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. $$s=\sqrt{\frac{n\left[\Sigma\left(f \cdot x^{2}\right)\right]-[\Sigma(f \cdot x)]^{2}}{n(n-1)}}$$ $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Blood Platelet } \\ \text { Count of Males } \end{array} & \text { Frequency } \\ \hline 0-99 & 1 \\ \hline 100-199 & 51 \\ \hline 200-299 & 90 \\ \hline 300-399 & 10 \\ \hline 400-499 & 0 \\ \hline 500-599 & 0 \\ \hline 600-699 & 1 \\ \hline \end{array} $$

Step by step solution

01

- Calculate Class Midpoints

The first step is to find the midpoints for each class. The midpoint (class midpoint) is calculated by adding the lower and upper class boundaries and then dividing by 2.\[\begin{array}{|c|c|c|}\hline \text{Class} & \text{Lower Bound} & \text{Upper Bound} & \text{Midpoint} \hline 0-99 & 0 & 99 & (0 + 99)/2 = 49.5 \hline 100-199 & 100 & 199 & (100 + 199)/2 = 149.5 \hline 200-299 & 200 & 299 & (200 + 299)/2 = 249.5 \hline 300-399 & 300 & 399 & (300 + 399)/2 = 349.5 \hline 400-499 & 400 & 499 & (400 + 499)/2 = 449.5 \hline 500-599 & 500 & 599 & (500 + 599)/2 = 549.5 \hline 600-699 & 600 & 699 & (600 + 699)/2 = 649.5 \hline\end{array}\]

02

- Calculate Class Midpoint Squared

Square the midpoints calculated in Step 1.\[\begin{array}{|c|c|c|c|}\hline \text{Class} & \text{Midpoint} & \text{Midpoint}^2 \hline 0-99 & 49.5 & 49.5^2 = 2450.25 \hline 100-199 & 149.5 & 149.5^2 = 22350.25 \hline 200-299 & 249.5 & 249.5^2 = 62250.25 \hline 300-399 & 349.5 & 349.5^2 = 122150.25 \hline 400-499 & 449.5 & 449.5^2 = 202050.25 \hline 500-599 & 549.5 & 549.5^2 = 302950.25 \hline 600-699 & 649.5 & 649.5^2 = 422850.25 \hline\end{array}\]

03

- Compute Frequencies of Midpoints and their Squares

Multiply the frequency by the midpoint and also by the midpoint squared for each class.\[\begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & \text{Frequency ( f )} & \text{Midpoint ( x )} & \text{f x } & \text{Midpoint}^2 & \text{f x^2 } \hline 0-99 & 1 & 49.5 & 1 \times 49.5 = 49.5 & 2450.25 & 1 \times 2450.25 = 2450.25 \hline 100-199 & 51 & 149.5 & 51 \times 149.5 = 7624.5 & 22350.25 & 51 \times 22350.25 = 1139862.75 \hline 200-299 & 90 & 249.5 & 90 \times 249.5 = 22455 & 62250.25 & 90 \times 62250.25 = 5602522.5 \hline 300-399 & 10 & 349.5 & 10 \times 349.5 = 3495 & 122150.25 & 10 \times 122150.25 = 1221502.5 \hline 400-499 & 0 & 449.5 & 0 & 202050.25 & 0 \hline 500-599 & 0 & 549.5 & 0 & 302950.25 & 0 \hline 600-699 & 1 & 649.5 & 1 \times 649.5 = 649.5 & 422850.25 & 1 \times 422850.25 = 422850.25 \hline\end{array}\]

04

- Calculate Total Frequency, Total of Frequency x and Total of Frequency x^2

Sum up the frequencies, frequency x and frequency x^2 .\[n = 1 + 51 + 90 + 10 + 0 + 0 + 1 = 153\sum (f x) = 49.5 + 7624.5 + 22455 + 3495 + 0 + 0 + 649.5 = 34273.5\sum (f x^2) = 2450.25 + 1139862.75 + 5602522.5 + 1221502.5 + 0 + 0 + 422850.25 = 8389188.25\]

