Question

The standard deviation of a variable x is\[10\]. Then, the standard deviation of \[50{\text{ }} + {\text{ }}5x\] is
1) \[{\mathbf{50}}\]
2) \[{\mathbf{550}}\]
3)\[\;{\mathbf{10}}\]
4) \[{\mathbf{500}}\]

Answer 1

Hint: To determine the value of standard deviation we will use the formula of variance according to which standard deviation is the square root of variance.
We will first take the given standard deviation of the given variable x and calculate its variance. Then with the help of calculated variance, we will determine the variance of \[50{\text{ }} + {\text{ }}5x\]. We will then take the square root of the variance of \[50{\text{ }} + {\text{ }}5x\] and find its standard deviation.

Formula Used:
Variance is given by the formula \[{\sigma ^2} = \frac{1}{n}\sum\limits_i^n {{{({x_i} - \mathop x\limits^\_ )}^2}} \]where xi is the observed value and x̄ is the mean value.

Complete step by step Solution:
Standard deviation for variable x\[ = 10\]
Thus, variance(x) = (standard deviation)$^2$ \[ = {(10)^2} = 100\]
Therefore finding variance \[\left( {50{\text{ }} + {\text{ }}5x} \right)\] and then taking its square root to find the standard deviation.
Thus, variance \[\left( {50{\text{ }} + {\text{ }}5x} \right)\]= variance \[\left( {5x} \right)\]
This is because variance does not depend on a change of origin. This means if we add or subtract a fixed number from data there is no effect on variance. Thus, variance depends on the scaling factor.
If the data is multiplied by a constant, the variance gets multiplied by the square of the constant. Thus,
\[ = {5^2}\;\]variance (x)
\[ = {5^2}\;{10^2}^\;\] (As standard deviation of x variable is given to be\[10\])
Therefore standard deviation of \[50 + 5x\] $ = 5(10)$\[ = {\text{ }}50\]

Hence, the correct option is (1).

Note: Standard deviation of \[50 + 5x\] cannot be solved as\[50 + 5\left( {10} \right) = 100\].The first variance is calculated then only the standard deviation is found. Variance can be zero when all data values given are the same. This signifies that there is no dispersion or spread of data. Thus, the observed value is the same as the mean value. There is as such no deviation observed from the mean value.