Question

1. Say 5% of circuit boards tested by a manufacturer are defective. Let Y be the number of defective boards in a random sample of size n = 25.

   1. What kind of random variable is Y ? In particular, write Y ~Distribution(p, n), where you fill in the correct distribution name and parameters p and n.

   2. Determine P (Y ≥ 5).

   3. DetermineP(1≤Y ≤4).

   4. What is the probability that none of the 25 boards are defective?

Answer 1

Y~Binomial(0.05,25)
p = 0.05 , n =25

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) = 0.2774 + 0.365 + 0.2305 + 0.093 + 0.0269
P(X <= 4) = 0.9928

P(Y > =5) = 1- P(Y < =4)
= 1 - 0.9928
= 0.0072

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 + 0.2305 + 0.093 + 0.0269
P(1 <= X <= 4) = 0.7154

c)

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) = 0.2774