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According to the Environmental Protection Agency, chloroform, which in its gaseous form is suspected to be a cancer-causing agent, is present in small quantities in all the country's 240,000 public water sources. If the mean and standard deviation of the amounts of chloroform present in water sources are 34 and 53 micrograms per liter \((\mu g / \mathrm{L})\), respectively, explain why chloroform amounts do not have a normal distribution.

Short Answer

Expert verified
Chloroform amounts cannot be negative, so the distribution isn't normal.

Step by step solution

01

Understand Normal Distribution Characteristics

Recall that a normal distribution is symmetric around the mean, and in such a distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three standard deviations.
02

Check Distribution Feasibility

Given the mean is 34 micrograms per liter and the standard deviation is 53 micrograms per liter, we expect negative values if it follows a normal distribution. Since negative values are not possible for chloroform amounts in the water, this suggests the distribution is not normal.
03

Evaluate Real-World Constraints

Chloroform amounts cannot be negative. In a normal distribution with such a large standard deviation relative to the mean, a significant portion of data points would unrealistically be negative, violating the physical constraints of the problem.
04

Conclude Why Distribution is Not Normal

The large standard deviation compared to the mean implies a spread that includes negative values, which are not plausible. This asymmetry and the impossibility of negative concentrations confirm the distribution is not normal.

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Key Concepts

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

Normal distribution properties
Normal distribution, often called the bell curve, is a statistical concept where data tends to be around a central value with no bias left or right. This distribution is symmetric, meaning that the mean, median, and mode are all equal. One of the key properties of a normal distribution is that the majority of the data—around 68%—lies within one standard deviation of the mean. Additionally, about 95% of data falls within two standard deviations, and 99.7% within three standard deviations. This is often referred to as the Empirical Rule or the 68-95-99.7 Rule.

However, normal distribution predominantly applies to continuous data without natural boundaries. For instance, data involving measurements or scores without negative possibilities could defy normal distribution properties deduced purely mathematically, as it would involve values that are not realistic in the given context.
Chloroform in water
Chloroform is a chemical compound that can be found in minimal amounts in water sources. It's crucial to monitor its concentration as it poses health risks; in gaseous forms, it's suspected to be carcinogenic. This compound's presence in water is measured in micrograms per liter \(\mu g / \mathrm{L}\), which provides a quantifiable measure of its concentration.

Due to environmental regulations, such as guidelines by the Environmental Protection Agency, chloroform levels are monitored closely in public water sources. Understanding its distribution is vital for regulatory compliance and public health safety. Importantly, chloroform concentrations cannot be negative. Any distribution model applied must account for this limitation, reflecting realistic and non-negative values.
Statistical distribution evaluation
Evaluating a statistical distribution involves analyzing how data points are spread out from the mean. For some data, such as the concentration of chloroform in water, we use this evaluation to understand if the data fits a normal distribution or another type.

In real-world applications, it's common to encounter data that doesn't satisfy the symmetric nature of a normal distribution due to constraints, such as the non-negativity of chloroform levels. If a distribution's standard deviation is large relative to the mean, it suggests that many of the values are spread far from the mean, leading to possible negative values, which are not possible for measurements like chloroform concentration.

Through visual analysis, such as histograms or statistical tests, we can evaluate whether the data's distribution aligns with theoretical models or if adjustments or alternative models are necessary.
Standard deviation in context
Standard deviation is a measure that captures how spread out the values in a data set are around the mean. In simpler terms, it tells us how much the individual data points differ from the average. In the scenario of chloroform in water, knowing the mean concentration and the standard deviation helps us understand the variability of chloroform amounts across different water sources. A high standard deviation relative to the mean suggests wide variability among data points. However, when dealing with quantities that cannot be negative, as in this case, a large standard deviation becomes problematic. It indicates the presence of many hypothetical values below zero, which physically cannot exist. Such a discrepancy signifies that the data does not follow a normal distribution and calls for potentially using other statistical models to more accurately represent the data's nature.

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Most popular questions from this chapter

A machine produces bearings with mean diameter 3.00 inches and standard deviation 0.01 inch. Bearings with diameters in excess of 3.02 inches or less than 2.98 inches will fail to meet quality specifications. a. Approximately what fraction of this machine's production will fail to meet specifications? b. What assumptions did you make concerning the distribution of bearing diameters in order to answer this question?

Resting breathing rates for college-age students are approximately normally distributed with mean 12 and standard deviation 2.3 breaths per minute. What fraction of all college-age students have breathing rates in the following intervals? a. 9.7 to 14.3 breaths per minute b. 7.4 to 16.6 breaths per minute c. 9.7 to 16.6 breaths per minute d. Less than 5.1 or more than 18.9 breaths per minute

The manufacturer of a new food additive for beef cattle claims that \(80 \%\) of the animals fed a diet including this additive should have monthly weight gains in excess of 20 pounds. A large sample of measurements on weight gains for cattle fed this diet exhibits an approximately normal distribution with mean 22 pounds and standard deviation 2 pounds. Do you think the sample information contradicts the manufacturer's claim? (Calculate the probability of a weight gain exceeding 20 pounds.)

The following results on summations will help us in calculating the sample variance \(s^{2}\). For any constant \(c\), a. \(\sum_{i=1}^{n} c=n c\). b. \(\sum_{i=1}^{n} c y_{i}=c \sum_{i=1}^{n} y_{i}\). c. \(\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)=\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} y_{i}\). Use (a), (b), and (c) to show that $$s^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}=\frac{1}{n-1}\left[\sum_{i=1}^{n} y_{i}^{2}-\frac{1}{n}\left(\sum_{i=1}^{n} y_{i}\right)^{2}\right].$$

A pharmaceutical company wants to know whether an experimental drug has an effect on systolic blood pressure. Fifteen randomly selected subjects were given the drug and, after sufficient time for the drug to have an impact, their systolic blood pressures were recorded. The data appear below:$$\begin{array}{lllll} 172 & 140 & 123 & 130 & 115 \\ 148 & 108 & 129 & 137 & 161 \\ 123 & 152 & 133 & 128 & 142 \end{array}$$ a. Approximate the value of \(s\) using the range approximation. b. Calculate the values of \(\bar{y}\) and \(s\) for the 15 blood pressure readings. c. Use Tchebysheff's theorem (Exercise 1.32 ) to find values \(a\) and \(b\) such that at least \(75 \%\) of the blood pressure measurements lie between \(a\) and \(b\) d. Did Tchebysheff's theorem work? That is, use the data to find the actual percent of blood pressure readings that are between the values \(a\) and \(b\) you found in part (c). Is this actual percentage greater than \(75 \%\) ?

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