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? 

(e) Calculate the expected value and standard deviation of X. (Round your standard deviation to two decimal places.)

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Answer 2

a)

Here, n = 25, p = 0.05, (1 - p) = 0.95 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 2).
P(X <= 2) = (25C0 * 0.05^0 * 0.95^25) + (25C1 * 0.05^1 * 0.95^24) + (25C2 * 0.05^2 * 0.95^23)
P(X <= 2) = 0.277 + 0.365 + 0.231
P(X <= 2) = 0.873

b)

Here, n = 25, p = 0.05, (1 - p) = 0.95 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >=5).
P(X <= 4) = (25C0 * 0.05^0 * 0.95^25) + (25C1 * 0.05^1 * 0.95^24) + (25C2 * 0.05^2 * 0.95^23) + (25C3 * 0.05^3 * 0.95^22) + (25C4 * 0.05^4 * 0.95^21)
P(X <= 4) = 0.277 + 0.365 + 0.231 + 0.093 + 0.027
P(X <= 4) = 0.993

P(x>=5) = 1 - P(x<=4)
= 1- 0.993 = 0.007

c)

Here, n = 25, p = 0.05, (1 - p) = 0.95, x1 = 1 and x2 = 4.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(1 <= X <= 4)
P(1 <= X <= 4) = (25C1 * 0.05^1 * 0.95^24) + (25C2 * 0.05^2 * 0.95^23) + (25C3 * 0.05^3 * 0.95^22) + (25C4 * 0.05^4 * 0.95^21)
P(1 <= X <= 4) = 0.365 + 0.231 + 0.093 + 0.027
P(1 <= X <= 4) = 0.716

d)

Here, n = 25, p = 0.05, (1 - p) = 0.95 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 25C0 * 0.05^0 * 0.95^25
P(X = 0) = 0.277

e)

Here, μ = n*p = 1.25,

σ = sqrt(np(1-p)) = 1.09