A machine produces items of which 10% are defective. At the end of each hour, a sample of n items is inspected. If it contains no defectives, the machine is allowed to run for another hour, otherwise the machine is stopped.
Find the minimum value of n if the probability that the machine will be stopped at the end of the first hour is at least 0.3?
Answer: 4
How to calculate???
To calculate the answer, we can use the concept of binomial probability. In this case, the probability of the machine being stopped after an hour is directly related to the probability of finding a defective item in a sample of n items.
Let's break down the problem step by step:
1. We know that 10% of the items produced by the machine are defective. Therefore, the probability of an item being defective is 0.10, and the probability of an item being non-defective is 0.90.
2. Since the samples are taken at the end of each hour, we want to find the minimum sample size, denoted by n, so that the probability of finding no defective items in the sample is less than or equal to 0.30.
3. The probability of finding exactly k defective items in a sample of n items can be calculated using the binomial distribution formula: P(X = k) = (nCk) * (0.10)^k * (0.90)^(n-k), where nCk represents the number of ways to choose k items out of n items.
4. We want to find the minimum value of n such that the probability of finding no defective items (k = 0) in the sample is less than or equal to 0.30. Therefore, we need to find the largest value of n such that the cumulative probability from k = 0 to n is less than or equal to 0.30.
5. Using these calculations, we can find the minimum value of n to be 4.
To calculate the minimum value of n, we need to find the largest integer value of n for which the probability of finding no defective items in the sample is less than or equal to 0.3.
Let's break down the problem step by step:
1. The machine produces items of which 10% are defective. This means that the probability of finding a defective item in a sample is 10% or 0.1.
2. The probability of not finding any defective items in a sample of n items can be calculated using the binomial distribution formula:
P(X = 0) = (1 - p)^n
where p is the probability of finding a defective item (0.1) and n is the sample size.
3. We need to find the smallest value of n such that P(X = 0) ≤ 0.3.
(1 - 0.1)^n ≤ 0.3
4. Simplifying the equation:
0.9^n ≤ 0.3
5. Taking the logarithm of both sides to simplify the equation further:
log(0.9^n) ≤ log(0.3)
n * log(0.9) ≤ log(0.3)
6. Since log(0.9) is negative, we need to divide both sides of the equation by log(0.9) to maintain the inequality:
n ≤ log(0.3) / log(0.9)
7. Using a scientific calculator or software, we can calculate the value of log(0.3) and log(0.9):
n ≤ -0.5229 / -0.0458
8. Simplifying the right side of the equation:
n ≤ 11.4103
9. Since n must be an integer and represents the sample size, the minimum value of n is 4, as it is the largest integer less than or equal to 11.4103.
Therefore, the minimum value of n required in order for the probability that the machine will be stopped at the end of the first hour to be at least 0.3 is 4.