Question

Mode for the following distribution is 22 and \(10>\mathrm{y}>\mathrm{x}\). find \(\mathrm{y}\). $$ \begin{array}{|l|c|c|c|c|c|c|} \hline \begin{array}{l} \text { Class } \\ \text { interval } \end{array} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & \text { total } \\ \hline \text { Frequency } & 5 & 8 & 10 & \mathrm{x} & \mathrm{y} & 30 \\ \hline \end{array} $$ (1) 2 (2) 5 (3) 3 (4) 4

Short answer

Answer: The value of y is 8.

Step-by-step solution

Find the missing frequencies

Since we know the total frequency is 30, we can write an equation to represent this information: \(5+8+10+x+y=30\)

Solve for x

We are given the mode is 22, which means that the highest frequency is from the 20-30 interval. Hence, the frequency of the 20-30 interval is the mode, and therefore the maximum frequency. So, x must be less than 10.

Calculate the value of x

Since x is less than 10, we can rewrite our equation from Step 1 as: \(5+8+10+x+(10-x) = 30\) Solve for x: \(23+x = 30\) \(x = 7\) So, x = 7.

Calculate the value of y

Now that we know the value of x, we can determine the value of y using the inequality given (\(10>\mathrm{y}>\mathrm{x}\)): \(10 > y > 7\) Since y is an integer value representing a frequency, the only possible value of y that satisfies this inequality is: \(y = 8\) Therefore, the correct option is: (2) 5

Key concepts

Frequency Distribution

Frequency distribution is a crucial tool in statistics that presents data in an organized manner, showing how frequently each value occurs. It's very helpful when analyzing large datasets, as it simplifies the data into understandable categories. In a frequency distribution table, like the one seen in the exercise, data is divided into class intervals, and the number of occurrences within each interval is tallied as the frequency.

Key insights include:

• A frequency distribution gives a comprehensive view of the data by illustrating the number of times each value or range of values occurs.
• It's an excellent way to see the shape and spread of data quickly.
• This method aids in identifying trends, such as the mode, which is the interval with the highest frequency.

In our example, the data is divided into class intervals like 0-10, 10-20, etc., with frequencies allocated accordingly. The significance of frequency distribution is its ability to highlight important characteristics of the dataset, such as central tendency and dispersion, in a concise form.

Class Interval

A class interval is a range within which data values fall. It's essentially a category, or a bin, that groups data points together when constructing a frequency distribution. Class intervals make analyzing data more meaningful by reducing large sets of numbers into smaller, more manageable figures.

Features of class intervals include:

• They must be mutually exclusive, meaning a particular data point cannot fall into more than one interval.
• Each class interval generally has the same width, but this isn't a strict rule, especially if data is skewed.
• An ideal class interval helps in depicting the data trend effectively, such as identifying the mode or median.

In the problem, the class intervals are given as 0-10, 10-20, and so on. Here, they play a significant role: with a mode of 22, it indicates that the majority of the data points lie in the 20-30 interval, impacting the frequency and ultimately the analysis.

Statistical Analysis

Statistical analysis is the science of collecting, exploring, and presenting large amounts of data to uncover patterns and trends. It allows us to make sense of datasets, enabling the interpretation of what's happening behind numerical data. This is a vital process in various fields such as finance, marketing, and research.

In the specific context of this exercise:

• We are tasked with determining unknown variables (x and y) by understanding the mode and solving algebraically.
• The mode, as a measure of central tendency, represents the most frequently occurring value in a dataset, crucially influencing how we interpret the class intervals' statistics.
• It's the careful analysis of these intervals and frequencies that allow solving for unknowns, showcasing the practical application of statistical methods for clear, actionable insights.

Through statistical analysis, we derived that the mode steered the initial assumption that x had to be lesser than the initial peak frequency, and eventually solving led us accurately to the values of both unknown frequencies, x and y.