Problem 44 - A Complete Resource Book in Mathematics for JEE Main

Question

In a series of \(2 n\) observations, half of them equal \(a\) and remaining half equal \(-a\). If the standard deviation of the observations is 2 , then \(|a|\) equals (A) \(\frac{1}{n}\) (B) \(\sqrt{2}\) (C) 2 (D) \(\frac{\sqrt{2}}{n}\)

Step-by-Step Solution

Answer

|a| equals 2.

Step 1: Understanding the Problem

We have a total of \(2n\) observations. Half of these observations are equal to \(a\), and the other half are equal to \(-a\). The standard deviation of these observations is given as 2. We need to find the absolute value of \(a\), denoted as \(|a|\).

Step 2: Mean of the Observations

Since there are \(n\) observations of \(a\) and \(n\) observations of \(-a\), the mean \( \mu \) of these observations is calculated as:\[ \mu = \frac{n \cdot a + n \cdot (-a)}{2n} = \frac{0}{2n} = 0 \]

Step 3: Standard Deviation Formula

The standard deviation \( \sigma \) is given by the formula:\[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \]where \( N = 2n \) and \( \mu = 0 \). So here it simplifies to:\[ \sigma = \sqrt{\frac{1}{2n} \sum_{i=1}^{2n} x_i^2} \]

Step 4: Sum of Squares of Observations

Since half of the observations are \(a\) and the other half are \(-a\), the sum of squares \( \sum x_i^2 \) is calculated as:\[ n \cdot a^2 + n \cdot a^2 = 2n \cdot a^2 \]

Step 5: Calculating Variance

Substitute \( \sum x_i^2 = 2n \cdot a^2 \) into the standard deviation formula:\[ \sigma = \sqrt{\frac{1}{2n} \cdot 2n \cdot a^2} = \sqrt{a^2} = |a| \]

Step 6: Equating to Given Standard Deviation

We know from the problem that the standard deviation \( \sigma = 2 \). Therefore, we have:\[ |a| = 2 \]

Key Concepts

Mean of Observations

In the given exercise, the mean of observations plays a critical role in understanding the properties of the data set. When we talk about the mean, it is essentially the average value of all observations in a data series. Here, we have a series with half the observations equal to \(a\) and the remaining half equal to \(-a\). To find the mean \( \mu \), we use the formula:

• First, add all the values together.
• Next, divide the sum by the total number of observations.

For this specific problem, it turns out that the sum \( n \cdot a + n \cdot (-a) \) equals zero because each \(a\) cancels out with \(-a\). Thus, the mean is \(0\), which greatly simplifies later calculations.

Sum of Squares

To understand standard deviation and variance, the sum of squares is a crucial component. It involves squaring each observation, then summing them up. Squaring the observations removes any sign, turning negative values positive and ensuring that large deviations from the mean contribute more.In this data set, half the observations are \(a\) and the other half are \(-a\). For the sum of squares, we calculate:

• Square each positive and negative \(a\), which both result in \(a^2\).
• There are \(n\) observations of \(a\) and \(n\) of \(-a\), so the sum is \(n\cdot a^2 + n\cdot a^2 = 2n \cdot a^2\).

This summation shows how the calculations balance out, essential when moving on to variance and standard deviation.

Variance Calculation

Variance measures how spread out or dispersed the observations are from the mean. Calculated using the sum of squares, variance is the average of these squared deviations from the mean.For our problem, the formula for variance \(\sigma^2\) employs the sum of squares:

• Insert the sum of squares from previous computations, \(2n \cdot a^2\).
• Divide by the number of observations \(2n\) to find variance: \( \sigma^2 = \frac{2n \cdot a^2}{2n} = a^2 \).

This variance shows directly how much observations each vary from the mean. Since the variance in this problem boils down to \(a^2\), equating it to the square of the given standard deviation helps quickly determine \(|a| = 2\). This step is crucial for solving standard deviation in uniform data sets.