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A history test was taken by 51 students. The scores ranged from 50 to 95 and were classified into 8 classes of width 6 units. The resulting frequency distribution appears below. Find \(s^{2}\) by applying the definition for \(s^{2}\). Then find s. $$ \begin{array}{|c|c|c|c|} \hline \text { Class } & \text { Class Mark } \mathrm{X}_{\mathrm{i}} & \text { Frequency } \mathrm{f}_{\mathrm{i}} & \mathrm{X}_{\mathrm{i}} \mathrm{f}_{\mathrm{i}} \\ \hline 1 & 51 & 2 & 102 \\ \hline 2 & 52 & 3 & 171 \\ \hline 3 & 63 & 5 & 315 \\ \hline 4 & 69 & 8 & 552 \\ \hline 5 & 75 & 10 & 750 \\ \hline 6 & 81 & 12 & 972 \\ \hline 7 & 87 & 10 & 870 \\ \hline 8 & 93 & 1 & 93 \\ \hline & & 51 & 3825= \\ & & & \\ \sum_{i=1} & X_{i} f_{i} \\ \hline \end{array} $$

Step by step solution

01

Calculate the mean (\(\overline{X}\))

To calculate the mean, first, find the sum of \(\sum_{i=1}^nf_iX_i\), which is given as 3825 in the problem. Next, divide this sum by the total number of students, 51. \(\overline{X} = \frac{\sum_{i=1}^nf_iX_i}{n} = \frac{3825}{51}\)

02

Calculate each \((X_i - \overline{X})^2\)

We will now compute \((X_i - \overline{X})^2\) for each class mark, \(X_i\): \[ \begin{array}{|c|c|c|c|} \hline Class & X_i & (X_i - \overline{X})^2\\\ \hline 1 & 51 & (51 - \frac{3825}{51})^2\\\ \hline 2 & 52 & (52 - \frac{3825}{51})^2\\\ \hline 3 & 63 & (63 - \frac{3825}{51})^2\\\ \hline 4 & 69 & (69 - \frac{3825}{51})^2\\\ \hline 5 & 75 & (75 - \frac{3825}{51})^2\\\ \hline 6 & 81 & (81 - \frac{3825}{51})^2\\\ \hline 7 & 87 & (87 - \frac{3825}{51})^2\\\ \hline 8 & 93 & (93 - \frac{3825}{51})^2\\\ \hline \end{array} \]

03

Multiply each \((X_i - \overline{X})^2\) with their respective frequencies

Now, we will multiply each \((X_i - \overline{X})^2\) with their respective frequencies, \(f_i\): \[ \begin{array}{|c|c|c|} \hline Class & (X_i - \overline{X})^2 & f_i(X_i - \overline{X})^2\\\ \hline 1 & (51 - \frac{3825}{51})^2 & 2(51 - \frac{3825}{51})^2\\\ \hline 2 & (52 - \frac{3825}{51})^2 & 3(52 - \frac{3825}{51})^2\\\ \hline 3 & (63 - \frac{3825}{51})^2 & 5(63 - \frac{3825}{51})^2\\\ \hline 4 & (69 - \frac{3825}{51})^2 & 8(69 - \frac{3825}{51})^2\\\ \hline 5 & (75 - \frac{3825}{51})^2 & 10(75 - \frac{3825}{51})^2\\\ \hline 6 & (81 - \frac{3825}{51})^2 & 12(81 - \frac{3825}{51})^2\\\ \hline 7 & (87 - \frac{3825}{51})^2 & 10(87 - \frac{3825}{51})^2\\\ \hline 8 & (93 - \frac{3825}{51})^2 & 1(93 - \frac{3825}{51})^2\\\ \hline \end{array} \]

04

Add up all the multiplied values

Next, we will sum all the computed values from the previous step to obtain the numerator of variance. \[ \sum_{i=1}^n f_i(X_i - \overline{X})^2 = 2(51 - \frac{3825}{51})^2 + 3(52 - \frac{3825}{51})^2 + \dots + 1(93 - \frac{3825}{51})^2 \]

05

Divide the sum by the total number of students (51) to obtain \(s^2\)

We will now divide the sum from the previous step by the total number of students, 51, to obtain the variance (\(s^2\)): \[ s^2 = \frac{\sum_{i=1}^n f_i(X_i - \overline{X})^2}{n} = \frac{2(51 - \frac{3825}{51})^2 + 3(52 - \frac{3825}{51})^2 + \dots + 1(93 - \frac{3825}{51})^2}{51} \]

06

Calculate the standard deviation (s) by taking the square root of the variance (\(s^2\))

Finally, to find the standard deviation (s), we will take the square root of the variance (\(s^2\)): \[ s = \sqrt{s^2} = \sqrt{\frac{2(51 - \frac{3825}{51})^2 + 3(52 - \frac{3825}{51})^2 + \dots + 1(93 - \frac{3825}{51})^2}{51}} \] Upon solving these steps, we will obtain the values for the variance (\(s^2\)) and the standard deviation (s) for the given frequency distribution.

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Key Concepts

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

Standard Deviation

Standard deviation is an important concept in statistics that measures the amount of variation or dispersion in a set of data values. It provides a way to understand how much individual data points differ from the mean.

• A small standard deviation means that the data points tend to be close to the mean.
• A large standard deviation indicates that the data points are spread out over a wide range of values.

Calculating standard deviation involves finding the variance first, which is the average of the squared differences from the Mean. Only after obtaining the variance, we compute the standard deviation by taking the square root of this variance. In mathematical terms, if the variance is represented as \(s^2\), then the standard deviation is \(s = \sqrt{s^2}\). While variance provides a squared measure, the standard deviation offers a more intuitive scale similar to the original data, making it easier for comparison and interpretation.

Class Interval

Class interval is a range of values within which data points fall in a frequency distribution. This is particularly useful for organizing large datasets into manageable groups.
Understanding class intervals allows students to simplify complex datasets.

• The width of each class interval illustrates the span between the upper and lower boundary values.
• In this exercise, the scores are grouped into intervals of width 6 units.

Each class interval must be exclusive and cover all possible scores. These intervals help in visualizing and analyzing the distribution of data, making it easier to identify patterns and trends.

Frequency Distribution

Frequency distribution is a way of organizing data to show the number of times each distinct value occurs. It is represented in a tabular form where each category of data is listed along with its frequency—the number of times it appears.
This structure is beneficial because it offers a clear summary of the data and allows for quick analysis.

• In the given problem, each class with their frequency indicates how many students scored within the class mark range.
• It helps to understand the concentration of data points around certain values.

The frequencies in a distribution should always total the number of observations in the set. This distribution can be used to calculate other statistical measures like mean, variance, and standard deviation meaningfully.

Mean Calculation

Mean calculation refers to finding the average of a set of numbers, which gives an indication of the central tendency of the data. To calculate the mean from a frequency distribution, multiply each class mark by its respective frequency, sum these products, and then divide by the total number of observations.
Mathematically, this can be expressed as: \[ \overline{X} = \frac{\sum_{i=1}^n f_iX_i}{n} \] where \(X_i\) is the class mark, \(f_i\) is the frequency of the class, and \(n\) is the total number of observations.
Understanding how to compute the mean is a foundational skill that helps in further statistical analysis, allowing comparisons between different datasets and aiding in identifying the deviation and dispersion of the data around this central value.