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JEE Main 2024
Statistics & Probability
Statistics
Easy

Question

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :

Options

Solution

Key Concepts and Formulas

To accurately solve problems involving the correction of statistical measures, we rely on the fundamental definitions of mean and variance:

  1. Mean (x\overline{x}): The average of a set of nn observations x1,x2,,xnx_1, x_2, \ldots, x_n. x=i=1nxin\overline{x} = \frac{\sum_{i=1}^{n} x_i}{n} From this, the sum of observations can be expressed as: xi=nx\sum x_i = n \cdot \overline{x}

  2. Variance (σ2\sigma^2): A measure of the spread of data points. The most practical computational formula is: σ2=i=1nxi2n(x)2\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\overline{x})^2 Rearranging this, the sum of squares of observations can be found: xi2=n(σ2+x2)\sum x_i^2 = n \cdot (\sigma^2 + \overline{x}^2)

  3. Correction Principle: When an incorrect observation (bb) is replaced by a correct observation (aa), the sums are adjusted as follows:

    • Corrected Sum of Observations: (xi)correct=(xi)incorrectb+a(\sum x_i)_{correct} = (\sum x_i)_{incorrect} - b + a
    • Corrected Sum of Squares of Observations: (xi2)correct=(xi2)incorrectb2+a2(\sum x_i^2)_{correct} = (\sum x_i^2)_{incorrect} - b^2 + a^2 The number of observations, nn, remains unchanged in such a replacement scenario.

Step-by-Step Solution

Let's systematically correct the given statistical measures.

Step 1: Extract Given (Incorrect) Information

  • What: Identify all initial parameters provided in the problem statement.
  • Why: These values form our baseline before any corrections are applied.
  • How:
    • Number of observations (nn) = 50
    • Incorrect Mean (xincorrect\overline{x}_{incorrect}) = 15
    • Incorrect Standard Deviation (σincorrect\sigma_{incorrect}) = 2
    • From the standard deviation, the Incorrect Variance (σincorrect2\sigma^2_{incorrect}) = 22=42^2 = 4.

Step 2: Calculate the Incorrect Sum of Observations

  • What: Determine the total sum of all observations as they were initially recorded, including the incorrect value.
  • Why: This sum is necessary to later subtract the incorrect observation and add the correct one.
  • How: Using the formula xi=nx\sum x_i = n \cdot \overline{x}: (xi)incorrect=nxincorrect(\sum x_i)_{incorrect} = n \cdot \overline{x}_{incorrect} (xi)incorrect=5015(\sum x_i)_{incorrect} = 50 \cdot 15 (xi)incorrect=750(\sum x_i)_{incorrect} = 750

Step 3: Calculate the Incorrect Sum of Squares of Observations

  • What: Determine the sum of the squares of all observations as initially recorded.
  • Why: This sum is crucial for calculating variance and will be adjusted to find the correct sum of squares.
  • How: Using the formula xi2=n(σ2+x2)\sum x_i^2 = n \cdot (\sigma^2 + \overline{x}^2): (xi2)incorrect=n(σincorrect2+(xincorrect)2)(\sum x_i^2)_{incorrect} = n \cdot (\sigma^2_{incorrect} + (\overline{x}_{incorrect})^2) (xi2)incorrect=50(4+(15)2)(\sum x_i^2)_{incorrect} = 50 \cdot (4 + (15)^2) (xi2)incorrect=50(4+225)(\sum x_i^2)_{incorrect} = 50 \cdot (4 + 225) (xi2)incorrect=50229(\sum x_i^2)_{incorrect} = 50 \cdot 229 (xi2)incorrect=11450(\sum x_i^2)_{incorrect} = 11450

