Acceptance sampling is an important quality control technique, where a batch of data is tested to determine if the proportion of units having a particular attribute exceeds a given percentage. Suppose that 13% of produced items are known to be nonconforming. Every week a batch of items is evaluated and the production machines are adjusted if the proportion of nonconforming items exceeds 17%. [You may find it useful to reference the z table.]
a. What is the probability that the production machines will be adjusted if the batch consists of 60 items? (Round final answer to 4 decimal places.)
b. What is the probability that the production machines will be adjusted if the batch consists of 120 items? (Round final answer to 4 decimal places.)
a. To find the probability that the production machines will be adjusted if the batch consists of 60 items, we need to find the probability that the proportion of nonconforming items exceeds 17%.
We can use the z-score formula to find the z-score corresponding to a proportion of 17%. The formula for the z-score is:
z = (p - P) / sqrt(P(1-P)/n)
Where p is the sample proportion (17% = 0.17), P is the population proportion (13% = 0.13), and n is the sample size (60).
Calculating the z-score:
z = (0.17 - 0.13) / sqrt(0.13(1-0.13)/60)
z = 0.04 / sqrt(0.13(0.87)/60)
z = 0.04 / sqrt(0.1121/60)
z = 0.04 / sqrt(0.0018683)
z = 0.04 / 0.0431661
z ≈ 0.9268
Using the z-table, we can find the probability associated with this z-score. Looking up z = 0.9268, we find that the corresponding probability is 0.8212.
Therefore, the probability that the production machines will be adjusted if the batch consists of 60 items is approximately 0.8212.
b. To find the probability that the production machines will be adjusted if the batch consists of 120 items, we follow the same steps as in part a, but with a sample size of 120.
Calculating the z-score:
z = (0.17 - 0.13) / sqrt(0.13(1-0.13)/120)
z = 0.04 / sqrt(0.13(0.87)/120)
z = 0.04 / sqrt(0.1121/120)
z = 0.04 / sqrt(0.0009342)
z = 0.04 / 0.0305367
z ≈ 1.3107
Using the z-table, we can find the probability associated with this z-score. Looking up z = 1.3107, we find that the corresponding probability is 0.9053.
Therefore, the probability that the production machines will be adjusted if the batch consists of 120 items is approximately 0.9053.
To solve this problem, we can use the normal approximation to the binomial distribution and the z-table.
a. To find the probability that the production machines will be adjusted if the batch consists of 60 items, we need to calculate the z-score for a proportion of 17% in a sample size of 60.
First, we calculate the mean (μ) and standard deviation (σ) for the sample proportion:
μ = n * p = 60 * 0.13 = 7.8
σ = √(n * p * (1 - p)) = √(60 * 0.13 * (1 - 0.13)) ≈ 2.828
Next, we calculate the z-score using the formula:
z = (x - μ) / σ
where x is the desired proportion, which is 17%.
z = (0.17 - 0.13) / 2.828 ≈ 0.04 / 2.828 ≈ 0.014
Using the z-table, we can find the probability associated with this z-score. The z-table gives us the area to the left of the z-score. Since we want the probability that the proportion exceeds 17%, we need to subtract the area to the left of the z-score from 1.
Using the z-table or a calculator with a z-table function, we find that the z-score of 0.014 corresponds to an area of approximately 0.5059. Therefore, the probability that the production machines will be adjusted if the batch consists of 60 items is approximately 1 - 0.5059 = 0.4941.
b. To find the probability that the production machines will be adjusted if the batch consists of 120 items, we follow the same steps as in part a.
μ = n * p = 120 * 0.13 = 15.6
σ = √(n * p * (1 - p)) = √(120 * 0.13 * (1 - 0.13)) ≈ 4.105
z = (x - μ) / σ
z = (0.17 - 0.13) / 4.105 ≈ 0.04 / 4.105 ≈ 0.0097
Using the z-table or a calculator with a z-table function, we find that the z-score of 0.0097 corresponds to an area of approximately 0.5038. Therefore, the probability that the production machines will be adjusted if the batch consists of 120 items is approximately 1 - 0.5038 = 0.4962.
So, the probabilities are:
a. 0.4941 (rounded to 4 decimal places)
b. 0.4962 (rounded to 4 decimal places)
To solve this problem, we can use the binomial distribution formula. Let's break down the problem step by step.
a) What is the probability that the production machines will be adjusted if the batch consists of 60 items?
To find this probability, we need to calculate the probability of observing a certain number of nonconforming items in a batch of 60 items. The probability of a nonconforming item is given as 13%, which means the probability of a conforming item is 87%.
We want to determine the probability of observing more than 17% nonconforming items in the batch. To do this, we need to sum up the probabilities of observing 18, 19, 20, and so on, up to 60 nonconforming items.
Using the binomial distribution formula, the probability of observing exactly k successes (nonconforming items) in n trials (items in the batch) is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n-k)
where C(n, k) is the binomial coefficient, p is the probability of success (proportion of nonconforming items), and (1 - p) is the probability of failure (proportion of conforming items).
In this case, n = 60, p = 0.13, and (1 - p) = 0.87.
Using a Z-table, we can find the cumulative probability of observing 17 or fewer nonconforming items in the batch. The Z-score corresponding to 17 is calculated as:
Z = (x - np) / sqrt(np(1-p))
where x is the number of successes (17), n is the number of trials (60), and p is the probability of success (0.13).
Z = (17 - 60 * 0.13) / sqrt(60 * 0.13 * 0.87)
Now, we can use the Z-table to find the cumulative probability associated with the Z-score. Subtracting this probability from 1 will give us the probability of observing more than 17 nonconforming items in the batch.
Using Z = (17 - 60 * 0.13) / sqrt(60 * 0.13 * 0.87) and the Z-table, we find the cumulative probability to be P(X ≤ 17) = 0.9322.
Therefore, the probability of observing more than 17 nonconforming items in the batch (and adjusting the machines) is 1 - P(X ≤ 17) = 1 - 0.9322 = 0.0678.
So, the probability that the production machines will be adjusted if the batch consists of 60 items is approximately 0.0678 (rounded to 4 decimal places).
b) What is the probability that the production machines will be adjusted if the batch consists of 120 items?
We'll follow a similar approach to calculate this probability. The only difference is that n (number of trials) will be 120 instead of 60.
Using the same formula and steps as before, we find the cumulative probability associated with observing 35 or fewer nonconforming items in the batch.
Z = (35 - 120 * 0.13) / sqrt(120 * 0.13 * 0.87)
Using the Z-table, we find P(X ≤ 35) = 0.9831.
Therefore, the probability of observing more than 35 nonconforming items in the batch (and adjusting the machines) is 1 - P(X ≤ 35) = 1 - 0.9831 = 0.0169.
So, the probability that the production machines will be adjusted if the batch consists of 120 items is approximately 0.0169 (rounded to 4 decimal places).