Question 1180785: A standard production makes on the average 97% at some factory. A randomly selected batch of products consisting of 200 units is checked. If 7 or more non-standard products will be among them, the batch is rejected. Find the probability that: a) there will be 4 non-standard products in the batch; b) the batch of products will be accepted.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem using the binomial distribution (and its normal approximation for part b):
**a) Probability of 4 non-standard products:**
1. **Probability of a non-standard product:** If 97% are standard, then 3% are non-standard. So, p = 0.03.
2. **Number of trials:** n = 200
3. **Number of non-standard products:** k = 4
4. **Binomial probability formula:**
P(k) = (nCk) * p^k * (1-p)^(n-k)
Where nCk is the binomial coefficient "n choose k" (n! / (k! * (n-k)!))
5. **Calculation:**
P(4) = (200C4) * (0.03)^4 * (0.97)^196
P(4) = (200! / (4! * 196!)) * 0.00000081 * 0.00465
P(4) ≈ 0.168
**b) Probability the batch will be accepted:**
The batch is accepted if there are *less than* 7 non-standard products (i.e., 0, 1, 2, 3, 4, 5, or 6 non-standard products). Calculating this directly using the binomial formula would involve summing the probabilities of each of these outcomes, which is tedious. Instead, we can use the normal approximation to the binomial distribution.
1. **Mean and standard deviation:**
* μ = n * p = 200 * 0.03 = 6
* σ = sqrt(n * p * (1-p)) = sqrt(200 * 0.03 * 0.97) ≈ 2.41
2. **Continuity correction:** Since we're approximating a discrete distribution (binomial) with a continuous one (normal), we use a continuity correction. We want P(x < 7), which we approximate as P(x < 6.5) in the normal distribution.
3. **Z-score:**
z = (x - μ) / σ = (6.5 - 6) / 2.41 ≈ 0.21
4. **Probability:** Using a standard normal table or calculator, find the probability P(z < 0.21). This is approximately 0.5832.
**Answers:**
* a) The probability of 4 non-standard products is approximately 0.168.
* b) The probability that the batch will be accepted is approximately 0.5832.
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