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JEE Main 2018
Statistics & Probability
Statistics
Hard

Question

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given?

Options

Solution

Key Concepts and Formulas

  1. Measures of Central Tendency: These statistics describe the central position of a dataset.

    • Mean (x\overline{x}): The average of all observations. For a dataset x1,x2,,xNx_1, x_2, \ldots, x_N, the mean is given by x=i=1NxiN\overline{x} = \frac{\sum_{i=1}^{N} x_i}{N}.
    • Median: The middle value of a dataset when arranged in ascending or descending order. If NN is odd, it's the (N+1)/2(N+1)/2-th term. If NN is even, it's the average of the N/2N/2-th and (N/2+1)(N/2+1)-th terms.
    • Mode: The value that appears most frequently in a dataset.
  2. Measures of Dispersion: These statistics describe the spread or variability of a dataset.

    • Variance (σ2\sigma^2): A measure of the average squared deviation of each data point from the mean. It is given by σ2=i=1N(xix)2N\sigma^2 = \frac{\sum_{i=1}^{N} (x_i - \overline{x})^2}{N}.
  3. Change of Origin: This refers to a transformation where a constant value is added to or subtracted from every observation in a dataset. If the original observations are xix_i, the new observations are xi=xi+cx_i' = x_i + c, where cc is a constant.

Step-by-Step Solution

Let the original scores of the students be x1,x2,,xNx_1, x_2, \ldots, x_N. The teacher gives grace marks of 10 to each student, so the constant c=10c = 10. The new scores will be x1,x2,,xNx_1', x_2', \ldots, x_N', where xi=xi+10x_i' = x_i + 10 for each student ii.

Step 1: Analyze the Effect on Mean (Option D)

  • What we are doing: We are calculating the new mean (x\overline{x}') after adding 10 marks to each score.
  • Why: To determine if the mean changes.
  • Mathematics: Let the original mean be x=i=1NxiN\overline{x} = \frac{\sum_{i=1}^{N} x_i}{N}. The new mean x\overline{x}' is calculated as: x=i=1NxiN\overline{x}' = \frac{\sum_{i=1}^{N} x_i'}{N} Substitute xi=xi+10x_i' = x_i + 10: x=i=1N(xi+10)N\overline{x}' = \frac{\sum_{i=1}^{N} (x_i + 10)}{N} We can split the summation: x=i=1Nxi+i=1N10N\overline{x}' = \frac{\sum_{i=1}^{N} x_i + \sum_{i=1}^{N} 10}{N} The sum of a constant 1010 for NN times is N×10N \times 10: x=i=1Nxi+N×10N\overline{x}' = \frac{\sum_{i=1}^{N} x_i + N \times 10}{N} Separate the fraction: x=i=1NxiN+N×10N\overline{x}' = \frac{\sum_{i=1}^{N} x_i}{N} + \frac{N \times 10}{N} Recognize i=1NxiN\frac{\sum_{i=1}^{N} x_i}{N} as the original mean x\overline{x}, and simplify the second term: x=x+10\overline{x}' = \overline{x} + 10
  • Reasoning: The new mean is the original mean plus 10. Thus, the mean changes. Therefore, option (D) is not the answer.

Step 2: Analyze the Effect on Median (Option A)

  • What we are doing: We are determining the new median (MM') after adding 10 marks to each score.
  • Why: To determine if the median changes.
  • Reasoning: The median is the middle value of an ordered dataset. When a constant value (10) is added to every score, the relative order of the scores does not change. If the original scores arranged in ascending order are x(1)x(2)x(N)x_{(1)} \le x_{(2)} \le \ldots \le x_{(N)}, then the new scores will be x(1)+10x(2)+10x(N)+10x_{(1)}+10 \le x_{(2)}+10 \le \ldots \le x_{(N)}+10. The middle value(s) in this new ordered list will simply be the original median value(s) plus 10.
  • Example: If the original median was MM, the new median MM' will be M+10M+10.
  • Conclusion: The median changes by adding 10 to the original median. Therefore, option (A) is not the answer.

