Question

The frequency distribution of the marks obtained by 28 students in a test carrying 40 marks is given below. $$ \begin{array}{|l|c|c|c|c|} \hline \text { Marks } & 0-10 & 10-20 & 20-30 & 30-40 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Students } \end{array} & 6 & \mathrm{x} & \mathrm{y} & 6 \\ \hline \end{array} $$ If the mean of the above data is 20 , then find the difference between \(\mathrm{x}\) and \(\mathrm{y}\). (1) 3 (2) 2 (3) 1 (4) 0

Short answer

Answer: The difference between the number of students with marks in the range of 10-20 and 20-30 is 0.

Step-by-step solution

Understanding the problem

We have a frequency distribution table with 4 intervals, and we know the total number of students is 28. We need to find the values of x and y, given that the mean of the data is 20.

Calculating the midpoints of each interval

Let's find the midpoint of each interval to represent the marks of the students. Midpoints are: $$\frac{0+10}{2}=5,$$ $$\frac{10+20}{2}=15,$$ $$\frac{20+30}{2}=25,$$ and $$\frac{30+40}{2}=35$$.

Setting up the mean formula

Now that we have the midpoints, we can set up the formula for the mean: $$\text{Mean}=\frac{\sum (\text{frequency} \times \text{midpoint})}{\text{Total number of students}}$$ We know that the mean is 20, and the total number of students is 28. So, we can write the equation: $$20=\frac{6 \times 5 + x \times 15 + y \times 25 + 6 \times 35}{28}$$

Finding the sum of x and y

Since there are 28 students in total, we can write the equation: $$6+x+y+6=28$$ Solving for x and y, we get: $$x+y=16$$

Solving the system of equations

Now, we have two equations to solve for x and y: 1. \(6+x+y+6=28\) 2. \(20=\frac{6 \times 5 + x \times 15 + y \times 25 + 6 \times 35}{28}\) On multiplying equation 2 by 28, we get: 560=6×5 + 15x + 25y + 6×35 Solving these two equations, we get x=8 and y=8.

Finding the difference between x and y

We now found the values of x and y, so we just need to find the difference between them: $$\text{Difference} = |x - y|=|8-8|=0$$ The correct answer is (4) 0.

Key concepts

Mean Calculation

Mean calculation is a statistical process that determines the average value in a data set. In the context of frequency distribution, such as in this exercise, it is very similar to calculating the arithmetic mean but with a slight twist. We must consider both the frequency of each interval and the precise midpoint of the intervals.

The process works like this:

• First, identify the midpoints for each interval in the frequency distribution table. This represents the central value that all data points in the interval approximate.
• Multiply each midpoint by the frequency of the interval. This gives us the 'total score' or contribution of each interval segment towards the overall mean.
• Add up all of these 'total scores' for a grand total sum of scores across the intervals.
• Finally, divide this grand total by the total number of data entries (students, in this case) to find the mean.

These steps allow us to weigh the values according to how often each interval appears, hence getting a fair representation of the average. This method works especially well for grouped frequency distributions.

Solving Equations

Solving equations involves finding the values of variables that make an equation true. In our problem, we encountered two key equations: one that helped determine the sum of students in certain intervals, and another that leveraged the mean score to find frequency unknowns.

Let's break it down:

• First, we set up two equations based on given conditions: the total count of students must match the sum of frequencies, and the provided mean equation.
• The first equation ensures that the total frequency equals the number of students—simple addition here.
• The second equation equates the actual mean (average score) to the sum of products of midpoints and their frequencies, further divided by the total number of students.
• With these equations, you can substitute or manipulate the variables, forming a system of equations that requires solving simultaneously to find missing frequencies.

This type of problem often requires basic algebra skills, helping to reinforce the understanding of relationships between different expressions.

Interval Midpoints

Interval midpoints are crucial when analyzing frequency distributions, as they provide a representative value for each interval. This is especially important when exact data points are not available, and you have ranges instead.

Here's why interval midpoints matter:

• Since precise scores are grouped into intervals, midpoints act as a typical, average value for that range.
• To find a midpoint, simply take the average of the interval boundaries. For instance, for an interval of 10-20, the midpoint is \( \frac{10+20}{2}=15 \).
• Midpoints translate the interval data into a single value that can easily be used in further calculations, such as computing means.
• This approach simplifies computations and helps in accurately depicting the distribution's characteristics.

Midpoints thus serve as proxies that stand in for real scores, acting as anchors in any following mathematical processing.