05

- Substitute Values into Standard Deviation Formula

Use the formula for the population standard deviation s . The formula is given by \[ s = \sqrt{ \frac{n [\sum (f x^2)] - [\sum (f x)]^2 }{n(n-1)}} \text{Substitute} = \sqrt{ \frac{153 \times 8389188.25 - 34273.5^2}{153(153 - 1) }}= \sqrt{ \frac{1284862154.25 - 1174647282.25}{153 \times 152 }}= \sqrt{ \frac{110214872}{23256}}= \sqrt{4740.55}= 68.86\]

06

- Compare with given standard deviations

Given standard deviations: 11.5, 8.9, 59.5, and 65.4. The computed standard deviation (68.86) is relatively close to 59.5 and 65.4, suggesting higher variability in blood platelet counts.

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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 summary of how often different values occur within a dataset. It is essential for understanding data because it condenses large datasets into a more manageable form. Frequency distribution tables typically include class intervals, frequencies, and, sometimes, cumulative frequencies. For example, in the blood platelet count data, each class interval (e.g., 0-99) shows the range of values and their corresponding frequency.

• Class Interval: Range of values
• Frequency: Number of occurrences of values within that range

The table is very straightforward:
0-99: 1 occurrence
100-199: 51 occurrences
200-299: 90 occurrences
By organizing data this way, it becomes easier to analyze patterns and extract meaningful statistics such as the standard deviation.

Standard Deviation Calculation

Standard deviation is a measure of how spread out the values in a dataset are. It helps to understand the variability or dispersion of the data points.
The formula used for calculating the standard deviation from a frequency distribution is:
\ \[ s = \sqrt{ \frac{n [\sum (f \cdot x^2)] - [\sum (f \cdot x)]^2 }{n(n-1)} } \ \]
Here’s a step-by-step approach to calculating it:

• First, calculate the class midpoints and their squares.
• Next, compute the product of the frequencies with both the midpoints and their squares.
• Sum these products to obtain \(\sum (f \cdot x)\) and \(\sum (f \cdot x^2)\).
• Substitute these values into the formula to get the standard deviation.

Following these steps ensures you properly account for the distribution of data points when calculating the standard deviation.

Class Midpoints

Class midpoints are the central value of each class interval. They are calculated by taking the average of the lower and upper boundaries of each class. For example, for the class interval of 0-99:
Midpoint = (0 + 99) / 2 = 49.5

• 0-99 âž” Midpoint = 49.5
• 100-199 âž” Midpoint = 149.5
• 200-299 âž” Midpoint = 249.5

Midpoints simplify subsequent calculations, such as determining the product of frequencies and midpoints. They are essential for approximating the values within each class interval.

Squared Midpoints

Once the class midpoints are determined, the next step is to square each of these midpoints. Squaring the midpoints involves multiplying the midpoint by itself:
For example, 49.5^2 = 2450.25

• 49.5 âž” 2450.25
• 149.5 âž” 22350.25
• 249.5 âž” 62250.25

Squared midpoints play an important role in variance and standard deviation calculations, as they help quantify the dispersion of data around the mean.

Frequency and Midpoint Multiplication

In this step, multiply each frequency by its corresponding midpoint and by the squared midpoint:

• Frequency \( \times \) Midpoint (e.g., 1 \( \times \) 49.5 = 49.5)
• Frequency \( \times \) Midpoint^2 (e.g., 1 \( \times \) 2450.25 = 2450.25)

These products are critical because:

• Summing the products of frequencies and midpoints (\( \sum (f \cdot x)\)) provides a weighted measure of data centrality.
• Summing the products of frequencies and squared midpoints (\( \sum (f \cdot x^2)\)) offers insight into data dispersion.

These values are then plugged into the standard deviation formula to compute the final measure of spread in the data.