Q.The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
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):
Concept: Corrected Mean and Standard Deviation — when an observation is removed or replaced, we first adjust the sum and sum of squares, then recompute.
When one observation is wrong, we first recover the original sum and sum of squares from the given mean and SD, then adjust for the correction — either by removing the wrong value or replacing it — and recompute the new mean and SD.
We are given 20 observations with mean xˉ=10 and standard deviation σ=2.
That means:
Total sum: ∑xi=20×10=200
Variance σ2=4, so ∑xi2 can be found from σ2=n∑xi2−(xˉ)2.
Let’s compute that first — it’s the foundation for both parts.
Step 1: Find the original sum of squares
From the variance formula:
σ2=n∑xi2−(xˉ)2
Plug in:
4=20∑xi2−100
So:
20∑xi2=104⇒∑xi2=2080
We now have:
n=20, ∑xi=200, ∑xi2=2080
The incorrect observation is 8.
Step 2: Case (i) — Omit the wrong item
If we remove the observation 8, the new number of observations is n′=19.
New sum:
∑xi′=200−8=192
New mean:
xˉ′=19192
New sum of squares:
∑(xi′)2=2080−82=2080−64=2016
New variance:
(σ′)2=192016−(19192)2
First compute (19192)2=36136864.
Now:
192016=19×192016×19=36138304
So:
(σ′)2=36138304−36136864=3611440
Thus:
σ′=3611440=191440
Simplify 1440=144×10=1210.
So:
σ′=191210
Tip
When omitting an observation, the denominator changes — don’t forget to divide by the new n, not the old one.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2026Set 2026-M1 markMCQ
Q.If for a distribution of 20 items, ∑(x−4)=10 and ∑(x−4)2=85 then the standard deviation is:
(A) 4
(B) 4.25
(C) 1.41
(D) 2
›Reveal solutionSolution
The standard deviation is found by first computing the mean from the given sum of deviations, then using the sum of squared deviations to find the variance, and finally taking the square root. The result is exactly 2, which corresponds to option (D).
Concept & Intuition
Standard deviation measures how spread out numbers are from the mean. Here we are given sums of deviations from a fixed number (4), not from the actual mean. The trick is to realize that ∑(x−4) tells us how far the actual mean is from 4, and ∑(x−4)2 gives us a stepping stone to the variance. We don’t need the original data — just these two summary statistics.
Find the actual mean xˉ.
The sum of deviations from a constant c is ∑(x−c)=∑x−nc.
Here c=4, n=20, and ∑(x−4)=10.
So:
∑x−20⋅4=10⇒∑x=10+80=90.
Hence the mean is:
xˉ=2090=4.5.
Relate ∑(x−4)2 to the variance.
Variance σ2 is defined as n1∑(x−xˉ)2.
We have ∑(x−4)2=85. Expand the square:
∑(x−4)2=∑[(x−xˉ)+(xˉ−4)]2.
Let d=xˉ−4=4.5−4=0.5. Then:
∑(x−4)2=∑(x−xˉ)2+2d∑(x−xˉ)+nd2.
But ∑(x−xˉ)=0 always, so the middle term vanishes. Thus:
Q.The mean and standard deviation of 100 items are 50 and 4, respectively then the sum of all squares of the items is
(A) 250000
(B) 251600
(C) 256100
(D) 265100
›Reveal solutionSolution
Use the variance formula σ2=n∑x2−xˉ2 and solve for ∑x2.
Step 1 — Write the variance formula
For n items with mean xˉ and standard deviation σ:
Q.The mean of five observations is 4 and their variance is 5.2 . If three of these observations are 1, 2 and 6, then the other two observations are
(A) 4, 7
(B) 2, 10
(C) 5, 6
(D) 2, 9
›Reveal solutionSolution
Using the given mean and variance, we set up equations for the sum and sum of squares of the five observations. Solving these yields the two missing numbers as 4 and 7, which corresponds to option (A).
Concept & Intuition
We have five numbers with a known mean and variance. The mean gives us the total sum; the variance gives us the sum of squares. Since three numbers are known, we can treat the two unknowns as variables and solve the system. This is a classic “reconstructing data from summary statistics” problem — the mean controls the center, the variance controls the spread.
Step-by-step solution
Use the mean to find the total sum.
Mean = 4, number of observations = 5.
Sum=5×4=20.
Let the two unknown observations be x and y. Then
1+2+6+x+y=20⇒9+x+y=20⇒x+y=11.(1)
Use the variance to find the sum of squares.
Variance = 5.2. Recall:
Variance=n∑xi2−(mean)2.
So
5.2=5∑xi2−42=5∑xi2−16.
Multiply by 5:
26=∑xi2−80⇒∑xi2=106.
The known squares: 12+22+62=1+4+36=41.
Thus
41+x2+y2=106⇒x2+y2=65.(2)
Solve the system (1) and (2).
From (1): y=11−x. Substitute into (2):