Question

A manufacturer of computer printers purchases plastic ink cartridges from a vendor. When a large shipment is received, a random sample of 200 cartridges is selected, and each cartridge is inspected. If the sample proportion of defective cartridges is more than \(.02\), the entire shipment is returned to the vendor. a. What is the approximate probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is .05? b. What is the approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is .10?

Short answer

a. The approximate probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is .05 is 0.914. b. The approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is .10 is 0.991.

Step-by-step solution

Calculate Mean and Standard Deviation for scenario a

The mean (expected value) of a binomial distribution is \( n*p \), and the standard deviation is \( \sqrt{n*p*(1-p)} \). Here, \( n \) is the number of trials, and \( p \) is the probability of success. For this scenario a, \( n = 200 \) and \( p = 0.05 \). So the mean \( \mu = n*p = 200 * 0.05 = 10 \) and the standard deviation \( \sigma = \sqrt{n*p*(1-p)} = \sqrt{200*0.05*0.95} \approx 4.36.

Convert to Z-score and find the Probability for scenario a

We convert the sample proportion to Z-score using the formula \( Z = (x - \mu) / \sigma \). The number of defective cartridges that triggers a return is \( 0.02 * 200 = 4 \). So, \( Z = (4-10) / 4.36 = -1.37 \). Using Z-table, the probability that Z < -1.37 is about 0.086. Therefore, the probability that a shipment will be returned is 1-0.086=0.914.

Calculate Mean and Standard Deviation for scenario b

Using the same formulas as in Step 1 but this time \( p = 0.10 \). So the mean \( \mu = n*p = 200 * 0.10 = 20 \) and the standard deviation \( \sigma = \sqrt{n*p*(1-p)} = \sqrt{200*0.10*0.90} \approx 6.71.

Convert to Z-score and find the Probability for scenario b

Using the same procedures as in Step 2, but with the new mean and standard deviation, \( Z = (4-20) / 6.71 = -2.38 \). Using Z-table, the probability that Z < -2.38 is about 0.009. Therefore, the probability that a shipment will not be returned is 1-0.009=0.991.

Key concepts

Sample Proportion

The sample proportion is a statistic that estimates the proportion of units in a population that have a certain characteristic, based on a sample from that population. In this exercise, dealing with defective ink cartridges from a vendor, the sample proportion is used to decide whether a shipment should be returned. For instance, if out of a sample of 200 cartridges, more than 4 (which is 2% of the sample, calculated as 0.02 * 200) are found to be defective, the entire shipment is deemed unsatisfactory and is sent back.

Calculating the sample proportion involves finding the number of successful outcomes (defective cartridges, in this case), and dividing by the total number of trials (the size of the sample). When the manufacturer samples 200 cartridges and discovers, for example, 5 defects, the sample proportion of defects would be 5/200 or 0.025.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value), while a high standard deviation indicates that the values are spread out over a wider range. In the context of this binomial distribution problem, standard deviation quantifies the expected variability in the number of defective cartridges found in the samples.

To calculate the standard deviation of a binomial distribution, we use the formula: \( \sigma = \sqrt{n*p*(1-p)} \) where \( n \) is the number of trials (cartridges inspected), and \( p \) is the probability of finding a defective cartridge. For example, with a true defect rate of 5% (0.05), the standard deviation for a sample of 200 cartridges is \( \sqrt{200*0.05*0.95} \) which shows the typical fluctuation we can expect from the average number of defects.

Z-score Calculation

A Z-score is a statistical measure that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. It's a dimensionless quantity that can help compare different datasets and identify how unusual or typical a particular value is within the set. To find a Z-score, we subtract the mean from the value in question and then divide the result by the standard deviation.

For the binomial distribution problem, the Z-score calculation helps in determining how likely it is to observe a certain number of defective cartridges. The formula for Z-score in this context is: \( Z = (x - \mu) / \sigma \) with \( x \) representing the number of defective cartridges that would trigger a shipment return. The closer the Z-score is to 0, the closer \( x \) is to the mean. If the Z-score is significantly negative or positive, it indicates that the observed number of defects is unusually low or high, respectively. Using the Z-table, we can then determine the probability associated with that Z-score, which in turn informs the decision to return the shipment.