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Suppose you want to create a mental picture of the relative frequency histogram for a large data set consisting of 1000 observations, and you know that the mean and standard deviation of the data set are 36 and \(3,\) respectively. a. If you are fairly certain that the relative frequency distribution of the data is mound-shaped, how might you picture the relative frequency distribution? (HINT: Use the Empirical Rule.) b. If you have no prior information concerning the shape of the relative frequency distribution, what can you say about the relative frequency histogram? (HINT: Construct intervals \(\bar{x} \pm k s\) for several choices of \(k .)\)

Short Answer

Expert verified
Please describe the mental picture of the relative frequency histogram for a mound-shaped distribution with a mean of 36 and a standard deviation of 3. Your mental picture of the relative frequency distribution for a mound-shaped histogram should be a bell-shaped curve with the majority of the data concentrated around the mean (36) and gradually decreasing as we move further away from the mean. Specifically, you should expect most of the data to be between 33 and 39 (68% of the data), with a few observations extending out to the 30-42 range (95% of the data), and very few beyond that (99.7% of the data within the 27-45 range).

Step by step solution

01

a. Mound-shaped distribution with Empirical Rule

Using the Empirical Rule, we know that for mound-shaped distributions: - About 68% of the data falls within 1 standard deviation of the mean, which means they fall in the interval \((36-3, 36+3)\) or \((33,39)\). - About 95% of the data falls within 2 standard deviations of the mean, which means they fall in the interval \((36-2*3, 36+2*3)\) or \((30,42)\). - About 99.7% of the data falls within 3 standard deviations of the mean, which means they fall in the interval \((36-3*3, 36+3*3)\) or \((27,45)\). So, your mental picture of the relative frequency distribution should be a mound-shaped histogram with the majority of the data concentrated around the mean (36), gradually decreasing as we move further away from the mean. Specifically, you should expect most of the data to be between 33 and 39, with a few observations extending out to the 30-42 range, and very few beyond that.
02

b. No information about the shape of distribution

In this case, since we do not have any prior information about the shape of the distribution, we can use intervals centered at the mean and extending for several multiples of the standard deviation. Let's consider three different intervals, where \(k\) represents the multiple of standard deviation: 1. \(\bar{x} \pm 1s\): The interval would be \((36-3, 36+3)\) or \((33,39)\). We don't know the percentage of data falling into this range, as the Empirical Rule doesn't apply here. 2. \(\bar{x} \pm 2s\): The interval would be \((36-2*3, 36+2*3)\) or \((30,42)\). Again, we don't know the percentage of the data falling into this range. 3. \(\bar{x} \pm 3s\): The interval would be \((36-3*3, 36+3*3)\) or \((27,45)\). We still don't know the percentage of the data falling into this range. As we can't make any specific claims about the percentage of data falling in each interval, the mental picture for this scenario would be more vague. You might imagine a histogram centered around the mean (36), but without the certainty of a mound-shaped pattern, we cannot predict the specific shape or distribution beyond the mean.

