Question 1180785
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.