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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 0.1922. b. The approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is .10 is 0.0039.

Step by step solution

01

Identify the Parameters

Identify the parameters for this problem. In this case, \(n = 200\) which is the sample size, \(p_1 = .05\) for part a, \(p_1 = .10\) for part b - is the true proportion of defective units in the population and \(p = .02\) is the benchmark proportion that the manufacturer uses to decide whether to accept or reject the shipment.
02

Find the Standard Deviation

Find the standard deviation for the sample proportion. The formula to find the standard deviation of a proportion is \(\sqrt{P(1-P)/n}\), where \(P\) is the true population proportion and \(n\) is the sample size. In part a, \(\sigma_a = \sqrt{.05*(1-.05)/200} = 0.0343\) and in part b, \(\sigma_b = \sqrt{.10*(1-.10)/200} = 0.03\) .
03

Apply the Normal Distribution

Compute the z-values for the sample proportion of \(p = .02\) using the normal approximation formula \(z = (p-P)/\sigma\), where \(p\) is our bench-mark sample proportion, \(P\) is the actual population proportion and \(\sigma\) is standard deviation. So, in part a, \(z_a = (.02-.05)/.0343 = -0.87\) and in part b, \(z_b = (.02-.10)/.03 = -2.66\).
04

Calculate the Probability

Using a Z-table or calculator to look up the z-values and find the probabilities associated with each. Probability for part a, \(P(Z < -0.87) = .1922\). This is the probability that the shipment is returned when the true proportion of defective cartridges is .05. For part b, \(P(Z < -2.66) = .0039\). This is the probability that the shipment is not returned when the true proportion of defective cartridges is .10.

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Key Concepts

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

Sample Proportion
Understanding the sample proportion is essential when working with statistics, especially in quality control scenarios like the exercise at hand. The sample proportion, represented as \( \hat{p} \), is a fraction that indicates the number of successes (or defective items in this context) over the total number of trials in a sample. To put this into perspective, if we take a random sample of 200 ink cartridges and find 4 defective ones, the sample proportion would be \( \hat{p} = \frac{4}{200} = 0.02 \).

The exercise mentions a threshold sample proportion of 0.02, which the manufacturer uses as a criterion to accept or reject a whole shipment. It's this proportion that informs the decision making, and it is imperative that students understand how to compute and interpret the sample proportion in relation to the true, unknown proportion of defects in the entire shipment.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetrical and bell-shaped. It is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). In most cases, the mean represents the 'center' of the distribution, while the standard deviation measures the 'spread' or variability around the mean.

In our exercise, whether a shipment is returned is based on the normal approximation to the sampling distribution of the sample proportion. This sampling distribution becomes approximately normal thanks to the Central Limit Theorem, which states that with a large enough sample size, the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of our problem, the standard deviation of the sample proportion (\( \sigma_{\hat{p}} \)) is calculated with the formula \( \sqrt{P(1-P)/n} \), where \( P \) is the true population proportion and \( n \) is the sample size. This value helps us understand the variability of the sample proportion from the true population proportion.

For example, if we have a true proportion of defective cartridges of 0.05, the standard deviation of the sample proportion is \( \sigma_a = \sqrt{.05(1-.05)/200} \), which conveys how much the sample proportion can be expected to vary from the true population proportion. A smaller standard deviation indicates that the sample proportion is likely to be closer to the true population proportion.
Z-Value
The z-value, or z-score, is a statistical measure that describes a value's relationship to the mean of a group of values, standardized to the standard deviation. It is calculated using the formula \( z = (\hat{p}-P)/\sigma_{\hat{p}} \), with \( \hat{p} \) representing the sample proportion, \( P \) the true population proportion, and \( \sigma_{\hat{p}} \) the standard deviation of the sample proportion.

In our exercise's scenario, the z-value tells us how many standard deviations away the sample proportion \( \hat{p} = 0.02 \) is from the true proportion \( P \), and thus, whether the shipment is likely to be returned. A negative z-value suggests that the sample proportion is below the mean, and we use the standard normal distribution to find the probability associated with that z-value, which informs the decision to return the shipment. The calculation of the z-value is a critical step towards understanding the probability that a particular event will occur.

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Most popular questions from this chapter

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation \(5 .\) a. What are the mean and standard deviation of the \(\bar{x}\) sampling distribution? Describe the shape of the \(\bar{x}\) sampling distribution. \(\mu_{\bar{\lambda}}=40 \sigma_{\bar{x}}=0.625\) b. What is the approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

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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): \(\begin{array}{lllll}2,3,3^{*} & 2,3,4 & 2,3,4^{*} & 2,3^{*}, 4 & 2,3^{*}, 4^{*}\end{array}\) \(2,4,4^{*} \quad 3,3^{*}, 4 \quad 3,3^{*}, 4^{*} \quad 3,4,4^{*} \quad 3^{*}, 4,4^{*}\) 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?

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