Answer to Question #105534 in Statistics and Probability for Sydney Kazembe

Question

In a food production process, packaged items are sampled as they come off a production line: a random sample of 5 items from each production batch is checked to see if each is tightly parked. A batch is accepted if all 5 sample items are satisfactory, and rejected if there are 3 or more unsatisfactory packages in it, otherwise a further sample is taken before making a decision. If in fact the packing machine is giving 80% of items properly parked, what is the probability that this second sample will be necessary? The second sample also consists of 5 items. What is the probability that, out of the two samples (10 items), there are 9 satisfactory? What are the assumptions behind your calculations?

Answer 1

Solution:

Using the Bernoulli formula, we find the probabilities that 5 out of 5 and 4 out of 5 samples will be of high quality.

$$p_{m,n}\approx \frac {1}{\sqrt{npq}} \varphi{(x)}$$

$$\varphi {(x)}=\frac {1}{\sqrt{2 \pi}} \exp ^{\frac {-x^2}{2}}$$

$$x=\frac {m-np}{\sqrt {npq}}$$

Let m=5, n=5, p=0.8, q=0.2

$$x \approx 1.12$$

$$p_{5,5}\approx 0.239$$

Let m=4, n=5, p=0.8, q=0.2

$$x \approx 0$$

$$p_{4,5} \approx 0.447$$

Since either the first or second option is possible, the probability that a second measurement is not required is

$$0.239+0.447=0.686$$

The likelihood that a second check will be necessary will be 1-0.686=0.314

We find the probability that out of 10 packs, 9 will be packed correctly

$$C_n ^m\times p^m\times q^{n-m}$$

$$C_{10} ^9\times 0.8^9\times 0.2^1=0.053$$

Since the machine produces fully packaged products in 80% of cases, the probability of receiving a packaged product is 0.8, therefore, the probability of an unpacked product is 1-0.8 = 0.2.

Answer: 0.314; 0.053