HOW TO FIND CORRECT MEAN AND STANDARD DEVIATION

Example 1 :

For a group of 100 candidates the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.

Solution :

Mean  =  x̄  =  60

Standard deviation  =  σ  =  15

Wrong scores  =  40 and 27

Correct scores  =  45 and 72

mean  =  Σx/n  =  60

Σx/100  =  60

Σx  =  6000 (wrong)

Correct Σx  =  6000 - wrong scores + correct scores

=  6000 - (40 + 27) + (45 + 72)

=  6000 - 67 + 117

Correct Σx  =  6050 (correct)

Correct mean  =  Correct Σx/n

=  6050/100

=  60.5

Standard deviation  =  σ  =  15

σ  =  √[(Σx²/n) - (Σx/n)²]

15  =  √[(Σx²/n) - (60)²]

225 + 3600  =  (Σx²/100)

Σx²  =  382500 (wrong)

Correct Σx²  =  382500 - 67² + 117²

  =  382500 - 4489 + 13689

 Correct Σx²   =  391700

σ  =  √[(391700/100) - (60.5)²]

σ  =  √(3917 - 3660.25)

σ  =  √256.75

σ  =  16.02

Example 2 :

The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.

Solution :

Mean  =  8 

Variance  =  16

Let "a" and "b" be the missing observations.

(2 + 4 + 10 + 12 + 14 + a + b)/7  =  8

42 + a + b  =  56

a + b  =  56 - 42

a + b  =  14  ---(1)

b = 14 - a

Variance  =  Σ(x - x̄)²/n

 Σ(x - x̄)²/7

=  (2-8)²+(4-8)²+(10-8)²+(12-8)²+(14-8)²+(a-8)²+ (b-8)²/ 7

(36 + 16 + 4 + 16 + 36 + (a-8)²+ (b-8)²)/ 7  =  16

108 + (a-8)²+ (b-8)²  =  112

(a-8)²+ (b-8)²  =  4  ----(2)

By applying the value of b in (2), we get

(a-8)²+ (14-a-8)²  =  4  ----(2)

(a-8)²+ (6-a)²  =  4 

a² + 64 - 16a + 36 + a² - 12a - 4  =  0

2a² -  28a + 96  =  0

a² -  14a + 48  =  0

(a - 6)(a- 8)  =  0

a = 6 and b = 8

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