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.