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A machine at Keats Corporation fills 64-ounce detergent jugs. The probability distribution of the amount of detergent in these jugs is normal with a mean of 64 ounces and a standard deviation of \(.4\) ounce. The quality control inspector takes a sample of 16 jugs once a week and measures the amount of detergent in these jugs. If the mean of this sample is either less than \(63.75\) ounces or greater than \(64.25\) ounces, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 16 jugs, the inspector will conclude that the machine needs an adjustment when actually it does not?

Short Answer

Expert verified
The probability that the inspector will conclude that the machine needs an adjustment when it does not is 0.0124, or 1.24%.

Step by step solution

01

Calculate Standard Error

Firstly, the standard error of the sample mean is given by the formula \(\sigma /\sqrt{n}\) where \(\sigma\) is the population standard deviation and \(n\) is the sample size. In this case, \(\sigma = 0.4\) and \(n = 16\), so the standard error (SE) is \(0.4 /\sqrt{16} = 0.1\) ounces.
02

Determine Z-scores

Next, calculate the Z-scores for the lower and upper bounds of the sample mean given. A Z-score indicates how many standard deviations a value is from the mean. The formula for a Z-score is \((X - \mu)/SE\), where \(X\) is a value from the dataset, \(\mu\) is the mean and SE is standard error. Here, the lower Z-score will be \((63.75 - 64)/0.1 = -2.5\), and the upper Z-score will be \((64.25 - 64)/0.1 = 2.5\). Therefore, we want the proportion of the area that is less than -2.5 or more than 2.5.
03

Calculate Probability

The values of -2.5 and 2.5 correspond to certain probabilities in a standard normal distribution. Looking these values up in Z-table, we find that the cumulative probability associated with -2.5 is about 0.0062 and with 2.5 it’s about 0.9938. The region outside these two is the probability of the inspector incorrectly concluding the machine needs adjustment. The area in the tails will be \(1 - 0.9938 + 0.0062 = 0.0124\). Therefore, the probability that based on the sample of 16 jugs, the inspector will conclude that the machine needs an adjustment when in fact it doesn’t is 0.0124 or 1.24%.

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Key Concepts

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

Understanding Normal Distribution
The concept of a normal distribution is central to probability and statistics. A normal distribution is a bell-shaped curve that is symmetrical around the mean. Most data points cluster around the center, and the probabilities for values taper off equally in both directions as they move away from the mean. This distribution can be characterized by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).

A mean of **64 ounces** in the detergent jugs means that most of the jugs will have close to 64 ounces of detergent. The standard deviation of **0.4 ounces** tells us how much variation there is from the mean. Since this process follows a normal distribution, it allows us to calculate probabilities using Z-scores, aiding in decision-making like quality control checks.
Role of Standard Error in Sampling
The standard error (SE) is a measure of the variability of the sample mean from the population mean. It provides insight into how much the sample mean is expected to fluctuate.

The formula for standard error is given by:\[SE = \frac{\sigma}{\sqrt{n}}\]
In our example, with a standard deviation (\(\sigma\)) of **0.4** and a sample size (\(n\)) of **16**, the SE becomes **0.1** ounces. This small SE indicates that the sample means are close estimates of the population mean. Thus, SE is crucial in determining how representative the sample is of the entire population. It aids in assessing the reliability of conclusions drawn from sample data.
Interpreting Z-scores
A Z-score tells us how many standard deviations a particular value is from the mean. Calculating Z-scores helps to determine whether a sample mean falls within a typical range, or if it's unusually high or low. The formula used is:\[Z = \frac{X - \mu}{SE}\]
Given that the lower bound is **63.75 ounces** and the upper bound is **64.25 ounces**, we calculated Z-scores of **-2.5** and **2.5** respectively. These scores indicate how extreme a sample mean is. A high absolute Z-score (like 2.5) signifies that the value is fairly far from the mean, which alerts us to potential issues if other factors confirm it. Z-scores are vital in determining the likelihood of different outcomes in a normally distributed dataset.
Applying Quality Control Measures
Quality control involves ensuring products meet certain standards and specifications. In this context, it's about maintaining the correct detergent amount in each jug. By setting thresholds at **63.75** and **64.25** ounces, the quality control process becomes more rigorous.

The probability calculated based on these thresholds (about **1.24%**) represents how likely it is for auditors to wrongly conclude adjustments are needed when they're not. This small probability shows a high confidence level, ensuring that the machine operates correctly. Quality control, using statistical tools like normal distribution and Z-scores, ensures efficiency and consistency in manufacturing, minimizing defects and meeting customer expectations.

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Most popular questions from this chapter

According to the American Automobile Association's 2012 annual report Your Driving Costs, the cost of owning and operating a four-wheel drive SUV is $$\$ 11,350$$ per year (USA TODAY, April 27, 2012). Note that this cost includes expenses for gasoline, maintenance, insurance, and financing for a vehicle that is driven 15,000 miles a year. Suppose that the distribution of such costs of owning and operating all fourwheel drive SUVs has a mean of $$\$ 11,350$$ with a standard deviation of $$\$ 2390 .$$ Let \(\bar{x}\) be the average of such costs of owning and operating a four-wheel drive SUV based on a random sample of 400 four-wheel drive SUVs. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\).

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