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JEE Main 2021
Statistics & Probability
Statistics
Hard

Question

Let in a series of 2n observations, half of them are equal to a and remaining half are equal to -a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a 2 + b 2 is equal to :

Options

Solution

1. Key Concepts and Formulas

To solve this problem efficiently, a solid understanding of the definitions of mean and standard deviation, along with their properties under linear data transformations, is essential.

  • Mean (x\overline{x}): Measure of Central Tendency For a set of NN observations x1,x2,,xNx_1, x_2, \dots, x_N, the mean is given by: x=i=1NxiN\overline{x} = \frac{\sum_{i=1}^{N} x_i}{N}

  • Standard Deviation (σ\sigma): Measure of Data Spread The standard deviation quantifies the dispersion of data points around the mean. A commonly used computational formula is: σ=i=1Nxi2N(x)2\sigma = \sqrt{\frac{\sum_{i=1}^{N} x_i^2}{N} - (\overline{x})^2} Note that σ\sigma is always non-negative.

  • Properties of Data Transformation (Change of Origin): If a constant bb is added to each observation xix_i to form a new set xi=xi+bx'_i = x_i + b:

    • New Mean: The new mean x\overline{x}' is the old mean plus the constant: x=x+b\overline{x}' = \overline{x} + b
    • New Standard Deviation: The new standard deviation σ\sigma' remains unchanged: σ=σ\sigma' = \sigma

2. Step-by-Step Solution

Let's apply these fundamental principles to systematically solve the problem.

Step 2.1: Analyze the Original Series of Observations

We are given 2n2n observations. Specifically, nn of these observations are equal to aa, and the remaining nn observations are equal to a-a. The total number of observations, N=2nN = 2n.

  • Calculate the Mean of the Original Series (xold\overline{x}_{\text{old}}): We use the formula x=xiN\overline{x} = \frac{\sum x_i}{N}. First, let's find the sum of all observations: i=12nxi=(a+a++a)n times+((a)+(a)++(a))n times\sum_{i=1}^{2n} x_i = \underbrace{(a + a + \dots + a)}_{n \text{ times}} + \underbrace{((-a) + (-a) + \dots + (-a))}_{n \text{ times}} i=12nxi=(na)+(n(a))=nana=0\sum_{i=1}^{2n} x_i = (n \cdot a) + (n \cdot (-a)) = na - na = 0 Now, calculate the mean: xold=02n=0\overline{x}_{\text{old}} = \frac{0}{2n} = 0 Why: The observations are perfectly symmetric around zero. For every positive value aa, there is a corresponding negative value a-a, which cancels out in the summation, resulting in a mean of zero.

  • Calculate the Standard Deviation of the Original Series (σold\sigma_{\text{old}}): We use the computational formula σ=xi2N(x)2\sigma = \sqrt{\frac{\sum x_i^2}{N} - (\overline{x})^2}. Since we found xold=0\overline{x}_{\text{old}} = 0, the formula simplifies to σold=xi2N\sigma_{\text{old}} = \sqrt{\frac{\sum x_i^2}{N}}.

    First, let's find the sum of the squares of the observations, xi2\sum x_i^2: i=12nxi2=(a2+a2++a2)n times+((a)2+(a)2++(a)2)n times\sum_{i=1}^{2n} x_i^2 = \underbrace{(a^2 + a^2 + \dots + a^2)}_{n \text{ times}} + \underbrace{((-a)^2 + (-a)^2 + \dots + (-a)^2)}_{n \text{ times}} Since (a)2=a2(-a)^2 = a^2, this simplifies to: i=12nxi2=(na2)+(na2)=2na2\sum_{i=1}^{2n} x_i^2 = (n \cdot a^2) + (n \cdot a^2) = 2n a^2 Now, substitute this into the standard deviation formula: σold=2na22n\sigma_{\text{old}} = \sqrt{\frac{2n a^2}{2n}} σold=a2\sigma_{\text{old}} = \sqrt{a^2} σold=a\sigma_{\text{old}} = |a| Why: The standard deviation must always be a non-negative value, as it measures spread. The square root of a2a^2 is always the absolute value of aa, i.e., a|a|.

Step 2.2: Apply Transformations and Use Given Information

A constant bb is added to each of the original observations. Let the new observations be xi=xi+bx'_i = x_i + b. We are given that the mean of this new set (xnew\overline{x}_{\text{new}}) is 5, and the standard deviation of this new set (σnew\sigma_{\text{new}}) is 20.

  • Determine the value of bb using the new mean: We use the property for the transformation of the mean: xnew=xold+b\overline{x}_{\text{new}} = \overline{x}_{\text{old}} + b. We are given xnew=5\overline{x}_{\text{new}} = 5 and we calculated xold=0\overline{x}_{\text{old}} = 0. Substituting these values: 5=0+b5 = 0 + b b=5b = 5 Why: Adding a constant bb to every observation shifts the entire dataset, and consequently its mean, by exactly bb. Since the original mean was 0, the new mean of 5 directly tells us the value of bb.

  • Determine the value of aa using the new standard deviation: We use the property for the transformation of the standard deviation: σnew=σold\sigma_{\text{new}} = \sigma_{\text{old}}. We are given σnew=20\sigma_{\text{new}} = 20 and we calculated σold=a\sigma_{\text{old}} = |a|. Substituting these values: 20=a20 = |a| This implies that aa can be either 2020 or 20-20. Why: Adding a constant to each data point represents a "change of origin." This operation shifts the entire distribution without changing the distances between data points or their spread around the mean. Therefore, the standard deviation remains unaffected.

Step 2.3: Calculate the Required Value

The problem asks for the value of a2+b2a^2 + b^2. We have found b=5b=5 and a=±20a=\pm 20.

Let's calculate a2a^2 and b2b^2: a2=(±20)2=400a^2 = (\pm 20)^2 = 400 b2=(5)2=25b^2 = (5)^2 = 25 Now, sum these values: a2+b2=400+25=425a^2 + b^2 = 400 + 25 = 425


3. Common Mistakes & Tips

  • Absolute Value: Always remember that x2=x\sqrt{x^2} = |x|, not just xx. This is crucial for standard deviation, which must be non-negative.
  • Transformation Rules: Be precise about how mean and standard deviation are affected by transformations. Adding a constant (xixi+bx_i \to x_i + b) changes the mean but not the standard deviation. Multiplying by a constant (xicxix_i \to c x_i) changes both the mean (by cc) and the standard deviation (by c|c|).
  • Symmetry in Data: For symmetric data sets (like a,aa, -a), quickly identifying a mean of zero can save time and simplify calculations.

4. Summary

This problem effectively tests the understanding of fundamental statistical measures and, more importantly, the properties of mean and standard deviation under a change of origin (adding a constant). By first calculating the mean and standard deviation of the original dataset, and then applying the transformation properties, we efficiently determined the values of aa and bb. The original mean was 0 and standard deviation was a|a|. After adding bb, the new mean became 0+b=50+b=5, giving b=5b=5. The new standard deviation remained a=20|a|=20. Finally, calculating a2+b2a^2+b^2 yielded 425.

The final answer is 425\boxed{425}, which corresponds to option (A).

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