation:

The random variable X is defined as the number of defective boards.

The probability that a circuit board is defective is, p = 0.05.

The sample of boards selected is of size, n = 25.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={25\choose x}0.05^{x}(1-0.05)^{25-x};\ x=0,1,2,3...$

(a)

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

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

$P(X\leq =x)=\sum\limits^{2}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.2774+0.3650+0.2305\\=0.8729$

Thus, the value of P (X ≤ 2) is 0.8729.

(b)

Compute the value of P (X ≥ 5) as follows:

P (X ≥ 5) = 1 - P (X < 5)

              $=1-\sum\limits^{4}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=1-0.9928\\=0.0072$

Thus, the value of P (X ≥ 5) is 0.0072.

(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)

                   $=\sum\limits^{4}_{x=1}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.3650+0.2305+0.0930+0.0269\\=0.7154$

Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.

(d)

Compute the value of P (X = 0) as follows:

$P(X=0)={25\choose 0}0.05^{0}(1-0.05)^{25-0}=1\times 1\times 0.277389=0.2774$

Thus, the probability that none of the 25 boards is defective is 0.2774.

(e)

Compute the expected value of X as follows:

$E(X)=np=25\times 0.05=1.25$

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{np(1-p)}=\sqrt{25\times 0.05\times (1-0.05)}=1.09$

Thus, the expected value and standard deviation of X are 1.25 and 1.09 respectively.