Question

A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.

a. 0.239 (binomial), 0.184 (Poisson)

b. 0.184 (binomial), 0.239 (Poisson)

c. 0.342 (binomial), 0.215 (Poisson)

d. 0.215 (binomial), 0.342 (Poisson)

Answer 1

The probability of having at most four defective bulbs in a string of 100 when using binomial distribution is about 0.8179, while using Poisson distribution the probability is approximately 0.8153, showing that both distributions yield similar results.

Step-by-step explanation:

When calculating the probability of having at most four defective bulbs in a string of 100, we can use both the binomial distribution and the Poisson distribution as methods of approximation. For the binomial distribution, with a probability of success (defective bulb) p = 0.03, and n = 100 trials (bulbs), the probability is found using the cumulative binomial probability function:

P(x ≤ 4) = binomcdf(100, 0.03, 4) ≈ .8179

The Poisson distribution is applicable here because the probability of success is small (p = 0.03) and the number of trials is large (n = 100). When μ = np = 3, the corresponding calculation for the Poisson distribution is:

P(x ≤ 4) = poissoncdf(3, 4) ≈ .8153

The Poisson approximation to the binomial distribution is considered good in this scenario because the difference in probabilities is very small, only 0.0026.