NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.Misc.Q.5
The mean and standard deviation of 20 observations is 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:
(1) If wrong item is omitted.
(2) If it is replaced by 12.
(1) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2
Mean, x = (1 ÷ n) × Σi=120 xi
=> 10 = (1 ÷ 20) × Σi=120 xi
=> Σi=120 xi = 200
That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
Correct mean = (correct sum ÷ 19) = (192 ÷ 19) = 10.1
Standard deviation σ = √ [(1 ÷ n) × Σi=1n xi2 – (Σi=1n xi)2 ÷ n2]
=> 2 = √ [(1 ÷ n) × incorrect Σi=1n xi 2 – (x)2]
=> 2 = √ [(1 ÷ 20) × incorrect Σi=1n xi2 – (10)2]
=> 4 = (1 ÷ 20) × incorrect Σi=1n xi2 – 100
=> incorrect Σi=1n xi2 = 2080
So, correct Σi=1n xi2 = incorrect Σi=1n xi2 – 8
= 2080 – 64
= 2016
So, Correct Standard deviation = √ [{correct Σ xi2 ÷ n} – (correct mean)2]
= √ [{2016 ÷ 19} – (10.1)2]
= √ [106.1 – 102.01]
= √4.09
= 2.02
(2) When 8 is replaced by 12,
Incorrect sum of observations = 200
So, Correct sum of observations = 200 – 8 + 12 = 204
Correct mean = correct sum ÷ 20 = 204 ÷ 20 = 10.2
Standard deviation σ = √ [{(1 ÷ n) × Σi=1n xi2} – {(Σi=1n xi)2 ÷ n2}]
=> 2 = √ [(1 ÷ n) × incorrect Σi=1n xi2 – (x )2]
=> 2 = √ [(1 ÷ 20) × incorrect Σi=1n xi2 – (10)2]
=> 4 = (1 ÷ 20) × incorrect Σi=1n xi2 – 100
=> (1 ÷ 20) × incorrect Σi=1n xi2 = 104
=> incorrect Σi=1n xi2 = 2080
So, correct Σi=1n xi2 = incorrect Σi=1n xi2 – 82 + 122
= 2080 – 64 + 144
= 2160
So, Correct Standard deviation = √ [(correct Σ xi2÷ n) – (correct mean)2]
= √ [(2160 ÷ 20) – (10.2)2]
= √ [108 – 104.04]
= √3.96
= 1.98