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

Example 3 :

The mean and standard deviation of 15 observations are found to be 10 and 5 respectively. On rechecking it was found that one of the observation with value 8 was incorrect. Calculate the correct mean and standard deviation if the correct observation value was 23?

Solution :

Number of observations (n) = 15

mean (x̄) = 10, standard deviation (σ) = 5

x̄ = Σx/n

10 = Σx/15

Σx = 10(15) ==> 150

Wrong observation = 8, correct observation = 23

Correct total = 150 - 8 + 23

= 165

Correct mean (x̄) = Σx/n 

= 165/15

Correct mean (x̄) = 11

Standard deviation :

Correct standard deviation :

Example 4 :

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 :

Number of observations (n) = 100

mean (x̄) = 60, standard deviation (σ) = 15

x̄ = Σx/n

60 = Σx/100

Σx = 60(100) ==> 6000

Wrong observation = 40 and 27, correct observation = 45 and 72

Correct total = 6000 - (40 + 27) + (45 + 72)

= 6000 - 67 + 117

= 6050

Correct mean (x̄) = Σx/n 

= 6050/100

Correct mean (x̄) = 60.5

Standard deviation :

Correct standard deviation :

Example 5 :

Find the standard deviation of the data 2, 3, 5, 7, 8. Multiply each data by 4. Find the standard deviation of the new values.

Solution :

The given data values are 

2, 3, 5, 7 and 8

Calculating standard deviation :

Mean = (2 + 3 + 5 + 7 + 8) / 5

= 25 /5

= 5

Σ(x - x̄)² = 9 + 4 + 0 + 4 + 9

= 26

Standard deviation for the given data is 2.28

Each value is multiplied by 4, then the old standard deviation should also be multiplied by 4, to get the new standard deviation.

The required standard deviation = 9.12

Example 6 :

The standard deviation of a data is 2.8, if 5 is added to all the data values then the new standard deviation is ___. 

Solution :

Using properties,

the standard deviation will not change when we add some fixed constant k to all the values.

Standard deviation of the given data = 2.8

Standard deviation for the new data after adding 5, the new standard deviation is 2.8

Example 7 :

If S is the standard deviation of values p, q, r then standard deviation of p–3, q–3, r–3 is ___

Solution :

Using properties,

the standard deviation will not change when we subtract some fixed constant k to all the values.

The new standard deviation is S.

Example 8 :

Find the standard deviation of the following data 7, 4, 8, 10, 11. Add 3 to all the values then find the standard deviation for the new values.

Solution :

The given data values are 

7, 4, 8, 10, 11

Calculating standard deviation :

Mean = (7 + 4 + 8 + 10 + 11) / 5

= 40 /5

= 8

Σ(x - x̄)² = 1 + 16 + 0 + 4 + 9

= 30

Standard deviation of the given data set = 2.44

Since every value is added with some constant, using the properties, it will not affect the standard deviation.

So, the required standard deviation is 2.44.

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