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Chapter 3: Problem 32

Find the mean of the data summarized in the frequency distribution. Also, compare the computed means to the actual means obtained by using the original list of data values, which are as follows: (Exercise 29) 36.2 years; (Exercise 30) 44. I years; (Exercise 31) 224.3; (Exercise 32) 255.I. $$\begin{aligned} &=\\\ &\begin{array}{|c|c|} \hline \begin{array}{c} \text { Blood Platelet } \\ \text { Count of Females } \\ (1000 \text { cells } / \mu L) \end{array} & \text { Frequency } \\ \hline 100-199 & 25 \\ \hline 200-299 & 92 \\ \hline 300-399 & 28 \\ \hline 400-499 & 0 \\ \hline 500-599 & 2 \\ \hline \end{array} \end{aligned}$$

Short Answer

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Computed Mean and Actual Mean comparison completed.

Step by step solution

01

- Determine the midpoints of each class interval

For each class interval, calculate the midpoint which is the average of the lower and upper bounds of the interval.
02

- Multiply midpoints by their corresponding frequencies

For each class interval, multiply the midpoint by the frequency of that interval to get the 'Weighted Frequency'.
03

- Sum all the weighted frequencies

Add up all the products obtained from Step 2 to get the total sum of the weighted frequencies.
04

- Sum all the frequencies

Add up all the frequencies to get the total number of data points.
05

- Calculate the mean

Divide the total sum of the weighted frequencies by the total sum of frequencies using the formula \(\text{Mean} = \frac{\text{Sum of weighted frequencies}}{\text{Sum of frequencies}}\).
06

- Compute the actual means using given data values

Sum the original data values provided in the question and divide by the number of values to get the actual mean.
07

- Compare the computed mean and actual mean

Explicitly compare the means obtained from the frequency distribution and the original data values.

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Key Concepts

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

Frequency Distribution
A frequency distribution is a summarized representation of data, showing the number of observations within particular intervals known as classes or categories. It's like organizing your bookshelf by genre and counting how many books you have in each genre. This helps in identifying patterns within the data and is especially useful when dealing with large datasets. Each class interval in a frequency distribution has a frequency, which tells us how many data points fall within that range.
Mean Calculation
The mean, often referred to as the average, of a data set is a measure of central tendency that gives us an idea of the 'center' or typical value in the dataset. In simple terms, the mean is calculated by adding up all the data points and dividing this total by the number of data points. However, when dealing with frequency distributions, we use a slightly different method. Instead of adding up each individual data point, we use class midpoints and their frequencies.
Weighted Frequencies
Weighted frequencies take into account both the class midpoints and their corresponding frequencies. For each class interval in a frequency distribution, we calculate the midpoint by averaging the lower and upper bounds of the interval. Then, we multiply these midpoints by the respective frequencies. This gives us a 'weighted frequency' for each class, which helps us in efficiently calculating the mean for a grouped data set. This way, we incorporate both the value and the frequency at which it occurs.
Comparative Analysis
Comparative analysis involves comparing the mean calculated from the frequency distribution with the mean obtained from the original data values directly. The goal is to check the accuracy and reliability of the mean derived from the frequency distribution. Sometimes, there might be slight differences due to rounding off in midpoints or variations in class widths. By performing a comparative analysis, we can understand how well the frequency distribution represents the original data and verify our calculations.

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Most popular questions from this chapter

The 20 brain volumes \(\left(\mathrm{cm}^{3}\right)\) from Data Set 8 "IQ and Brain Size" in Appendix B vary from a low of \(963 \mathrm{cm}^{3}\) to a high of \(1439 \mathrm{cm}^{3} .\) Use the range rule of thumb to estimate the standard deviation \(s\) and compare the result to the exact standard deviation of \(124.9 \mathrm{cm}^{3}\).

Let a population consist of the values 9 cigarettes, 10 cigarettes, and 20 cigarettes smoked in a day (based on data from the California Health Interview Survey). Assume that samples of two values are randomly selected with replacement from this population. (That is, a selected value is replaced before the second selection is made.) a. Find the variance \(\sigma^{2}\) of the population \(\\{9 \text { cigarettes, } 10 \text { cigarettes, } 20 \text { cigarettes }\\}\) b. After listing the nine different possible samples of two values selected with replacement, find the sample variance \(s^{2}\) (which includes division by \(n-1\) ) for each of them; then find the mean of the nine sample variances \(s^{2}\) c. For each of the nine different possible samples of two values selected with replacement, find the variance by treating each sample as if it is a population (using the formula for population variance, which includes division by \(n\) ); then find the mean of those nine population variances. d. Which approach results in values that are better estimates of \(\sigma^{2}:\) part (b) or part (c)? Why? When computing variances of samples, should you use division by \(n\) or \(n-1 ?\) e. The preceding parts show that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\). Is \(s\) an unbiased estimator of \(\sigma ?\) Explain.

z Scores Lebron James, one of the most successful basketball players of all time, has a height of 6 feet 8 inches, or \(203 \mathrm{cm} .\) Based on statistics from Data Set 1 "Body Data" in Appendix B, his height converts to the \(z\) score of 4.07 . How many standard deviations is his height above the mean?

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-I, where we found measures of center. Here we find measures of variation.) Then answer the given questions. Listed below are prices in dollars for one night at different hotels located on Las Vegas Boulevard (the "Strip"). How useful are the measures of variation for someone searching for a room? $$212 \quad 77 \quad 121 \quad 104 \quad 153 \quad 264 \quad 195 \quad 244$$

The geometric mean is often used in business and economics for finding average rates of change, average rates of growth, or average ratios. To find the geometric mean of \(n\) values (all of which are positive), first multiply the values, then find the \(n\) th root of the product. For a 6 -year period, money deposited in annual certificates of deposit had annual interest rates of \(5.154 \%, 2.730 \%, 0.488 \%, 0.319 \%, 0.313 \%,\) and \(0.268 \% .\) Identify the single percentage growth rate that is the same as the five consecutive growth rates by computing the geometric mean of \(1.05154,1.02730,1.00488,1.00319,1.00313,\) and 1.00268.
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