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Key Concepts

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

Relative Frequency Histogram
A relative frequency histogram is a powerful visual tool used to display the frequency of data points in a dataset relative to the total number of observations. Unlike a regular frequency histogram which shows the absolute number of occurrences, the relative frequency histogram shows the proportion or percentage of observations within each interval or bin.
This type of histogram is quite useful because it allows you to quickly and effectively compare datasets of different sizes by just focusing on the proportions, making it easier to understand the distribution pattern.
  • In our problem, the relative frequency histogram could be imagined using the Empirical Rule, creating clearly delineated bins such as n(33,39), (30,42), and (27,45) based on the multiples of the standard deviation from the mean.
  • This forms part of a mound-shaped distribution where the bulk of the observations cluster around the mean value, with less frequency as you move further away from it.
Mound-Shaped Distribution
Mound-shaped distribution, often referenced as the normal distribution, is one where the data is symmetrically distributed around the mean, forming a bell shape when plotted on a graph. This distribution is perfectly symmetrical and typically seen in natural phenomena.
According to the Empirical Rule:
  • Approximately 68% of data falls within one standard deviation.
  • About 95% is within two standard deviations.
  • Almost 99.7% is within three standard deviations.
In the exercise, a mound-shaped distribution is inferred if data behaves according to these proportions. The relative frequency histogram then would typically show a peak at the mean and tapering tails as you move to either side. In this specific case, with a mean of 36 and a standard deviation of 3, data likely concentrates around 33 to 39, then reduces as you move towards 30 and 42, shrinking even further out to 27 and 45.
Standard Deviation
Standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of values. It indicates how spread out the data points are from the mean. The higher the standard deviation, the more spread out the data.
In the context of this problem:
  • The standard deviation is given as 3, showcasing how far the data typically deviates from the mean of 36.
  • When visualizing data using a mound-shaped distribution, understanding the standard deviation is crucial as it helps determine the intervals where most of the data points lie (using the Empirical Rule).
The Empirical Rule applies directly here, helping us use the standard deviation to visualize how the data plots relative to the mean, predicting the concentration of values within set intervals. Thus, a standard deviation gives us insight into the data's compactness and can aid in comparing variability across different datasets.
Mean
The mean, commonly known as the average, is a central measure in statistics that sums up all data points and divides by the number of points. It's an important statistic that summarizes a dataset with one value.
  • In this problem, the mean is 36, representing the central tendency of the data.
  • The mean is the point around which the standard deviation is measured; thus, it sits at the peak of the mound-shaped distribution.
While offering a quick snapshot of the central value, it's important to note the mean might be influenced by extreme values, so knowing the distribution pattern (like a normal or skewed distribution) is also important for interpreting the mean accurately. It serves as the anchor point for plotting the distribution in a relative frequency histogram when using the Empirical Rule to predict data spread.

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Most popular questions from this chapter

Refer to Exercise \(2.17 .\) The percentage of iron oxide in each of five pottery samples collected at the Island Thorns site was: $$ \begin{array}{lllll} 1.28 & 2.39 & 1.50 & 1.88 & 1.51 \end{array} $$ a. Use the range approximation to find an estimate of s, using an appropriate divisor from Table 2.6 . b. Calculate the standard deviation \(s\). How close did your estimate come to the actual value of \(s\) ?

A company interested in lumbering rights for a certain tract of slash pine trees is told that the mean diameter of these trees is 14 inches with a standard deviation of 2.8 inches. Assume the distribution of diameters is roughly mound-shaped. a. What fraction of the trees will have diameters between 8.4 and 22.4 inches? b. What fraction of the trees will have diameters greater than 16.8 inches?

You are given \(n=8\) measurements: 3,2,5,6,4 4,3,5 a. Find \(\bar{x}\). b. Find \(m\). c. Based on the results of parts a and b, are the measurements symmetric or skewed? Draw a dotplot to confirm your answer.

A set of data has a mean of 75 and a standard deviation of \(5 .\) You know nothing else about the size of the data set or the shape of the data distribution. a. What can you say about the proportion of measurements that fall between 60 and \(90 ?\) b. What can you say about the proportion of measurements that fall between 65 and \(85 ?\) c. What can you say about the proportion of measurements that are less than \(65 ?\)

Suppose that some measurements occur more than once and that the data \(x_{1}, x_{2}, \ldots, x_{k}\) are arranged in a frequency table as shown here: $$ \begin{array}{cc} \text { Observations } & \text { Frequency } f_{i} \\ \hline x_{1} & f_{1} \\ x_{2} & f_{2} \\ \cdot & \cdot \\ \cdot & \cdot \\ \cdot & \cdot \\ x_{k} & f_{k} \end{array} $$ The formulas for the mean and variance for grouped data are \(\bar{x}=\frac{\sum x_{i} f_{i}}{n}\) $$ \text { where } n=\Sigma f_{i} $$ and $$ s^{2}=\frac{\sum x_{i}^{2} f_{i}-\frac{\left(\sum x_{i} f_{i}\right)^{2}}{n}}{n-1} $$ Notice that if each value occurs once, these formulas reduce to those given in the text. Although these formulas for grouped data are primarily of value when you have a large number of measurements, demonstrate their use for the sample \(1,0,0,1,3,1,3,2,3,0,\) 0,1,1,3,2 a. Calculate \(\bar{x}\) and \(s^{2}\) directly, using the formulas for ungrouped data. b. The frequency table for the \(n=15\) measurements is as follows: $$ \begin{array}{ll} x & f \\ \hline 0 & 4 \\ 1 & 5 \\ 2 & 2 \\ 3 & 4 \end{array} $$ Calculate \(\bar{x}\) and \(s^{2}\) using the formulas for grouped data. Compare with your answers to part a.

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