Step 4: Establish Relationships for Correct and Incorrect Observations

  • What: Define variables for the correct and incorrect observations and form equations based on the given information.
  • Why: We need to find the specific values of these observations to properly adjust our sums.
  • How: Let aa be the correct observation and bb be the incorrect observation.
    • Given: "the sum of correct and incorrect observations is 70". a+b=70(Equation 1)a + b = 70 \quad \text{(Equation 1)}
    • Given: the correct mean (xcorrect\overline{x}_{correct}) is 16. We can express the correct sum of observations in two ways:
      • Using the correct mean: (xi)correct=nxcorrect=5016=800(\sum x_i)_{correct} = n \cdot \overline{x}_{correct} = 50 \cdot 16 = 800.
      • Using the correction principle: (xi)correct=(xi)incorrectb+a=750b+a(\sum x_i)_{correct} = (\sum x_i)_{incorrect} - b + a = 750 - b + a. Equating these two expressions for the correct sum: 800=750b+a800 = 750 - b + a Rearranging this gives our second equation: ab=800750a - b = 800 - 750 ab=50(Equation 2)a - b = 50 \quad \text{(Equation 2)}

Step 5: Determine the Values of Correct and Incorrect Observations

  • What: Solve the system of linear equations from Step 4.
  • Why: This yields the exact values of aa and bb, which are essential for making the final corrections.
  • How:
    1. a+b=70a + b = 70
    2. ab=50a - b = 50 Adding Equation 1 and Equation 2: (a+b)+(ab)=70+50(a + b) + (a - b) = 70 + 50 2a=1202a = 120 a=60a = 60 Substitute a=60a = 60 into Equation 1: 60+b=7060 + b = 70 b=10b = 10 So, the correct observation is 60, and the incorrect observation is 10.

Step 6: Calculate the Correct Sum of Squares of Observations

  • What: Adjust the incorrect sum of squares using the values of aa and bb.
  • Why: This corrected sum is a direct input for the correct variance calculation.
  • How: Using the correction principle (xi2)correct=(xi2)incorrectb2+a2(\sum x_i^2)_{correct} = (\sum x_i^2)_{incorrect} - b^2 + a^2: (xi2)correct=11450(10)2+(60)2(\sum x_i^2)_{correct} = 11450 - (10)^2 + (60)^2 (xi2)correct=11450100+3600(\sum x_i^2)_{correct} = 11450 - 100 + 3600 (xi2)correct=11350+3600(\sum x_i^2)_{correct} = 11350 + 3600 (xi2)correct=14950(\sum x_i^2)_{correct} = 14950

Step 7: Calculate the Final Correct Variance

  • What: Compute the correct variance using the corrected sum of squares, the correct mean, and the number of observations.
  • Why: This is the final value requested by the problem.
  • How: Using the variance formula σ2=xi2n(x)2\sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2: σcorrect2=(xi2)correctn(xcorrect)2\sigma^2_{correct} = \frac{(\sum x_i^2)_{correct}}{n} - (\overline{x}_{correct})^2 σcorrect2=1495050(16)2\sigma^2_{correct} = \frac{14950}{50} - (16)^2 σcorrect2=299256\sigma^2_{correct} = 299 - 256 σcorrect2=43\sigma^2_{correct} = 43

Common Mistakes & Tips

  • Squaring vs. Not Squaring: Remember to subtract bb and add aa for the sum of observations (xi\sum x_i), but subtract b2b^2 and add a2a^2 for the sum of squares of observations (xi2\sum x_i^2). This is a common point of error.
  • Number of Observations (nn): In scenarios where an observation is replaced, the number of observations (nn) remains constant. Only if observations are added or removed without replacement does nn change.
  • Arithmetic Accuracy: Double-check all calculations, especially squaring numbers and divisions, as a small error can lead to an incorrect final result.

Summary

This problem required us to correct the mean and variance of a dataset after identifying an incorrect observation. We systematically used the formulas for mean and variance to first calculate the incorrect sums (xi\sum x_i and xi2\sum x_i^2). Then, by utilizing the given correct mean and the relationship between the correct and incorrect observations, we solved for the specific values of the incorrect and correct observations. Finally, these values were used to adjust the sums of observations and sums of squares of observations, leading to the accurate calculation of the correct variance.

The final answer is 43\boxed{\text{43}}, which corresponds to option (A).

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