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. $$s=\sqrt{\frac{n\left[\Sigma\left(f \cdot x^{2}\right)\right]-[\Sigma(f \cdot x)]^{2}}{n(n-1)}}$$ $$ \begin{array}{|c|c|} \hline \text { Age (yr) of Best Actress } & \\ \text { When Oscar Was Won } & \text { Frequency } \\ \hline 20-29 & 29 \\ \hline 30-39 & 34 \\ \hline 40-49 & 14 \\ \hline 50-59 & 3 \\ \hline 60-69 & 5 \\ \hline 70-79 & 1 \\ \hline 80-89 & 1 \\ \hline \end{array} $$

Step by step solution

01

- Identify class midpoints

The class midpoint for each class interval is calculated as the average of the lower and upper bounds of the interval. For example, the midpoint for the 20-29 interval is \(\frac{20+29}{2} = 24.5\). Repeat this for all class intervals.

02

- Calculate midpoints and frequencies

Calculate the midpoints for all intervals and list them along with their frequencies: - 20-29: Midpoint = 24.5, Frequency = 29 - 30-39: Midpoint = 34.5, Frequency = 34 - 40-49: Midpoint = 44.5, Frequency = 14 - 50-59: Midpoint = 54.5, Frequency = 3 - 60-69: Midpoint = 64.5, Frequency = 5 - 70-79: Midpoint = 74.5, Frequency = 1 - 80-89: Midpoint = 84.5, Frequency = 1

03

- Compute total number of sample values

Sum the frequencies to find the total number of sample values, \(n\). \(n = 29 + 34 + 14 + 3 + 5 + 1 + 1 = 87\)

04

- Compute \( \Sigma (f \cdot x) \) and \( \Sigma (f \cdot x^2) \)

Calculate the sum of the products of frequency and midpoint for each class, and the sum of the products of frequency and the square of the midpoint for each class: \( \Sigma (f \cdot x) = 29 \cdot 24.5 + 34 \cdot 34.5 + 14 \cdot 44.5 + 3 \cdot 54.5 + 5 \cdot 64.5 + 1 \cdot 74.5 + 1 \cdot 84.5 \ = 710.5 + 1173 + 623 + 163.5 + 322.5 + 74.5 + 84.5 = 3151\) \( \Sigma (f \cdot x^{2}) = 29 \cdot 24.5^2 + 34 \cdot 34.5^2 + 14 \cdot 44.5^2 + 3 \cdot 54.5^2 + 5 \cdot 64.5^2 + 1 \cdot 74.5^2 + 1 \cdot 84.5^2 = 17420.25 + 40903.5 + 27711.5 + 8907.75 + 20811.25 + 5552.25 + 7140.25 = 128446.75\)

05

- Apply standard deviation formula

Use the formula to find the standard deviation: \(\sigma = \sqrt{\frac{n[\Sigma(f \cdot x^{2})] - [\Sigma(f \cdot x)]^{2}}{n(n-1)}} \) Substitute the values calculated in previous steps: \( n = 87 \), \( \Sigma (f \cdot x) = 3151 \), and \( \Sigma (f \cdot x^2) = 128446.75 \). \[s = \sqrt{\frac{87 \[128446.75\] - [3151]^2}{87 \cdot 86}} = \sqrt{\frac{11194792.25 - 9931411}{7482}} \approx \sqrt{\frac{1269381.25}{7482}} \approx \sqrt{169.67} \approx 13.03 \]

06

- Compare standard deviations

Compare the computed standard deviation with the standard deviations obtained by using Formula 3-4 for the original list of data values from Exercises 37-40: - Exercise 37: 11.5 years - Exercise 38: 8.9 years - Exercise 39: 59.5 years - Exercise 40: 65.4 years

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Key Concepts

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

statistics

Statistics is an integral part of understanding data and making informed decisions. It provides tools and methods to summarize, analyze, and interpret numerical information. One vital statistical measure is the standard deviation, which quantifies the amount of variation or dispersion in a dataset. Understanding standard deviation helps us comprehend how spread out the data values are from the mean (average).
In problems involving frequency distributions, as seen in the provided exercise, we resort to various statistical measures for detailed analysis. These measures include calculating class midpoints, sums of squares, and ultimately the standard deviation. When we compute the standard deviation, we're assessing the consistency of our data, allowing us to make better predictions and decisions based on the dataset's behavior.

frequency distribution

A frequency distribution is a way to display data that shows the number of observations (frequency) per interval or category. This distribution helps in identifying patterns and trends within the data.
In the given exercise, the frequency distribution is shown through age intervals of Oscar-winning actresses. Each age range (e.g., 20-29, 30-39) represents a class, and the corresponding number of winners in that range is the frequency. These frequencies help us calculate important summary statistics, including the midpoints and subsequently the standard deviation.
By structuring data into a frequency distribution, we can simplify complex datasets and make them more understandable, paving the way for more in-depth statistical analysis.

class midpoint

The class midpoint is a crucial concept in frequency distribution. It represents the middle value of a class interval, calculated as the average of the lower and upper bounds of the interval. This midpoint acts as a representative value for all data points in that class.
In the given exercise, the midpoint for the first class (20-29) is calculated as \( (20 + 29) / 2 = 24.5 \). Midpoints are similarly determined for all other classes. These values are essential when computing other statistics, such as the \(\Sigma(f \cdot x)\) and \(\Sigma(f \cdot x^2)\).
The class midpoints simplify the computation process and help in summarizing the data effectively. This summary is especially useful when dealing with large datasets.

sum of squares

The 'sum of squares' is a fundamental concept in statistics, referring to the sum of the squared deviations of individual values from the mean or other central value. In the context of the provided exercise, we compute the sum of squares in two ways: \(\Sigma(f \cdot x)\) and \(\Sigma(f \cdot x^2)\).
- \(\Sigma(f \cdot x)\) is the sum of the products of frequencies and class midpoints.
- \(\Sigma(f \cdot x^2)\) is the sum of the products of frequencies and the squares of the class midpoints.
These sums are then utilized in the standard deviation formula. They help quantify the total variability within the dataset. Understanding and calculating the sum of squares provide insight into the data's dispersion and are crucial in formulating statistical interpretations.