Answer:
97.4% probability
Step-by-step explanation:
Since there is only two possible outcome of inspecting the catridge : either defective or not, we can solve the problem using the binomial probability distribution (approximated to normal).
To approximate the binomial probability to normal, we find the expected value and the standard deviation of the probability of exactly x sucesses on n repeated trials with p probability
Expected values E(X) = np
Standard deviation √V(X) = √np(1-p)
using the z-score formula (X - μ)/σ, we can solve normally distributed problem.
Where μ is the mean and σ is the standard deviation. The Z-score is is the measure of how much a sample is from the mean. The p-value associated with this z-score which is the probability that the value of the measure is smaller than X can be checked on the z-score table.
FOR THIS QUESTION,
a random sample of 200 cartridges is selected
n = 200
Therefore If there are more than 0.02 × 200 = 4 defective, the sample will be returned.
To determine the approximate probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is 0.05
μ = E(X)
= 0.05 × 200 = 10
σ = √V(X) = √np(1-p)
= √200 × 0.05 × 0.95
= 3.08
This probability is 1 subtracted by the pvalue of Z when X =4
Z = (X - μ)/σ
Z= (4 - 10)/3.08
Z = -1.95
Z = -1.95 has a P-value of 0.026 (on the z-score table)
This means that there is a 1 - 0.026 = 0.974
= 97.4% probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is 0.05.
- its the answer