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

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

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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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