Suppose that the probability that an item produced by a certain machine will be defective is 0.2. A sample of 12 items is chosen. Let X be the number of defectives among the selected 12 items.
(a) Find the probability of at most 1 defective item.
(b) Find the probability of at least 1 defective item
(c) Find the probability of 3 or 4 defective items
(d) Find the expected value and standard deviation.
To find the probabilities and expected value for the given scenario, we will use the concept of the binomial distribution since we are dealing with a fixed number of trials (12 items) and a binary outcome (defective or non-defective).
The probability of success (finding a defective item) is 0.2, and the probability of failure (finding a non-defective item) is 1 - 0.2 = 0.8.
(a) Probability of at most 1 defective item (X ≤ 1):
To find this probability, we need to calculate P(X = 0) + P(X = 1).
P(X = 0) = (12 choose 0) * (0.2^0) * (0.8^12) = 1 * 1 * (0.8^12) = 0.0687 (rounded to 4 decimal places)
P(X = 1) = (12 choose 1) * (0.2^1) * (0.8^11) = 12 * 0.2 * (0.8^11) = 0.2060 (rounded to 4 decimal places)
P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.0687 + 0.2060 = 0.2747 (rounded to 4 decimal places)
Therefore, the probability of at most 1 defective item is approximately 0.2747.
(b) Probability of at least 1 defective item (X ≥ 1):
To find this probability, we need to calculate 1 - P(X = 0).
P(X = 0) is already calculated as 0.0687 (rounded to 4 decimal places).
P(X ≥ 1) = 1 - P(X = 0) = 1 - 0.0687 = 0.9313 (rounded to 4 decimal places)
Therefore, the probability of at least 1 defective item is approximately 0.9313.
(c) Probability of 3 or 4 defective items (X = 3 or X = 4):
To find this probability, we need to calculate P(X = 3) + P(X = 4).
P(X = 3) = (12 choose 3) * (0.2^3) * (0.8^9) = 220 * 0.008 * (0.8^9) = 0.2362 (rounded to 4 decimal places)
P(X = 4) = (12 choose 4) * (0.2^4) * (0.8^8) = 495 * 0.0016 * (0.8^8) = 0.1540 (rounded to 4 decimal places)
P(X = 3 or X = 4) = P(X = 3) + P(X = 4) = 0.2362 + 0.1540 = 0.3902 (rounded to 4 decimal places)
Therefore, the probability of having 3 or 4 defective items is approximately 0.3902.
(d) Expected value and standard deviation:
The expected value (mean) of a binomial distribution is given by E(X) = n * p, where n is the number of trials and p is the probability of success.
E(X) = 12 * 0.2 = 2.4 (rounded to 1 decimal place)
The standard deviation (SD) of a binomial distribution is given by SD(X) = sqrt(n * p * q), where q is the probability of failure.
SD(X) = sqrt(12 * 0.2 * 0.8) ≈ 1.3856 (rounded to 4 decimal places)
Therefore, the expected value is approximately 2.4, and the standard deviation is approximately 1.3856.