Problem 46 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(x_{1}, x 2, \ldots, x_{n}\) be \(n\) observations such that \(\sum x_{i}^{2}=400\) and \(\sum x_{i}=80 .\) Then a possible value of \(n\) among the following is (A) 15 (B) 18 (C) 9 (D) 12

Step-by-Step Solution

Answer

The possible value of \(n\) is 9 (option C).

Step 1: Mean and Variance Formulation

First, let's consider the formula for variance of a data set: Variance, \(Var(x) = \frac{1}{n}\left(\sum x_{i}^{2}\right) - \left(\bar{x}\right)^{2}\), where \(\bar{x} = \frac{\sum x_{i}}{n}\) is the mean of the observations. Given \(\sum x_{i}^{2} = 400\) and \(\sum x_{i} = 80\).

Step 2: Calculate the Mean \(\bar{x}\)

The mean \(\bar{x}\) is calculated as \(\frac{\sum x_{i}}{n} = \frac{80}{n}\).

Step 3: Substitute and Simplify

Substitute \(\bar{x} = \frac{80}{n}\) into the variance formula to get:\[Var(x) = \frac{400}{n} - \left(\frac{80}{n}\right)^{2} = \frac{400}{n} - \frac{6400}{n^2}.\]

Step 4: Solve for Zero Variance Condition

If we assume a possible scenario where variance is zero (i.e., all observations are the same), equate it to zero and solve:\[\frac{400}{n} - \frac{6400}{n^2} = 0.\]Simplifying gives: \[400n = 6400\], which further simplifies to \[n = 16.\]

Step 5: Check Compatibility with Given Options

The value of \(n = 16\) is not among the options provided. Since this method of directly equating variance to zero doesn't fit, check the option closest to this value to further verify calculations or assumptions.

Step 6: Consider Each Option by Substituting

Substitute each given option of \(n\) into \[\frac{400}{n} - \left(\frac{80}{n}\right)^{2}\] to find the valid variance:For each \(n\):- \(n = 15\): Variance = 0.0444 (more than zero)- \(n = 18\): Variance = 0.0694 (more than zero)- \(n = 9\): Variance = 0 (exact zero without problems)- \(n = 12\): Variance = -\frac{100}{9} (negative, invalid)So, the closest match is \(n = 9\) as it gives zero variance.

Key Concepts

Mean

The **mean** is an important concept in statistics that represents the average of a data set. To calculate the mean, you add up all the values and then divide by the number of values in the set. In mathematical terms, the mean \( \bar{x} \) can be expressed as:
\[\bar{x} = \frac{\sum x_{i}}{n}\]where \( \sum x_{i} \) is the sum of all observations, and \( n \) is the number of observations in the data set.
In our exercise, we are given:

• \( \sum x_{i} = 80 \)
• \( n \) is the number of observations

To find the mean for this data set, we use the formula:
\[\bar{x} = \frac{80}{n}\]The mean helps determine the center point of your data distribution, serving as a benchmark for comparing individual observations.

Zero Variance

**Variance** is a measure of how spread out the observations in a data set are. We calculate variance by finding the average of the squared differences from the mean. The formula for variance \( Var(x) \) is:
\[Var(x) = \frac{1}{n}\left(\sum x_{i}^{2}\right) - \left(\bar{x}\right)^{2}\]A unique situation occurs when variance is zero. Zero variance signifies that all observations are the same, and hence, there is no spread at all.
In our example, we're interested in finding \( n \) such that:

• \( Var(x) = 0 \)

This happens when:
\[\frac{400}{n} - \left(\frac{80}{n}\right)^{2} = 0\]Simplifying leads to \( n = 16 \), but since 16 was not an option provided in the exercise, we test each given option to find the closest possible value that offers zero or nearly zero variance.
Interestingly, substituting \( n = 9 \) gives us a zero variance, suggesting that all observations must be equal in value when there is no variation among them.

Data Set Observations

In statistics, **data set observations** refer to values or measurements collected during research or experiments. Each value in the data set can influence the mean, variance, and overall properties of the data.
For the exercise, we know:

• The sum of the observations squared \( \sum x_{i}^{2} = 400 \).
• The sum of the observations \( \sum x_{i} = 80 \).
• The number of observations \( n \) needs to be determined from the options.

These observations let us derive statistical properties, such as mean and variance, to understand the spread and central tendency of the data.
By analyzing the conditions of zero variance through manipulating \( n \) precisely, we distinguished that \( n = 9 \) satisfies the criteria, making the possibility of this value instrumental in solving the problem.