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The nicotine content in a single cigarette of a particular brand has a distribution with mean \(0.8 \mathrm{mg}\) and standard deviation \(0.1 \mathrm{mg}\). If 100 of these cigarettes are analyzed, what is the probability that the resulting sample mean nicotine content will be less than \(0.79\) ? less than \(0.77 ?\)

Short Answer

Expert verified
The probability that the sample mean nicotine content will be less than 0.79 mg is 0.1587, and less than 0.77 mg is 0.0013.

Step by step solution

01

Compute the standard error

To carry out these calculations, we first need to know the standard deviation of the sample mean, which is also called the standard error (SE). It can be found using the formula \(SE = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the number of observations. Substituting the given values, we get \(SE = \frac{0.1}{\sqrt{100}} = 0.01\) mg.
02

Compute the standard (Z) score

Next, calculate the standard (Z) score for each specified value. The standard score indicates how many standard errors a particular value is from the population mean. It is calculated as \(Z = \frac{\overline{x} - \mu}{SE}\), where \(\overline{x}\) is the specified value and \(\mu\) is the population mean. For \(\overline{x} = 0.79\) and \(0.77\), we get \(Z = \frac{0.79 - 0.8}{0.01} = -1.0\) and \(Z = \frac{0.77 - 0.8}{0.01} = -3.0\), respectively.
03

Find the probability using the standard normal distribution

According to the standard normal distribution, the probability for \(Z = -1.0\) is \(0.1587\) and for \(Z = -3.0\) is \(0.0013\). Hence, the probability that the sample mean nicotine content will be less than \(0.79\) mg is \(0.1587\), and less than \(0.77\) mg is \(0.0013\).

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Key Concepts

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

Standard Error
When dealing with sample data, the standard error (SE) plays a critical role in understanding the variance of the sample mean compared to the population mean. Essentially, it provides an estimate of how much the sample mean would vary if you repeatedly took samples from the same population. This concept is crucial when analyzing large data sets where actual computation of each data point's deviation from the average is impractical.

The formula for standard error is:
  • \(SE = \frac{\sigma}{\sqrt{n}}\)
Here, \(\sigma\) represents the standard deviation of the population, and \(n\) is the number of observations or sample size.

Using this formula, you can determine the precision of the sample mean. A smaller standard error indicates the sample mean is a more accurate reflection of the population mean. In our example, with a population standard deviation of \(0.1\) mg for nicotine content and a sample size of \(100\), the standard error is calculated as \(0.01\) mg.
Z Score
The Z score is a statistical measurement that describes a value's position within the context of the mean and standard deviation of a dataset. It helps in determining how unusual or typical a particular score is, given a normal distribution. A Z score allows comparisons between different data sets or scores from the same data set.

The formula to calculate the Z score is:
  • \(Z = \frac{\overline{x} - \mu}{SE}\)
Where \(\overline{x}\) is the observed sample mean, \(\mu\) is the population mean, and SE is the standard error.

In our nicotine example, to determine how far \(0.79\) mg and \(0.77\) mg are from the population mean of \(0.8\) mg, we calculate Z scores of \(-1.0\) and \(-3.0\) respectively. These Z scores suggest that \(0.79\) mg is one standard deviation less than the mean, and \(0.77\) mg is three deviations less, indicating a more uncommon outcome.
Standard Normal Distribution
The standard normal distribution is the idealized form of a probability distribution where all samples score in a bell curve shape. This specific type of normal distribution has a mean of 0 and a standard deviation of 1. It serves as a fundamental reference for many statistical operations, like hypothesis testing and confidence intervals.

By converting individual scores into Z scores, any normal distribution can be transformed into a standard normal distribution. This transformation allows the use of standard tables or software to find probabilities of obtaining specific values under the curve.

For instance, in our exercise, by using the standard normal distribution, we calculated the probabilities associated with Z scores \(-1.0\) and \(-3.0\). From these calculations, we found the probability of having a sample mean nicotine content of less than \(0.79\) mg is \(0.1587\), and less than \(0.77\) mg is \(0.0013\), showing how unusual these outcomes might be under normal conditions.

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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. 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 \(0.7 ?\)

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=\) 2 minutes and \(\sigma=0.8\) minute, respectively. a. If \(\bar{x}\) is the sample mean time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

Explain the difference between \(\sigma\) and \(\sigma_{\bar{x}}\) and between \(\mu\) and \(\mu_{\bar{x}}^{-\text {. }}\)

What is the difference between \(\bar{x}\) and \(\mu\) ? between s and \(\sigma\) ?

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm} ?\)

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