Question 1192123: When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: randomly select and test 46 batteries and determine whether each is within specifications. The entire shipment will be accepted if at most 2 of the batteries do not meet specifications. Suppose a certain shipment contains 7,000 batteries, and 2% of them do not meet specifications. What is the probability that this entire shipment of 7,000 batteries will be accepted? (Round your answer to three decimal places; add trailing zeros as needed.)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem using the binomial probability formula:
**1. Define the variables:**
* n = 46 (sample size)
* p = 0.02 (probability of a battery not meeting specifications)
* k = number of batteries not meeting specifications (0, 1, or 2 for acceptance)
**2. Calculate the probabilities:**
We need to calculate the probability of 0, 1, or 2 batteries not meeting specifications and then add these probabilities together. The binomial probability formula is:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where nCx is the number of combinations of n items taken x at a time (also written as "n choose x").
* P(0) = (46C0) * (0.02)^0 * (0.98)^46 ≈ 0.396
* P(1) = (46C1) * (0.02)^1 * (0.98)^45 ≈ 0.372
* P(2) = (46C2) * (0.02)^2 * (0.98)^44 ≈ 0.168
**3. Add the probabilities:**
P(acceptance) = P(0) + P(1) + P(2) ≈ 0.396 + 0.372 + 0.168 ≈ 0.936
**Answer:**
The probability that the entire shipment will be accepted is approximately 0.936.
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