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

(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.

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} * p^(r)* (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} * 0.05^(3)* (1-0.05)^(25-3)`

=
`2300 * 0.05^(3)* 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} * 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)`

=
`1 * 1 * 0.95^(25)+25 * 0.05^(1)* 0.95^(24)+300 * 0.05^(2)* 0.95^(23)+2300 * 0.05^(3)* 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} * 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)`

=
`25 * 0.05^(1)* 0.95^(24)+300 * 0.05^(2)* 0.95^(23)+2300 * 0.05^(3)* 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} * 0.05^(0)* (1-0.05)^(25-0)`

=
`1 * 1* 0.95^(25)`

= 0.277

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

E(X) =
`n * p`

=
`25 * 0.05` = 1.25

The standard deviation of X is given by;

S.D.(X) =
`√(n * p * (1-p))`

=
`√(25 * 0.05 * (1-0.05))`

= 1.089