While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct

NCERT Exemplar · Q23

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

Telangana TsbieLong· 3mImportance★★★★★est

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Concept understanding — Corrected Mean And Standard Deviation

Corrected Mean and Standard Deviation

Imagine a teacher who has computed the average marks of a class and the standard deviation, only to discover afterwards that one mark was entered wrongly, or that a student must be added or removed. Recomputing everything from the raw list of 100 marks would be tedious. Corrected mean and standard deviation is the technique for updating these two summary measures when the data changes by a small amount — without going back to the full dataset.

Note

In the CBSE Class 11 syllabus we always use the population definitions: mean $\bar{x} = \dfrac{1}{n}\sum x_i$ and variance $\sigma^2 = \dfrac{1}{n}\sum(x_i-\bar{x})^2$ . The denominator is $n$ , never $n-1$ .

The Two Quantities Everything Rests On

Both the mean and the standard deviation can be rebuilt from just two running totals:

the sum of the observations, $\sum x_i$
the sum of the squares of the observations, $\sum x_i^2$

Using the population formulas, a very useful rearrangement is:

\sigma^2 = \frac{1}{n}\sum x_i^2 - \bar{x}^2

so that from a known mean and standard deviation we can recover both totals:

\sum x_i = n\,\bar{x}, \qquad \sum x_i^2 = n\left(\sigma^2 + \bar{x}^2\right)

Updating for a Change in the Data

Once you hold $\sum x_i$ and $\sum x_i^2$ , every kind of correction is just simple bookkeeping.

Remove an observation $a$ : $\sum x_i \to \sum x_i - a$ , $\sum x_i^2 \to \sum x_i^2 - a^2$ , and $n \to n-1$ .
Add an observation $b$ : $\sum x_i \to \sum x_i + b$ , $\sum x_i^2 \to \sum x_i^2 + b^2$ , and $n \to n+1$ .
Replace a wrong value $a$ by the correct value $b$ : do both at once — $\sum x_i \to \sum x_i - a + b$ and $\sum x_i^2 \to \sum x_i^2 - a^2 + b^2$ , with $n$ unchanged.

Then recompute:

\bar{x}_{\text{new}} = \frac{\sum x_i\ (\text{updated})}{n\ (\text{updated})}, \qquad \sigma_{\text{new}} = \sqrt{\frac{\sum x_i^2\ (\text{updated})}{n\ (\text{updated})} - \bar{x}_{\text{new}}^2}

Worked Example (a mis-recorded value)

Problem. The mean and standard deviation of $100$ observations were found to be $40$ and $5.1$ . Later it was found that one observation was wrongly read as $50$ instead of its correct value $40$ . Find the correct mean and standard deviation.

Step 1 — recover the totals.

\sum x_i = 100 \times 40 = 4000

From $\sigma^2 = \dfrac{1}{n}\sum x_i^2 - \bar{x}^2$ with $\sigma = 5.1$ :

\sum x_i^2 = n\left(\sigma^2 + \bar{x}^2\right) = 100\,(26.01 + 1600) = 162601

Step 2 — correct the totals (replace $50$ by $40$ ; $n$ stays $100$ ):

\sum x_i = 4000 - 50 + 40 = 3990

\sum x_i^2 = 162601 - 50^2 + 40^2 = 162601 - 2500 + 1600 = 161701

Step 3 — corrected mean and standard deviation.

$\bar{x}_{\text{new}} = \frac{3990}{100} = 39.9$ …

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Correcting the total and the sum of squares for the misread value gives correct mean $= 42.3$ and correct variance $= 43.81$ .

Step-by-step solution

For $n$ observations,

\text{Variance} = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2

1. Wrong total. With $n = 10$ and wrong mean $45$ : $\ \sum x_i = 45\times 10 = 450$ .

2. Wrong sum of squares. From the wrong variance $16$ :

16 = \frac{\sum x_i^2}{10} - 45^2 \ \Rightarrow\ \frac{\sum x_i^2}{10} = 16 + 2025 = 2041 \ \Rightarrow\ \sum x_i^2 = 20410.

