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The tar contents of 8 brands of cigarettes selected at random from the latest list released by the Federal Trade Commission are as follows: 7.3,8.6,10.4 \(16.1,12.2,15.1,14.5,\) and 9.3 milligrams. Calculate (a) the mean; (b) the variance.

Short Answer

Expert verified
The mean tar content is 11.6875 milligrams and the variance is 7.5734375 square milligrams.

Step by step solution

01

Calculate the Sum of the Data

First, sum up all the data points. This gives: \(7.3 + 8.6 + 10.4 + 16.1 + 12.2 + 15.1 + 14.5 + 9.3 = 93.5\) milligrams.
02

Calculate the Mean

Next, calculate the mean by dividing the sum from Step 1 by the number of data points. Here, there are 8 data points, so the mean is: \(93.5 / 8 = 11.6875\) milligrams.
03

Compute the Squared Differences from the Mean

Subtract the mean from each data point, square the result, and sum them all. This yields: \([(7.3 - 11.6875)^2 + (8.6 - 11.6875)^2 + ... + (9.3 - 11.6875)^2 = 60.5875\] square milligrams.
04

Calculate the Variance

Finally, compute the variance by dividing the result from Step 3 by the total number of data points: \(60.5875 / 8 = 7.5734375\) square milligrams.

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Key Concepts

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

Mean Calculation
The mean, commonly known as the average, is a fundamental aspect of descriptive statistics and is vital in data analysis. To find the mean of a data set, you sum up all the individual numbers and then divide that total by the number of data points. For example, in the context of tar contents in cigarette brands, if you have the values 7.3, 8.6, 10.4, 16.1, 12.2, 15.1, 14.5, and 9.3 milligrams, you first add them together to get 93.5 milligrams.

Since there are 8 brands in this sample, you divide 93.5 milligrams by 8, which gives you the mean tar content of 11.6875 milligrams. In simple terms, if each brand had the same amount of tar, 11.6875 milligrams would be that common value. Understanding mean calculation is crucial as it represents the central tendency of a data set, giving a quick snapshot of the data's middle ground.
Variance Calculation
While the mean provides a measure of the center of the data, the variance tells us about the spread of the data, or how much the data differs from the mean. Calculating variance involves several steps. After finding the mean of the data set, you subtract the mean from each data point to find the deviation. Then, you square these deviations to make them positive, add them all together, and finally, divide by the number of data points to normalize the sum.

For our cigarette tar content example, you would take each of the eight data points, subtract the mean (11.6875 milligrams), square these differences, and sum them to 60.5875 square milligrams. Dividing by the number of data points, which is 8, yields a variance of 7.5734375 square milligrams. This variance is a quantitative expression of variability in the data set, indicating how much the individual data points differ from the calculated mean.
Data Analysis
Data analysis is the broader process of inspecting, cleansing, transforming, and modeling data with the goal of discovering useful information, suggesting conclusions, and supporting decision-making. In the context of descriptive statistics, it often begins with calculating basic measures like the mean and variance, which provide a simple summary of the data.

Analyzing the tar content of cigarettes, we could use descriptive statistics to compare brands, identify trends in tar levels, or inform consumers. For instance, if certain brands consistently have tar contents significantly above the mean, these could be flagged as higher risk. Additionally, if the variance is high, it indicates a broad range of tar contents among different brands, which might lead to further investigation into the manufacturing processes. Ultimately, data analysis involving these statistical methods helps translate raw data into meaningful insights.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X\),, be a random sample from a distribution that can take on only positive values. Use the central limit theorem to produce an argument that if \(n\) is sufficiently large, then \(Y=X_{1} X_{2} \cdot X_{n}\) has approximately a log normal distribution.

The heights of 1000 students are approximately normally distributed with a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. If 200 random samples of size 25 are drawn from this population and the means recorded to the nearest tenth of a centimeter, determine (a) the mean and standard deviation of the sampling distribution of \(\bar{X}\); (b) the number of sample means that fall between 172.5 and 175.8 centimeters inclusive; (c) the number of sample means falling below 172.0 centimeters.

Two alloys \(A\) and \(B\) are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two in terms of maximum load capacity in tons. This is the maximum that can be tolerated without breaking. It is known that the two standard deviations in load capacity are equal at 5 tons each. An experiment is conducted in which 30 specimens of each alloy (A and B) are tested and the results gave $$\bar{x}_{A}=49.5, \quad \bar{x}_{B}=45.5 ; \quad \bar{X} A-\bar{x}_{B}=4$$ The manufacturers of alloy \(A\) are convinced that this evidence shows conclusively that \(\mu_{A}>\mu_{B}\) and strongly supports their alloy. Manufacturers of alloy \(B\) claim that the experiment could easily have given \(x_{A}-x_{B}=\) 4 even if the two population means arc: equal. In other words, "tilings are inconclusive!" (a) Make an argument that manufacturers of alloy \(B\) are wrong. Do it by computing $$P\left(\bar{X}_{A_{-}} X_{B}^{-}>4 \mid \mu_{A}=\mu_{B}\right).$$ (b) Do you think these data strongly support alloy \(A\) ?

A manufacturing firm claims that the batteries used in their electronic games will last an average of 30 hours. To maintain this average. L6 batteries are tested each month. If the computed /-value falls between \(-t_{0.025}\) and \(t_{0.025},\) the firm is satisfied with its claim. What conclusion should the firm draw from a sample that has a mean \(x=27.5\) hours and a standard deviation \(a=5\) hours? Assume the distribution of battery lives to be approximately normal.

The scores on a placement test given to college freshmen for the past five years are approximately normally distributed with a mean \(\mu=71\) and a variance \(a^{2}=8\). Would you still consider \(\sigma^{2}=8\) to be a valid value of the variance if a random sample of 20 students who take this placement test this year obtain a value of \(s^{2}=20 ?\)

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