Solveeit Logo

Question

Question: The variance of \(20\) observations is \(5\). If each observation is multiplied by two, find the sta...

The variance of 2020 observations is 55. If each observation is multiplied by two, find the standard deviation and variance of the resulting observations.

Explanation

Solution

In the given question, we are provided with some information about a series where we are given the number of terms and their variance. We have to find the standard deviation and variance of the series obtained by making the changes in the original series as suggested in the question.

Complete step by step answer:
The standard deviation is a measure of the amount of variation or dispersion of a set of values. So, we have a series of observations of 2020 observations whose variance is 55. We know, standard deviation is,
σ=i=1n(xiμ)2n\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}} .
Where, n=n = total observations, xi={x_i} = each term and μ=\mu = mean of observations.
We also know that variance of observations is the square of the standard deviation. So, let the given series of observations is, x1{x_1}, x2{x_2}, x3{x_3}, x4{x_4}, …., x20{x_{20}}.

Now, on multiplying 22 to each of the observations, the series becomes, 2x12{x_1}, 2x22{x_2}, 2x32{x_3}, 2x42{x_4}, …., 2x202{x_{20}}.
Therefore, the mean of the new series becomes, μ1=2x1+2x2+2x3+.....+2x2020{\mu _1} = \dfrac{{2{x_1} + 2{x_2} + 2{x_3} + ..... + 2{x_{20}}}}{{20}}
μ1=2(x1+x2+x3+....+x20)20\Rightarrow {\mu _1} = \dfrac{{2\left( {{x_1} + {x_2} + {x_3} + .... + {x_{20}}} \right)}}{{20}}
[Taking 22 common from the numerator]
μ1=2μ\Rightarrow {\mu _1} = 2\mu
[μ=\mu = older mean]
Now, using this new mean in the formula of standard deviation, for new series, we get,
σ1=i=120(2xi(2μ))220{\sigma _1} = \sqrt {\dfrac{{\sum\limits_{i = 1}^{20} {{{\left( {2{x_i} - \left( {2\mu } \right)} \right)}^2}} }}{{20}}}

Taking, 22 common from numerator, inside the root, we get,
σ1=(2)2i=1n(xiμ)2n\Rightarrow {\sigma _1} = \sqrt {\dfrac{{{{(2)}^2}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}}
Now, taking 22 out of the root, we get,
σ1=2i=1n(xiμ)2n\Rightarrow {\sigma _1} = 2\sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}}
σ1=2σ\Rightarrow {\sigma _1} = 2\sigma
[σ=\sigma = older standard deviation]
Now, we are given, the variance of the older series is 55. So, the standard deviation of the older series must be the square root of the variance. So, we get,
σ=5\sigma = \sqrt 5
Therefore, the standard deviation of the new series is,
σ1=2σ=2(5)=25{\sigma _1} = 2\sigma = 2\left( {\sqrt 5 } \right) = 2\sqrt 5
Also, the new variance must be the square of the new standard deviation.

So, we get variance =(25)2=20 = {\left( {2\sqrt 5 } \right)^2} = 20.

Note: From the above problem we can say that the standard deviation remains constant or unchanged if any constant is added or subtracted from all the observations. But the standard deviation gets multiplied or divided by the constant if all the observations are multiplied or divided by the same constant.