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:
(i) If wrong item is omitted  (ii) If it is replaced by12

Solution:

Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

x = 1/n ∑²⁰_{i = 1}x_i

10 = 1/20 ∑²⁰_{i = 1}x_i

∑²⁰_{i = 1}x_i= 200

That is, incorrect sum of observations = 200

Correct sum of observations = 200 - 8 = 192

Therefore, correct mean = correct sum/19 = 192/19 = 10.1

Standard deviation,

σ = √1/n ∑ⁿ_{i = 1} x_i² - 1/n² (∑ⁿ_{i = 1} x_i)² = √1/n ∑ⁿ_{i = 1} x_i² - (x)²

2 = √1/20 Incorrect ∑ⁿ_{i = 1} x_i² - (- 10)²

4 = 1/20 Incorrect ∑ⁿ_{i = 1} x_i² - 100

= Incorrect ∑ⁿ_{i = 1} x_i² = 2080

Hence,

Correct ∑ⁿ_{i = 1} x_i² = 4 = Incorrect ∑ⁿ_{i = 1} x_i² - (8)²

= 2080 - 64

= 2016

Correct standard deviation = √Correct ∑ⁿ_{i = 1} x_i²/n - (correct mean)²

= √2016/19 - (192/19)²

= √1440/361

= √3.988

= 1.99

(ii) When 8 is replaced by 12 ,

Incorrect sum of observations = 200

Correct sum of observations = 200 - 8 + 12 = 204

Hence, correct mean = correct sum/20 = 204/20 = 10.2

Standard deviation,

σ = √1/n ∑ⁿ_{i = 1} x_i² - 1/n² (∑ⁿ_{i = 1} x_i)² = √1/n ∑ⁿ_{i = 1} x_i² - (x)²

2 = √1/20 Incorrect ∑ⁿ_{i = 1} x_i² - (- 10)²

4 = 1/20 Incorrect ∑ⁿ_{i = 1} x_i² - 100

= Incorrect ∑ⁿ_{i = 1} x_i² = 2080

Hence,

Correct ∑ⁿ_{i = 1} x_i² = 4 = Incorrect ∑ⁿ_{i = 1} x_i² - (8)² + (12)²

= 2080 - 64 + 144

= 2160

Correct standard deviation = √Correct ∑ⁿ_{i = 1} x_i²/n - (correct mean)²

= √2060/20 - (10.2)²

= √108 - 104.04

= √3.96

= 1.98

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NCERT Solutions Class 11 Maths Chapter 15 Exercise ME Question 5

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:(i) If wrong item is omitted (ii) If it is replaced by12

Summary:

(i) Correct mean and correct standard deviation are 10.1 and 1.99 (ii) Correct mean and correct standard deviation are 10.2 and 1.98