Step 3: Analyze the Effect on Mode (Option B)

  • What we are doing: We are determining the new mode (MoMo') after adding 10 marks to each score.
  • Why: To determine if the mode changes.
  • Reasoning: The mode is the score that appears most frequently. If a score MoMo was the most frequent score in the original dataset, then after adding 10 to every score, the new score Mo+10Mo+10 will be the most frequent score in the transformed dataset. It will appear the same number of times as MoMo did previously, and no other score will appear with higher frequency.
  • Example: If the original mode was MoMo, the new mode MoMo' will be Mo+10Mo+10.
  • Conclusion: The mode changes by adding 10 to the original mode. Therefore, option (B) is not the answer.

Step 4: Analyze the Effect on Variance (Option C)

  • What we are doing: We are calculating the new variance (σ2\sigma'^2) after adding 10 marks to each score.
  • Why: To determine if the variance changes. This is a measure of dispersion.
  • Mathematics: The original variance is σ2=i=1N(xix)2N\sigma^2 = \frac{\sum_{i=1}^{N} (x_i - \overline{x})^2}{N}. The new variance σ2\sigma'^2 is calculated using the new scores xix_i' and the new mean x\overline{x}': σ2=i=1N(xix)2N\sigma'^2 = \frac{\sum_{i=1}^{N} (x_i' - \overline{x}')^2}{N} From Step 1, we know that xi=xi+10x_i' = x_i + 10 and x=x+10\overline{x}' = \overline{x} + 10. Substitute these into the formula: σ2=i=1N((xi+10)(x+10))2N\sigma'^2 = \frac{\sum_{i=1}^{N} ((x_i + 10) - (\overline{x} + 10))^2}{N} Simplify the term inside the parenthesis: σ2=i=1N(xi+10x10)2N\sigma'^2 = \frac{\sum_{i=1}^{N} (x_i + 10 - \overline{x} - 10)^2}{N} The +10+10 and 10-10 cancel out: σ2=i=1N(xix)2N\sigma'^2 = \frac{\sum_{i=1}^{N} (x_i - \overline{x})^2}{N} This expression is exactly the formula for the original variance σ2\sigma^2: σ2=σ2\sigma'^2 = \sigma^2
  • Reasoning: The new variance is equal to the original variance. This means the variance remains unchanged. Variance measures the spread of data points around their mean. When all data points and the mean are shifted by the same constant, their relative distances from the mean remain unchanged, thus the spread does not change.
  • Conclusion: The variance does not change even after adding grace marks. Therefore, option (C) is the correct answer.

Common Mistakes & Tips

  • Distinguish Change of Origin from Change of Scale:
    • Change of Origin (adding/subtracting a constant cc): Measures of central tendency (mean, median, mode) change by adding/subtracting cc. Measures of dispersion (variance, standard deviation, range, quartile deviation) do not change.
    • Change of Scale (multiplying/dividing by a constant kk): Measures of central tendency change by being multiplied/divided by kk. Measures of dispersion also change: variance is multiplied/divided by k2k^2, and standard deviation is multiplied/divided by k|k|.
  • Intuition for Dispersion: Imagine a cluster of points on a number line. If you slide the entire cluster (every point) to the right or left by a fixed amount, the spread or dispersion of the points within the cluster does not change, only their absolute position on the number line does.
  • Standard Deviation: Since variance is unchanged, its square root, the standard deviation, also remains unchanged.

Summary

When a constant value is added to or subtracted from every observation in a dataset (a "change of origin"), measures of central tendency (mean, median, mode) shift by that constant value. However, measures of dispersion, such as variance and standard deviation, remain unaffected because the relative distances between data points and their mean do not change. In this problem, adding 10 grace marks to each student's score is a change of origin. Therefore, the variance of the scores will not change.

The final answer is C\boxed{C}

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