Q.The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
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 xˉ=n1∑xi and variance σ2=n1∑(xi−xˉ)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, ∑xi
the sum of the squares of the observations, ∑xi2
Using the population formulas, a very useful rearrangement is:
σ2=n1∑xi2−xˉ2
so that from a known mean and standard deviation we can recover both totals:
∑xi=nxˉ,∑xi2=n(σ2+xˉ2)
Updating for a Change in the Data
Once you hold ∑xi and ∑xi2, every kind of correction is just simple bookkeeping.
Remove an observation a: ∑xi→∑xi−a, ∑xi2→∑xi2−a2, and n→n−1.
Add an observation b: ∑xi→∑xi+b, ∑xi2→∑xi2+b2, and n→n+1.
Replace a wrong value a by the correct value b: do both at once — ∑xi→∑xi−a+b and ∑xi2→∑xi2−a2+b2, with n unchanged.
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.
∑xi=100×40=4000
From σ2=n1∑xi2−xˉ2 with σ=5.1:
∑xi2=n(σ2+xˉ2)=100(26.01+1600)=162601
Step 2 — correct the totals (replace 50 by 40; n stays 100):
After omitting the three incorrect observations, the corrected mean =20 and the corrected standard deviation =97894≈3.04.
Step 1 — Recover the totals from the given summary
We are told the original group has n=100 observations with mean xˉ=20 and standard deviation σ=3. Using the population formulas (NCERT Class 11, divisor n):
∑xi=nxˉ=100×20=2000.
For the sum of squares, start from the variance definition and rearrange:
σ2=n∑xi2−xˉ2⇒32=100∑xi2−202⇒9=100∑xi2−400.
So
∑xi2=100×(9+400)=100×409=40900.
Step 2 — Remove the three incorrect observations
The wrong values recorded were 21,21,18. Their contribution to each total is:
∑(removed)=21+21+18=60,
∑(removed)2=212+212+182=441+441+324=1206.
Omitting them leaves n′=100−3=97 observations with corrected totals:
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 xˉ=40, wrong standard deviation σ=5.1, one value recorded as 50 should be 40.
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.
For any dataset x1,x2,…,xn, ∑∣xi−median∣≤∑∣xi−c∣ for any c.
This is a standard result in statistics. Hence, Statement II is true.
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. …