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Question: Standard deviation of n observations \({a_1},{a_2},{a_3},..............{a_n}\)is \(\sigma \)then the...

Standard deviation of n observations a1,a2,a3,..............an{a_1},{a_2},{a_3},..............{a_n}is σ\sigma then the standard deviation of the observations λa1,λa2,λa3,..............λan\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}is
A. λσ\lambda \sigma
B. −λσ- \lambda \sigma
C. ∣λ∣σ\left| \lambda \right|\sigma
D. λnσ{\lambda ^n}\sigma

Explanation

Solution

According to given in the question we have to obtain the standard deviation of the observations λa1,λa2,λa3,..............λan\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}so, first of all we have to let the mean the original data.
Now, we have to use the formula to find the variance as given below:

Formula used:
σ2=∑i=1i=n(xi−x‾)2n....................(a){\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \overline x )}^2}} }}{n}....................(a)
Where, n is the number of total observations and x‾\overline x is the mean for the given observations.
Now, we have to multiply by the constant to the mean which we let and after that we have to let the standard deviation for the new observations obtained. So, we have to use the formula (a) again to find the variance.

Complete step-by-step solution:
Step 1: First of all we have to let the mean of the original data as μ\mu
Step 2: Now, we have to find the variance with the help of the formula (a) as mentioned in the solution hint. Hence,
⇒σ2=∑i=1i=n(xi−μ)2n ⇒σ=∑i=1i=n(xi−μ)2n.................(1) \Rightarrow {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \mu )}^2}} }}{n} \\\ \Rightarrow \sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \mu )}^2}} }}{n}} .................(1)
Step 3: Now, we have to let that new standard deviation for the new observations λa1,λa2,λa3,..............λan\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n} is σ′{\sigma '}

⇒σ′=∑i=1i=n(λxi−λμ)2n ⇒σ′=∑i=1i=nλ2(xi−μ)2n ⇒σ′=∣λ∣∑i=1i=n(xi−μ)2n..................(2) \Rightarrow {\sigma '} = \sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{{(\lambda {x_i} - \lambda \mu )}^2}} }}{n}} \\\ \Rightarrow {\sigma '} = \sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{\lambda ^2}{{({x_i} - \mu )}^2}} }}{n}} \\\ \Rightarrow {\sigma '} = \left| \lambda \right|\sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \mu )}^2}} }}{n}} ..................(2)

Step 4: Now, on substituting the equation (1) as obtained in the step 2 in the equation (2)
⇒σ′=∣λ∣σ\Rightarrow {\sigma '} = \left| \lambda \right|\sigma

Hence, with the help of formula (a) we have obtained the standard deviation of the observations λa1,λa2,λa3,..............λan\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n} is =∣λ∣σ= \left| \lambda \right|\sigma

Note: Variance is a measurement of the spread between a given number of the data set that is, it measures how far each number in the set is from the mean and hence, from every other number in the set.
The standard deviation is the measurement of how to spread out numbers and its symbol is σ\sigma