Problem 42 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A data consists of \(\mathrm{n}\) observations: \(x_{1}, x_{2}, \ldots, x_{n} .\) If \(\sum_{i=1}^{\mathrm{n}}\left(x_{i}+1\right)^{2}=9 \mathrm{n}\) and \(\sum_{i=1}^{\mathrm{n}}\left(x_{i}-1\right)^{2}=5 \mathrm{n}\) then the standard deviation of this data is: [Jan. 09, 2019 (II)] (a) 2 (b) \(\sqrt{5}\) (c) 5 (d) \(\sqrt{7}\)

Step-by-Step Solution

Answer

The standard deviation is \(\sqrt{5}\).

Step 1: Expand the Given Squared Terms

First, let's expand \((x_i + 1)^2\) and \((x_i - 1)^2\). \((x_i + 1)^2 = x_i^2 + 2x_i + 1\)\((x_i - 1)^2 = x_i^2 - 2x_i + 1\)

Step 2: Sum the Expanded Terms

Using the expansions from Step 1, sum over all \(n\) observations for each expression.\[\sum_{i=1}^{n} (x_i^2 + 2x_i + 1) = \sum_{i=1}^{n} x_i^2 + 2\sum_{i=1}^{n} x_i + n = 9n\]\[\sum_{i=1}^{n} (x_i^2 - 2x_i + 1) = \sum_{i=1}^{n} x_i^2 - 2\sum_{i=1}^{n} x_i + n = 5n\]

Step 3: Set Up Two Equations

Use the sums from Step 2 to set up two equations:1. \(\sum_{i=1}^{n} x_i^2 + 2\sum_{i=1}^{n} x_i + n = 9n\)2. \(\sum_{i=1}^{n} x_i^2 - 2\sum_{i=1}^{n} x_i + n = 5n\)

Step 4: Subtract and Solve for Data Sum

Subtract the second equation from the first:\(\left(\sum_{i=1}^{n} x_i^2 + 2\sum_{i=1}^{n} x_i + n\right) - \left(\sum_{i=1}^{n} x_i^2 - 2\sum_{i=1}^{n} x_i + n\right) = 9n - 5n\) This simplifies to:\(4\sum_{i=1}^{n} x_i = 4n\)So, \(\sum_{i=1}^{n} x_i = n\). This indicates the mean is \(1\).

Step 5: Solve for Sum of Squares

Substitute \(\sum_{i=1}^{n} x_i = n\) into either equation:For example, from the first equation: \(\sum_{i=1}^{n} x_i^2 + 2n + n = 9n\)This gives \(\sum_{i=1}^{n} x_i^2 = 6n\).

Step 6: Compute Variance and Standard Deviation

Use the formula for variance, \(\sigma^2 = \frac{1}{n}\left(\sum_{i=1}^{n} x_i^2 - \frac{(\sum_{i=1}^{n} x_i)^2}{n}\right)\):Given \(\sum_{i=1}^{n} x_i = n\), \((\sum_{i=1}^{n} x_i)^2 = n^2\), and already found \(\sum_{i=1}^{n} x_i^2 = 6n\), substitute into:\(\sigma^2 = \frac{1}{n}(6n - \frac{n^2}{n})\) \(\sigma^2 = 6 - 1 = 5\)Thus, the standard deviation is \(\sqrt{5}\).

Key Concepts

Mean

The mean of a dataset is one of the simplest yet most essential statistical concepts. It represents the average value of the data points. To calculate the mean, you add up all the values in your dataset and then divide by the number of values there.
For instance, if you have 5 observations: 1, 3, 5, 7, and 9, the mean would be \[ \text{Mean} = \frac{1 + 3 + 5 + 7 + 9}{5} = \frac{25}{5} = 5. \]In the context of the problem, after solving the equations, we find that the sum of our observations, \( \sum x_i \), equals \( n \). This directly implies that the mean value of the dataset is \( 1 \), since \[ \text{Mean} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{n}{n} = 1. \]The mean helps us understand what the 'center' of the dataset is, providing a single figure that summarizes the entire dataset.

Variance

Variance measures how much each number in a dataset varies from the mean and, consequently, for the dataset as a whole. A smaller variance indicates that the data points tend to be closer to the mean. The formula to compute variance is: \[ \sigma^2 = \frac{1}{n} \left( \sum_{i=1}^{n} x_i^2 - \frac{(\sum_{i=1}^{n} x_i)^2}{n} \right). \]In the given problem, we derived that the \( \sum_{i=1}^{n} x_i^2 = 6n \), which when used in the variance formula along with \( \sum_{i=1}^{n} x_i = n \), helps calculate variance as: \[ \sigma^2 = \frac{1}{n} \left( 6n - \frac{n^2}{n} \right) = \frac{1}{n} (6n - n) = \frac{5n}{n} = 5. \]Variance, thus, quantifies the spread of the data points and is fundamentally linked with standard deviation, helping to gauge variability.

Sum of Squares

The sum of squares is a crucial component in statistical computations, including calculating variance and standard deviation. It sums the squares of differences between each data point and the mean. This computation is useful as it exacerbates larger differences, making it easier to assess data variability.
In this exercise, we manipulated equations to find \( \sum_{i=1}^{n} x_i^2 = 6n \).This quantity is important because it lays the groundwork for calculating variance and, ultimately, standard deviation.
Generally, the sum of squares can be expressed as:\[ \text{Sum of squares} = \sum_{i=1}^{n} (x_i - \text{mean})^2, \]which involves squaring each observation's deviation from the mean and summing them.In our case, since we knew the sum of squares in terms of observations, it facilitated computing variance and led us to the final standard deviation. This not only highlights individual discrepancies from the mean, but it also emphasizes overall distribution behavior of data.