Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 11
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While Calculating the Mean and Variance of 10 Readings, a Student Wrongly Used the Reading of 52 for the Correct Reading 25. He Obtained the Mean and Variance as 45 and 16 Respectively. - Mathematics

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ConceptVariance and Standard Deviation Standard Deviation of a Discrete Frequency Distribution

Question

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Solution

Given:

Number of observations, n = 10

Mean,

\[\bar {x}\]  = 45  Variance,
\[\sigma^2\]   = 16
Now,
Incorrect mean,
\[\bar {x}\]  = 45 
\[\Rightarrow \frac{\text{ Incorrect } \sum_{} x_i}{10} = 45\]
\[ \Rightarrow \text{ Incorrect }  \sum_{} x_i = 450\]
∴ Correct \[\sum_{} x_i = 450 - 52 + 25 = 423\]
⇒ Correct mean = \[\frac{\text{ Correct } \sum_{} x_i}{10} = \frac{423}{10} = 42 . 3\]
Incorrect variance,
\[\sigma^2\]  = 16
\[\Rightarrow 16 = \frac{\text{ Incorrect }  \sum_{} x_i^2}{10} - \left( 45 \right)^2 \]
\[ \Rightarrow \text{ Incorrect } \sum_{} x_i^2 = 10\left( 16 + 2025 \right) = 20410\]
∴ Correct 
\[\sum_{} x_i^2 = 20410 - \left( 52 \right)^2 + \left( 25 \right)^2 = 20410 - 2704 + 625 = 18331\] 
Now,
Correct variance = \[\frac{18331}{10} - \left( 42 . 3 \right)^2 = 1833 . 1 - 1789 . 29 = 43 . 81\]
 

 

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APPEARS IN

 RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 32: Statistics
Ex. 32.6 | Q: 9 | Page no. 42
Solution While Calculating the Mean and Variance of 10 Readings, a Student Wrongly Used the Reading of 52 for the Correct Reading 25. He Obtained the Mean and Variance as 45 and 16 Respectively. Concept: Variance and Standard Deviation - Standard Deviation of a Discrete Frequency Distribution.
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