The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations

Solution:

Let the observations be x₁, x₂, x₃, x₄, x₅ and x₆.

It is given that the mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are y_i, then

y_i = 3x_i i.e., x_i = 1 y_i, for i = 1 to 6

Therefore, new mean,

y = (y₁ + y₂ + y₃ + y₄ + y₅ + y₆)/6

= 3( x₁ + x₂ + x₃ + x₄ + x₅ + x₆)/6

= 3 × 8 ....from (1)

= 24

(σ) = 1/6 ∑⁶_{i = 1} (x₁ - x)²

(4²) = 1/6 ∑⁶_{i = 1} (x₁ - x)²

∑⁶_{i = 1} (x - x)² = 96 ....(2)

From (1) and (2) , it can be observed that,

y = 3x and x = 1/3 y

Substituting the values of x₁ and x in (2), we obtain

∑⁶_{i = 1} (1/3 y_i - 1/3 y)² = 96

∑⁶_{i = 1} (y_i - y)² = 864

Therefore, variance of new observations is (1/6 x 864) = 144

Hence, the standard deviation of new observations is 144 = 12

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NCERT Solutions Class 11 Maths Chapter 15 Exercise ME Question 3

The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Summary:

Therefore, the new mean and new standard deviation are 24 and 12