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Chapter 8: Problem 31

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

Expert verified
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

01

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.
02

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.
03

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.
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The thickness (in millimeters) of the coating applied to disk drives is a characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness \((x)\) has a normal distribution with a mean of \(3 \mathrm{~mm}\) and a standard deviation of \(0.05\) \(\mathrm{mm}\). Suppose that the process will be monitored by selecting a random sample of 16 drives from each shift's production and determining \(\bar{x}\), the mean coating thickness for the sample. a. Describe the sampling distribution of \(\bar{x}\) (for a sample of size 16 ). b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(3 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 3 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(3 \pm 3 \sigma_{\bar{x}}\). (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{\bar{x}}\) is called a process control chart.) c. Referring to Part (b), what is the probability that a sample mean will be outside \(3 \pm 3 \sigma_{\bar{x}}\) just by chance (i.e., when there are no unusual circumstances)? d. Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coating thickness of \(3.05 \mathrm{~mm}\). What is the probability that a problem will be detected when the next sample is taken? (Hint: This will occur if \(\bar{x}>3+3 \sigma_{\bar{x}}\) or \(\bar{x}<3-3 \sigma_{\bar{x}}\) when \(\mu=\) 3.05.) b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(3 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 3 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(3 \pm 3 \sigma_{\bar{x}}\). (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{\bar{x}}\) is called a process control chart.)

Suppose that \(20 \%\) of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than \(.25 .\) What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only \(.20 ?\)

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there is a weight limit of \(2500 \mathrm{lb}\). Assume that the average weight of students, faculty, and staff on campus is \(150 \mathrm{lb}\), that the standard deviation is \(27 \mathrm{lb}\), and that the distribution of weights of individuals on campus is approximately normal. If a random sample of 16 persons from the campus is to be taken: a. What is the expected value of the sample mean of their weights? b. What is the standard deviation of the sampling distribution of the sample mean weight? c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of \(2500 \mathrm{lb} ?\) d. What is the chance that a random sample of 16 persons on the elevator will exceed the weight limit?

Consider the following population: \(\\{2,3,3,4,4\\}\). The value of \(\mu\) is \(3.2\), but suppose that this is not known to an investigator, who therefore wants to estimate \(\mu\) from sample data. Three possible statistics for estimating \(\mu\) are Statistic \(1:\) the sample mean, \(\bar{x}\) Statistic 2 : the sample median Statistic 3 : the average of the largest and the smallest values in the sample A random sample of size 3 will be selected without replacement. Provided that we disregard the order in which the observations are selected, there are 10 possible samples that might result (writing 3 and \(3^{*}, 4\) and \(4^{*}\) to distinguish the two 3 's and the two 4 's in the population): $$\begin{array}{rlllll} 2,3,3^{*} & 2,3,4 & 2,3,4^{*} & 2,3^{*}, 4 & 2,3^{*}, 4^{*} \\ 2,4,4^{*} & 3,3^{*}, 4 & 3,3^{*}, 4^{*} & 3,4,4^{*} & 3^{*}, 4,4^{*} \end{array}$$ For each of these 10 samples, compute Statistics 1,2, and 3\. Construct the sampling distribution of each of these statistics. Which statistic would you recommend for estimating \(\mu\) and why?

An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable \(x\) with a mean value of \(50 \mathrm{lb}\) and a standard deviation of \(20 \mathrm{lb}\). If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With \(n=100\), the total weight exceeds the limit when the average weight \(\bar{x}\) exceeds \(6000 / 100\).)
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