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Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. In words, define the random variable \(\overline{X}\) .

Short Answer

Expert verified
The random variable \( \overline{X} \) is the mean weight of a sample of 20 heads of lettuce.

Step by step solution

01

Define the Random Variable

A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In this context, we need to consider what \( \overline{X} \) represents.
02

Specify \\( \overline{X} \\\) Contextually

In this context, \( \overline{X} \) represents the average weight of a randomly selected sample of 20 heads of lettuce. \( \overline{X} \) is used to determine the central tendency or mean of the sample data.
03

Express \\( \overline{X} \\\) in Words

In words, \( \overline{X} \), the random variable, can be defined as the mean weight of a sample of 20 heads of lettuce selected at random from the population.

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Key Concepts

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

Sample Mean
In statistics, the sample mean is a fundamental concept that helps us to understand the average of a set of values drawn from a population. Let's break this down to make it clearer:
  • The "sample" refers to a subset chosen from a larger group, known as the population.
  • The "mean" is a measure of central tendency, giving us an idea of where the middle of the data tends to be.
When we talk about the sample mean, we consider it as the arithmetic average of the sample sizes. It is denoted by \( \overline{X} \) (pronounced "X-bar").
To compute the sample mean, sum up all the sample data values and then divide by the number of values. In mathematical terms, if you have a sample \( x_1, x_2, ..., x_n \), the sample mean \( \overline{X} \) is calculated as:\[ \overline{X} = \frac{x_1 + x_2 + \cdots + x_n}{n} \] The sample mean is crucial because it serves as an estimate of the population mean, providing insights into the overall characteristics of the population from which the sample was drawn.
Normal Distribution
The normal distribution is a common continuous probability distribution, often called a "bell curve" due to its shape. It is an essential concept in statistics for various reasons:
  • It describes how data values are distributed for many natural phenomena.
  • Many statistical tests and methodologies rely on the assumption of normality.
A normal distribution is symmetric around its mean, and its features are defined by two key parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).
This distribution has the following characteristics:
  • The mean, median, and mode are all equal and located at the center of the distribution.
  • About 68% of data points lie within one standard deviation from the mean.
  • Approximately 95% fall within two standard deviations, and 99.7% within three.
The scenario in the original exercise mentions that the population distribution of head weight is normal. This implies that the weights of the lettuce heads should distribute evenly around the mean weight, which can help in understanding and predicting data behavior.
Population Standard Deviation
Population standard deviation is a measure that quantifies the amount of variability or dispersion in a set of data. It is different from sample standard deviation, particularly with how we treat the data:
  • The population standard deviation (denoted as \( \sigma \)) considers every data point in the entire population, rather than just the sample.
  • This measure helps to understand how much individual data points deviate from the mean of the population.
To calculate the population standard deviation for a set of values \( x_1, x_2, ..., x_N \), where \( N \) is the size of the population, use the following formula:\[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \]Here, \( \mu \) is the population mean.
Understanding population standard deviation is crucial in statistical analysis because it provides insights into the reliability and stability of the data. In the original exercise, knowing the population standard deviation helps illustrate how data samples compare to the broader population, aiding in assessing distribution normality and variation intensity.

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Most popular questions from this chapter

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. In words, define the random variable \(X\) .

Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Identify the following: a. \(\overline{x}=\) b. \(s_{x}=\) C. \(n=\) d. \(n-1=\)

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. Using the same \(p^{\prime}\) and \(n=80,\) how would the error bound change if the confidence level were increased to 98\(\% ?\) Why?

Forbes magazine published data on the best small firms in 2012 . These were firms that had been publicly traded for at least a year, have a stock price of at least \(\$ 5\) per share, and have reported annual revenue between \(\$ 5\) million and \(\$ 1\) billion. The Table 8.13 shows the ages of the corporate \(\mathrm{CEOs}\) for a random sample of these firms. $$\begin{array}{|c|c|c|c|c|}\hline 48 & {58} & {51} & {61} & {56} \\ \hline 59 & {74} & {63} & {53} & {50} \\ \hline 59 & {60} & {60} & {57} & {46} \\\ \hline 55 & {63} & {57} & {47} & {55} \\ \hline 57 & {43} & {61} & {62} & {49} \\ \hline 67 & {67} & {55} & {55} & {49} \\ \hline\end{array}$$ Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution.

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.

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