Question

records show that 3.8% of the fuses delivered to an arsenal are defective. in a random sample of 150 fuses, what is the probability that the number of defective fuses is:

a.) less than 5

b.) more than 10

c.) between 5 and 10

Answer 1

a)

Here, n = 150, p = 0.038, (1 - p) = 0.962 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 < 5) = (150C0 * 0.038^0 * 0.962^150) + (150C1 * 0.038^1 * 0.962^149) + (150C2 * 0.038^2 * 0.962^148) + (150C3 * 0.038^3 * 0.962^147) + (150C4 * 0.038^4 * 0.962^146)
P(X < 5) = 0.003 + 0.0177 + 0.0522 + 0.1017 + 0.1477
P(X < 5) = 0.3223

b)

Here, n = 150, p = 0.038, (1 - p) = 0.962 and x = 10
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 10).
P(X <= 10) = (150C0 * 0.038^0 * 0.962^150) + (150C1 * 0.038^1 * 0.962^149) + (150C2 * 0.038^2 * 0.962^148) + (150C3 * 0.038^3 * 0.962^147) + (150C4 * 0.038^4 * 0.962^146) + (150C5 * 0.038^5 * 0.962^145) + (150C6 * 0.038^6 * 0.962^144) + (150C7 * 0.038^7 * 0.962^143) + (150C8 * 0.038^8 * 0.962^142) + (150C9 * 0.038^9 * 0.962^141) + (150C10 * 0.038^10 * 0.962^140)
P(X <= 10) = 0.003 + 0.0177 + 0.0522 + 0.1017 + 0.1477 + 0.1703 + 0.1626 + 0.1321 + 0.0933 + 0.0581 + 0.0324
P(X <= 10) = 0.9711

P(X> 10) = 1 - P(x<-10)
= 1 - 0.9711
= 0.0289

c)

Here, n = 150, p = 0.038, (1 - p) = 0.962, x1 = 5 and x2 = 10.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(5 <= X <= 10)
P(5 <= X <= 10) = (150C5 * 0.038^5 * 0.962^145) + (150C6 * 0.038^6 * 0.962^144) + (150C7 * 0.038^7 * 0.962^143) + (150C8 * 0.038^8 * 0.962^142) + (150C9 * 0.038^9 * 0.962^141) + (150C10 * 0.038^10 * 0.962^140)
P(5 <= X <= 10) = 0.1703 + 0.1626 + 0.1321 + 0.0933 + 0.0581 + 0.0324
P(5 <= X <= 10) = 0.6488