3. Correct the total. Replace the wrong $52$ with the correct $25$ :

$\sum x_i^{\text{corr}} = 450 - 52 + 25 = 423.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$\begin{array}{c|ccccccc} x_i & 2 & 9 & 8 & 3 & 5 & 7 \\ f_i & 5 & 3 & 1 & 6 & 6 & 1 \\ \end{array}$
›Reveal solutionSolution

The mean deviation from the median equals $2$ , so the correct option is (A).

Solution

Order the data by $x_i$ and build cumulative frequencies ( $N=\sum f_i=22$ ):

$x_i$ 2 3 5 7 8 9
$f_i$ 5 6 6 1 1 3
c.f. 5 11 17 18 19 22

Median: $N/2=11$ ; the cumulative frequency first reaches $11$ at $x=3$ , so the median is $M=3$ .

Weighted absolute deviations $f_i\,|x_i-3|$ : …
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$Q. The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively. Later on it was found that one of the observations was taken as 50 in the place of 40. If the wrong entry is replaced by the correct one, then the sum of the squares of all the observations is (A) 162701 (B) 163501 (C) 162601 (D) 161701$
›Reveal solutionSolution

Correct the sum of squares after replacing 50 by 40: it drops to $161701$ (option D).

Given: $n=100$ , wrong mean $\bar x = 40$ , wrong standard deviation $\sigma = 5.1$ , one value recorded as $50$ should be $40$ .

Step 1 — Sum of squares with the wrong entry.
$\sigma^2 = \frac{\sum x_i^2}{n} - \bar x^2 \implies \sum x_i^2 = n(\sigma^2 + \bar x^2)$
$\sum x_i^2 = 100\,(5.1^2 + 40^2) = 100\,(26.01 + 1600) = 100 \times 1626.01 = 162601$ …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$Q. Statement (I): The range of the ungrouped data does not change even if certain intermediate observations are removed. Statement (II): The value of the mean deviation of an ungrouped data about the median is always less than or equal to the value of the mean deviation computed about any other measure of central tendency. Statement (III): For a grouped data, range is approximated as the difference between the lower limit of the largest class and the upper limit of the smallest class. (A) Statements I and II are true but statement III is false (B) Statements II and III are true but statement I is false (C) Statement III and I are true but statement II is false (D) Statements I, II and III are true$
›Reveal solutionSolution
Range depends only on the extreme values, so removing intermediate data leaves it unchanged. Mean deviation is minimized about the median. For grouped data, range is approximated using class boundaries, not limits. Statement III is false, making option (A) correct.

The three statements test your grasp of basic descriptive statistics — range, mean deviation, and how they behave with ungrouped versus grouped data. Let’s examine each one carefully.

Statement I: Range of ungrouped data does not change if certain intermediate observations are removed.

Range is defined as the difference between the maximum and minimum values in a dataset. Only these two extremes matter. Removing any observation that is neither the maximum nor the minimum — that is, any “intermediate” value — leaves both extremes untouched. So the range stays exactly the same.

Tip

Think of range as the span from the smallest to the largest. If you delete a number in the middle, the endpoints don’t move.

This statement is true.

Statement II: Mean deviation about the median is always ≤ mean deviation about any other measure of central tendency.

Mean deviation (MD) measures the average absolute distance of data points from a central value. The median has a special property: it minimizes the sum of absolute deviations. For any other value — mean, mode, or any arbitrary number — the sum of absolute deviations is larger or equal. Since MD is just that sum divided by the number of observations, the same inequality holds.
$x_1, x_2, \dots, x_n$

Statement III: For grouped data, range is approximated as the difference between the lower limit of the largest class and the upper limit of the smallest class. …
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$x_i$	2	3	5	7	8	9
$f_i$	5	6	6	1	1	3
c.f.	5	11	17	18	19	22

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

Step-by-step solution

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