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JEE Main 2021
Statistics & Probability
Statistics
Hard

Question

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If μ\mu and σ2\sigma^2 denote the mean and variance of the correct observations respectively, then 15(μ+μ2+σ2)15\left(\mu+\mu^2+\sigma^2\right) is equal to __________.

Answer: 2521

Solution

Solution

Key Concepts and Formulas

When dealing with statistical data, especially problems involving corrections due to errors in observation, a solid understanding of the fundamental definitions of mean and variance is crucial.

Step-by-Step Solution

Step 1: Calculate the Initial (Incorrect) Sums We are given the initial (incorrect) statistical measures:

Our first task is to determine the sum of observations (xi x_i') and the sum of squares of observations (xi2 x_i'^2) based on these incorrect figures. These intermediate sums are necessary for making the corrections.

Step 2: Determine the Corrected Sums The problem states that an observation was incorrectly read as 10 instead of 12. To correct our statistical calculations, we must adjust both the sum of observations and the sum of squares of observations. For each sum, we subtract the value that was incorrectly included and add the value that should have been included.

Step 3: Calculate the Correct Mean and Variance With the corrected sums of observations (xi=182 x_i = 182) and squares of observations (xi2=2339 x_i^2 = 2339), we can now compute the true mean ($$) and variance (2^2) for the n=15n=15 observations.

Step 4: Evaluate the Required Expression We need to calculate the value of 15(+2+2)15(+^2+^2). This is where the key identity 2+2=xi2n^2 + ^2 = \sum x_i^2{n} significantly simplifies the process.

Substitute the identity into the expression: 15(+2+2)=15(+xi2n)15(+^2+^2) = 15( + \sum x_i^2{n}) Now, substitute the expressions for and $\sum x_i^2{n}$ using our corrected values:15(182{15} + 2339{15})We can factor out $1{15}$ from the terms inside the parenthesis:15 1{15} (182 + 2339) 1 (182 + 2339) 182 + 2339 = 2521Byemployingtheidentity,weavoideddirectcalculationswiththefractionalvaluesofBy employing the identity, we avoided direct calculations with the fractional values of and 2^2, leading to a much cleaner and less error-prone computation.

Common Mistakes & Tips

Summary

The problem required us to find the value of an expression involving the correct mean and variance after an observation error. We systematically approached this by first calculating the initial incorrect sum of observations and sum of squares of observations. Following this, we performed the crucial step of correcting these sums by removing the erroneous data point's contribution and adding the correct one. Finally, using these corrected sums, we efficiently evaluated the target expression 15(+2+2)15(+^2+^2) by leveraging the powerful identity 2+2=xi2n^2 + ^2 = \sum x_i^2{n}, which significantly simplified the calculation and led us to the final integer answer.

The final answer is 2521.

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