91Ó°ÊÓ

Refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where \(x\) represents the class midpoint, \(f\) represents the class frequency, and \(n\) represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4. $$s=\sqrt{\frac{n\left[\Sigma\left(f \cdot x^{2}\right)\right]-[\Sigma(f \cdot x)]^{2}}{n(n-1)}}$$ $$ \begin{array}{c|c} \hline \begin{array}{c} \text { Blood Platelet } \\ \text { Count of } \\ \text { Females } \end{array} & \text { Frequency } \\ \hline 100-199 & 25 \\ \hline 200-299 & 92 \\ \hline 300-399 & 28 \\ \hline 400-499 & 0 \\ \hline 500-599 & 2 \\ \hline \end{array} $$

Step by step solution

01

- Identify Class Midpoints and Frequencies

First, calculate the midpoints for each class interval. The midpoint is the average of the lower and upper boundaries of each class interval.\[ \text{Midpoint}_i = \frac{\text{Lower Bound}_i + \text{Upper Bound}_i}{2} \]For each class: 100-199, 200-299, 300-399, 400-499, 500-599.

02

- Calculating Midpoints

Calculate each midpoint:\[ \text{Midpoint}_1 = \frac{100 + 199}{2} = 149.5 \]\[ \text{Midpoint}_2 = \frac{200 + 299}{2} = 249.5 \]\[ \text{Midpoint}_3 = \frac{300 + 399}{2} = 349.5 \]\[ \text{Midpoint}_4 = \frac{400 + 499}{2} = 449.5 \]\[ \text{Midpoint}_5 = \frac{500 + 599}{2} = 549.5 \]

03

- Find Frequency Midpoint Products

Multiply the midpoints by their respective frequencies (\(f \times x\)):\[ f_1 \times x_1 = 25 \times 149.5 = 3737.5 \]\[ f_2 \times x_2 = 92 \times 249.5 = 22954 \]\[ f_3 \times x_3 = 28 \times 349.5 = 9786 \]\[ f_4 \times x_4 = 0 \times 449.5 = 0 \]\[ f_5 \times x_5 = 2 \times 549.5 = 1099 \]

04

- Find Frequency Midpoint^2 Products

Multiply the midpoints squared by their respective frequencies (\(f \times x^2\)):\[ f_1 \times x_1^2 = 25 \times (149.5)^2 = 25 \times 22350.25 = 558756.25 \]\[ f_2 \times x_2^2 = 92 \times (249.5)^2 = 92 \times 62250.25 = 5727023 \]\[ f_3 \times x_3^2 = 28 \times (349.5)^2 = 28 \times 122151.25 = 3420235 \]\[ f_4 \times x_4^2 = 0 \times (449.5)^2 = 0 \]\[ f_5 \times x_5^2 = 2 \times (549.5)^2 = 2 \times 301950.25 = 603900.5 \]

05

- Calculate Sums

Compute the sum of all the frequencies (\(n\)), the sum of the products (\(\Sigma(f \times x)\)), and the sum of the squared products (\(\Sigma(f \times x^2)\)):\[ n = 25 + 92 + 28 + 0 + 2 = 147 \]\[ \Sigma(f \times x) = 3737.5 + 22954 + 9786 + 0 + 1099 = 37576.5 \]\[ \Sigma(f \times x^2) = 558756.25 + 5727023 + 3420235 + 0 + 603900.5 = 10394914.75 \]

06

- Calculate Standard Deviation

Use the formula to calculate the standard deviation:\[ s = \sqrt{\frac{n[\Sigma(f \times x^2)] - [\Sigma(f \times x)]^2}{n(n-1)}} \]Substitute the values:\[ s = \sqrt{\frac{147[10394914.75] - [37576.5]^2}{147(146)}} \]Calculate the numerator and the denominator:\[ \text{Numerator} = 147 \times 10394914.75 - 37576.5^2 = 1528005382.25 - 1411970476.25 = 116034906 \]\[ \text{Denominator} = 147 \times 146 = 21462 \]Finally,\[ s = \sqrt{\frac{116034906}{21462}} \approx \sqrt{5406.4} \approx 73.5 \]

07

- Compare Standard Deviations

Compare the computed standard deviation (73.5) with those given in Exercises 37-40: 11.5 years, 8.9 years, 59.5, and 65.4. The computed standard deviation (73.5) is significantly higher.

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Key Concepts

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

statistical formulas

Statistical formulas are essential tools in educational math, helping us make sense of data by providing a means to summarize and understand different aspects of data sets. In this example, we calculated the standard deviation using a specific formula. The standard deviation is a measure showing how spread out the data points are from the mean. It requires calculating midpoints, summing frequencies, and working with the squares of these intervals.
Your main formula here is:
\[s = \sqrt{\frac{n[\Sigma(f \times x^2)] - [\Sigma(f \times x)]^2}{n(n-1)}}\]
This formula might seem complex initially, but breaking it into clear steps, as shown in the exercise, makes it more approachable. Summing values and frequency products are necessary to find reliable descriptive statistics like standard deviation.

frequency distributions

Frequency distributions are a method used in statistics to display how often different values or ranges of values occur within a data set. In our example, the frequency distribution table shows the number of observations within predefined intervals (e.g., blood platelet counts between 100-199, 200-299, etc.).

Here is a reminder of the distribution used in this exercise:

• 100-199: 25 occurrences
• 200-299: 92 occurrences
• 300-399: 28 occurrences
• 400-499: 0 occurrences
• 500-599: 2 occurrences

Using frequency distributions allows us to organize data to observe patterns. For calculating the standard deviation, the frequencies help weight the midpoints appropriately.

class midpoints

Class midpoints are a key part of summarizing grouped data. A midpoint represents the center value of a class interval. They are calculated by averaging the lower and upper boundary values of each interval.
For instance, to find the midpoint of the class interval 100-199, we use:\[ \text{Midpoint}_1 = \frac{100 + 199}{2} = 149.5 \]
These midpoints are used in conjunction with frequencies to find the mean and variances required for computing the standard deviation. They act as representative values around which the data within that interval is centered.

educational math

Educational math involves using mathematical concepts to solve real-world problems effectively. By learning to calculate standard deviations and understand distributions, students gain valuable skills to analyze data sets. Such statistics are crucial for interpreting data in various academic fields.
Remember that every step in mathematical education builds on previous knowledge. Understanding midpoints, frequency distributions, and using statistical formulas helps solidify foundational math skills for more complex future challenges.

Your proficiency in these areas will boost your analytical capabilities and prepare you for more sophisticated tasks in statistics and beyond.