
Standard deviation of n observations ${a_1},{a_2},{a_3},..............{a_n}$is $\sigma $then the standard deviation of the observations $\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}$is
A. $\lambda \sigma $
B. $ - \lambda \sigma $
C. $\left| \lambda \right|\sigma $
D. ${\lambda ^n}\sigma $
Answer
506.4k+ views
Hint: According to given in the question we have to obtain the standard deviation of the observations $\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}$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:
${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \overline x )}^2}} }}{n}....................(a)$
Where, n is the number of total observations and $\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 $\mu $
Step 2: Now, we have to find the variance with the help of the formula (a) as mentioned in the solution hint. Hence,
$
\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 $\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}$ is ${\sigma '}$
\[
\Rightarrow {\sigma '} = \sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{{(\lambda {x_i} - \lambda \mu )}^2}} }}{n}} \\
\Rightarrow {\sigma '} = \sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{\lambda ^2}{{({x_i} - \mu )}^2}} }}{n}} \\
\Rightarrow {\sigma '} = \left| \lambda \right|\sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \mu )}^2}} }}{n}} ..................(2)
\]
Step 4: Now, on substituting the equation (1) as obtained in the step 2 in the equation (2)
$ \Rightarrow {\sigma '} = \left| \lambda \right|\sigma $
Hence, with the help of formula (a) we have obtained the standard deviation of the observations $\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}$ is $ = \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 $\sigma $
Now, we have to use the formula to find the variance as given below:
Formula used:
${\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \overline x )}^2}} }}{n}....................(a)$
Where, n is the number of total observations and $\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 $\mu $
Step 2: Now, we have to find the variance with the help of the formula (a) as mentioned in the solution hint. Hence,
$
\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 $\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}$ is ${\sigma '}$
\[
\Rightarrow {\sigma '} = \sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{{(\lambda {x_i} - \lambda \mu )}^2}} }}{n}} \\
\Rightarrow {\sigma '} = \sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{\lambda ^2}{{({x_i} - \mu )}^2}} }}{n}} \\
\Rightarrow {\sigma '} = \left| \lambda \right|\sqrt {\dfrac{{\sum\limits_{i = 1}^{i = n} {{{({x_i} - \mu )}^2}} }}{n}} ..................(2)
\]
Step 4: Now, on substituting the equation (1) as obtained in the step 2 in the equation (2)
$ \Rightarrow {\sigma '} = \left| \lambda \right|\sigma $
Hence, with the help of formula (a) we have obtained the standard deviation of the observations $\lambda {a_1},\lambda {a_2},\lambda {a_3},..............\lambda {a_n}$ is $ = \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 $\sigma $
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
