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Chapter 15: Problem 32

Glucose Testing (continued). Shelia's measured glucose level one hour after having a sugary drink varies according to the Normal distribution with \(\mu=122 \mathrm{mg} / \mathrm{dL}\) and \(\sigma=12 \mathrm{mg} / \mathrm{dL}\). What is the level \(L\) such that there is probability only \(0.05\) that the mean glucose level of four test results falls above L? (Hint: This requires a backward Normal calculation. See page 89 in Chapter 3 if you need a review.)

Short Answer

Expert verified
The level \(L\) is approximately 131.87 mg/dL.

Step by step solution

01

Understand the Normal Distribution

We are given that Shelia's glucose levels follow a Normal distribution with mean \(\mu = 122\) mg/dL and standard deviation \(\sigma = 12\) mg/dL. We need to find the level \(L\) such that there is a 5% probability that the average glucose level of four tests exceeds \(L\).
02

Determine the Sampling Distribution

Since we are dealing with the mean of four tests, we use the sampling distribution of the sample mean. The mean of the sample mean is the same as the population mean, \(\mu = 122\), and the standard deviation of the sample mean \(\sigma_{\bar{x}}\) is given by \(\sigma / \sqrt{n}\), where \(n\) is the number of samples. Here, \(n = 4\), so \(\sigma_{\bar{x}} = 12 / \sqrt{4} = 6\) mg/dL.
03

Find the Z-Score Corresponding to the 5% Upper Tail

We need to find a Z-score such that 5% of the probability lies above it. Using a standard Normal distribution table or inverse methods, we find that the Z-score corresponding to the top 5% (or the 95th percentile) is approximately \(1.645\).
04

Calculate the Level \(L\) Using the Z-Score

Using the Z-score formula for the sample mean, \(Z = (\bar{X} - \mu) / \sigma_{\bar{x}}\), where \(\bar{X}\) is the mean value we are looking for, we set it equal to the Z-score calculated (\(1.645\)) and solve for \(\bar{X}\):\[ 1.645 = (L - 122) / 6 \]Solving for \(L\), we get:\[ L = 1.645 \times 6 + 122 = 131.87 \]
05

Conclude the Value of \(L\)

Thus, the level \(L\) is approximately \(131.87\) mg/dL. This means that there is a 5% probability that the mean glucose level of four test results falls above \(131.87\) mg/dL.

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Key Concepts

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

Standard Deviation
Standard deviation is a measure of the dispersion or spread in a set of values. In simpler terms, it tells us how much the individual data points differ from the average value. For example, in Shelia's glucose testing scenario, the standard deviation is given as \( \sigma = 12 \) mg/dL.
This implies that the glucose levels typically deviate 12 mg/dL from the mean glucose level, which is \( \mu = 122 \) mg/dL. A small standard deviation indicates that most data points are close to the mean, while a larger one means they are spread out over a wider range.
Understanding standard deviation is crucial when analyzing variability, as it helps identify how consistent or varied your data is.
Sample Mean
The sample mean, denoted as \( \bar{X} \), is the average of a set of data points collected from a sample. For Shelia, the goal was to determine the sample mean of her glucose levels from four tests.
The sample mean is calculated by adding all the sample data points together and dividing by the number of data points. So if her test results were four readings, you would sum these readings and then divide by four.
In probability and statistics, the sample mean is essential as it serves as an estimate of the population mean.
This value allows us to infer details about the entire group based on a small, manageable selection.
Z-Score
A Z-score tells you how many standard deviations away a particular data point is from the mean of the data set.
In essence, it helps us understand the position of a specific data point within a distribution. For Shelia's glucose level problem, we needed to find a Z-score that indicates the top 5% cutoff.
This statistic is crucial to determine how unusual or typical a result is compared to the average. The formula for calculating the Z-score is: \[Z = \frac{(X - \mu)}{\sigma}\]where \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
For sampling scenarios, Z-scores help in finding critical values and understanding extremes.
Sampling Distribution
Sampling distribution refers to the probability distribution of a statistic obtained from a large number of samples drawn from a specific population. In Shelia's scenario, the sampling distribution of the sample mean was considered, as it shows how the mean of multiple glucose tests fluctuate.
One notable characteristic is that the mean of the sampling distribution is equal to the population mean; hence, it remains \( \mu = 122 \) mg/dL.
The standard deviation of the sampling distribution, known as the standard error, is calculated by \( \sigma / \sqrt{n} \), where \( n \) is the sample size. For Shelia's case, the standard error is \( 6 \) mg/dL.
Sampling distribution is foundational to making inferences as it allows us to estimate the probabilities of outcomes across different samples.

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Most popular questions from this chapter

An October 20, 2019, poll of Canadian adults who were registered voters found that \(31.6 \%\) said they would vote conservative in the October 2019 elections. Election records show that \(\mathbf{3 4 . 4} \%\) actually vot ed conservative. The boldface number is a a. sampling distribution. b. statistic. c. parameter.

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Playing the Numbers: The House Has a Business. Unlike Joe (see the previous exercise), the operators of the numbers racket can rely on the law of large numbers. It is said that the New York City mobster Casper Holstein took as many as 25,000 bets per day in the Prohibition era. That's 150,000 bets in a week if he takes Sunday off. Casper's mean winnings per bet are \(\$ 0.40\) (he pays out an average of 60 cents per dollar bet to people like Joe and keeps the other 40 cents). His standard deviation for single bets is about \(\$ 18.96\), the same as Joe's. a. What are the mean and standard deviation of Casper's average winnings \(x\) on his 150,000 bets? b. According to the central limit theorem, what is the approximate probability that Casper's average winnings per bet are between \(\$ 0.30\) and \(\$ 0.50\) ? After only a week, Casper can be pretty confident that his winnings will be quite close to \(\$ 0.40\) per bet.

Pollutants in Auto Exhausts. In 2017, the entire fleet of light-duty vehicles sold in the United States by each manufacturer was required to emit an average of no more than 86 milligrams per mile (mg/mi) of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) over the useful life (150,000 miles of driving) of the vehicle. NOX + NMOG emissions over the useful life for one car model vary Normally with mean \(80 \mathrm{mg} / \mathrm{mi}\) and standard deviation \(4 \mathrm{mg} / \mathrm{mi}\). a. What is the probability that a single car of this model emits more than \(86 \mathrm{mg} / \mathrm{mi}\) of NOX + NMOG? b. A company has 25 cars of this model in its fleet. What is the probability that the average NOX + NMOG level \(x\) of these cars is above \(86 \mathrm{mg} / \mathrm{mi}\) ?

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