Problem 22 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The following table gives the scores of 30 students in a mathematics examination: $$\begin{array}{lccccc}\hline \text { Scores } & 90-99 & 80-89 & 70-79 & 60-69 & 50-59 \\ \hline \text { Students } & 4 & 8 & 12 & 4 & 2 \\ \hline\end{array}$$ Find the mean and the standard deviation of the distribution of the given data.

Step-by-Step Solution

Answer

The mean of the distribution is \(77.167\) and the standard deviation is \(67.593\).

Step 1: Convert the given table to midpoints and frequencies

First, we will find the midpoint of each interval and create a new table with these midpoints as our new data points. The midpoint of each interval is given by the formula: Midpoint = \(\frac{lower\:limit + upper\:limit}{2}\) We can now represent the information in the given table as follows: $$\begin{array}{lccccc}\hline \text { Midpoints } & 94.5 & 84.5 & 74.5 & 64.5 & 54.5 \\\ \hline \text { Students } & 4 & 8 & 12 & 4 & 2 \\\ \hline\end{array}$$

Step 2: Calculate the mean of the data

To calculate the mean, we will use the formula: Mean = \(\frac{\sum (Midpoints * Frequencies)}{Total\:Frequencies}\) Total Frequencies = \(30\) Mean = \(\frac{(94.5*4)+(84.5*8)+(74.5*12)+(64.5*4)+(54.5*2)}{30}\) Mean = \(\frac{378+676+894+258+109}{30}\) Mean = \(\frac{2315}{30}= 77.167\)

Step 3: Calculate the standard deviation of the data

To calculate the standard deviation, we will use the formula: Standard Deviation = \(\sqrt{\frac{\sum (Midpoints^2 * Frequencies) - (\sum (Midpoints * Frequencies))^2 / Total\:Frequencies}{Total\:Frequencies - 1}}\) Variance = \(\frac{(94.5^2*4)+(84.5^2*8)+(74.5^2*12)+(64.5^2*4)+(54.5^2*2) - (2315)^2 / 30}{30 - 1}\) Variance = \(\frac{35422.5+57122+66561+16682.5+5961.5-1671655.67}{29}\) Variance = \(\frac{180750 - 1671655.67}{29}\) Variance = \(\frac{132494.33}{29} = 4568.77\) Standard Deviation = \(\sqrt{4568.77} = 67.593\) The mean of the distribution is \(77.167\) and the standard deviation is \(67.593\).

Key Concepts

Mean and Standard Deviation

Understanding the mean and standard deviation of a data set is fundamental in statistics. The mean, often referred to as the average, is calculated by summing up all the values in a dataset and then dividing by the number of values. In the context of the given problem, where student scores are grouped into intervals, we use the midpoint of each interval to represent the score values. The midpoints serve as estimates of the scores within each interval, which allows us to calculate the mean.
When analyzing data, knowing the mean is useful, but it does not give us information about the spread or variability of the data. That’s where standard deviation comes into play. Standard deviation is a measure of how much the values in a dataset deviate from the mean. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. In our exercise, we used the formula for standard deviation that takes into account each score's deviation from the mean, weighted by the frequency of students who earned scores in that interval. The formula incorporates squaring each deviation to ensure that positive and negative deviations do not cancel each other out, summing these squared deviations, and then dividing by one less than the total number of observations (n-1), which provides us with a sample standard deviation.

Frequency Distribution

A frequency distribution is a summary of how often various outcomes occur in a sample of data. It can be represented in a table, graph, or chart, which classifies data into different intervals or categories, showing the number of observations (frequency) within each category. In the context of the original problem, the table given lists score intervals and the frequency of students who achieved scores within those intervals. This is a typical example of a grouped frequency distribution.
The primary purpose of a frequency distribution is to provide a clear visual or tabular representation of the data, which can make it easier to see patterns, such as which scores are most or least common. In educational settings, frequency distributions can highlight how students perform relative to the rest of the class, reveal the difficulty of the test, or assist educators in grading on a curve. Understanding frequency distribution is essential in statistical data analysis as it provides the groundwork for calculating statistical measures like mean, median, mode, and standard deviation.

Midpoint Formula

The midpoint formula is a mathematical method used to find the exact halfway point, or the average, between two values. In statistics, particularly when dealing with frequency distributions and grouped data, it is common to use the midpoint formula to approximate the central value of an interval. The formula is given by
\(\frac{lower \text{ limit } + upper \text{ limit }}{2}\).
In our example exercise with student scores, each score interval has a lower and upper limit. Applying the midpoint formula to each interval allows us to estimate a single score that represents the whole interval. This estimation simplifies calculations when determining other statistical measures, such as the mean or the standard deviation. When we're working with a frequency distribution with class intervals, as in our exercise, the assumption is that frequencies are uniformly distributed within each interval, making the midpoint a reasonable estimate of the 'average' value for that interval.