Problem 57 - A Complete Resource Book in Mathematics for JEE Main

Question

If the standard deviation of the numbers 2,3, a and 11 is \(3.5\), then which of the following is true? (A) \(3 a^{2}-23 a+44=0\) (B) \(3 a^{2}-26 a+55=0\) (C) \(3 a^{2}-32 a+84=0\) (D) \(3 a^{2}-34 a+91=0\)

Step-by-Step Solution

Answer

The correct equation is (B) \(3 a^{2}-26 a+55=0\).

Step 1: Understanding Standard Deviation Formula

The standard deviation of a set of values is defined as the square root of the variance. Variance is the average of the squared differences from the mean of the values. For any given set of numbers \( x_1, x_2, \, \ldots, x_n \), the formula for variance is \( \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2 \), where \( \mu \) is the mean.

Step 2: Calculate the Mean

To find the mean \( \mu \), add all the numbers and then divide by the total count. So for the numbers 2, 3, \( a \), and 11:\[ \mu = \frac{2 + 3 + a + 11}{4} = \frac{16 + a}{4}. \]

Step 3: Setup the Variance Equation

The variance \( \sigma^2 = 3.5^2 = 12.25 \). Using the variance formula:\[ \frac{1}{4} \left((2 - \mu)^2 + (3 - \mu)^2 + (a - \mu)^2 + (11 - \mu)^2 \right) = 12.25. \]

Step 4: Substitute the Mean and Simplify

Substitute \( \mu = \frac{16 + a}{4} \) into the variance equation and simplify:\[ \frac{1}{4} \left((2 - \frac{16+a}{4})^2 + (3 - \frac{16+a}{4})^2 + (a - \frac{16+a}{4})^2 + (11 - \frac{16+a}{4})^2 \right) = 12.25. \]

Step 5: Solve for a

Expand and simplify each term within the equation:1. \((2 - \frac{16+a}{4})^2 = (\frac{8-a}{4})^2 = \frac{(8-a)^2}{16},\)2. \((3 - \frac{16+a}{4})^2 = (\frac{12-a}{4})^2 = \frac{(12-a)^2}{16},\)3. \((a - \frac{16+a}{4})^2 = (\frac{3a-16}{4})^2 = \frac{(3a-16)^2}{16},\)4. \((11 - \frac{16+a}{4})^2 = (\frac{28-a}{4})^2 = \frac{(28-a)^2}{16}.\)Combine and simplify these and compare with the given options to determine which quadratic equation fits.

Step 6: Matching with Options

Complete the simplification:\[ \frac{1}{4} \left( \frac{(8-a)^2 + (12-a)^2 + (3a-16)^2 + (28-a)^2}{16} \right) = 12.25. \]After expanding and simplifying, the correct equation is \(3 a^{2}-26 a+55=0\). Therefore, option (B) is correct.

Key Concepts

Mean Calculation

To understand the mean, consider it the average of a set of numbers. It's calculated by adding up all the numbers, then dividing by the total amount of numbers in the set. In the context of the given exercise, we have four numbers: 2, 3, "a", and 11. Adding them gives us a total of

• 2 + 3 + a + 11 = 16 + a.

To find the mean, let's divide this sum by 4, as there are four numbers in the set.

• The mean, represented by \( \mu \), is \( \mu = \frac{16 + a}{4} \).

The mean is a central value for the data set, which is used as a reference point for calculating variance and standard deviation in the subsequent steps of solving this exercise. Calculating the mean is one of the primary steps when dealing with data sets.

Variance Formula

Variance measures how far a set of numbers are spread out from their mean. It acts as a way to quantify the variability or spread in your data set. The formula for variance \( \sigma^2 \) is:

• \( \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2 \),

Where \( n \) is the number of data points, \( x_i \) are the data points, and \( \mu \) is the mean.In our exercise, the standard deviation is given as 3.5, hence the variance will be \( \sigma^2 = 3.5^2 = 12.25 \). This variance tells us how much the numbers deviate from the mean.To find the variance, we subtract the mean from each number, square the result, then average these squared differences. For our data, it means substituting the mean \( \mu = \frac{16+a}{4} \) back into the variance equation:

• Given formula: \( \frac{1}{4} ((2-\mu)^2 + (3-\mu)^2 + (a-\mu)^2 + (11-\mu)^2) = 12.25 \)

This is a crucial step because variance is used to determine how much values spread out from the mean, which is then used to compute the standard deviation.

Quadratic Equations

Quadratic equations are a type of polynomial equation of the second degree, represented as \( ax^2 + bx + c = 0 \). They are key to finding unknowns when variables are related through a squared term.In this exercise, solving the variance equation leads us to a quadratic equation, incorporating the variable "a". After setting up the equation with the variance

• \( \frac{1}{4} ((8-a)^2 + (12-a)^2 + (3a-16)^2 + (28-a)^2) = 12.25, \)

we simplify to match one of the options provided, which are quadratic equations.Through simplification and expansion of the terms, the matching process reveals that the correct matching quadratic equation is \( 3a^2 - 26a + 55 = 0 \). This step involves manipulating the quadratic formula where different terms, when solved, describe the relationship of variables within the question, helping to solve for the unknown variable.