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Question: A group of \[10\] observations has mean \[5\] and S.D.\[2\sqrt 6 \]. Another group of \[20\] observa...

A group of 1010 observations has mean 55 and S.D.262\sqrt 6 . Another group of 2020 observations has mean 55 and S.D. 323\sqrt 2 , then the S.D. of combined group of 3030 observations is
(A) 5\sqrt 5
(B) 252\sqrt 5
(C) 353\sqrt 5
(D) None of these

Explanation

Solution

In this question they have given mean and standard deviation of two types of observation. We have to find mean and standard deviation of another observation related to the given observations. By using combined mean and combined standard deviation we will find the required result.

Formula used: The formula of combined mean,
x12=n1.x1+n2.x2n1+n2{\overline x _{12}} = \dfrac{{{n_1}.{{\overline x }_1} + {n_2}.{{\overline x }_2}}}{{{n_1} + {n_2}}}
The formula of combined Standard deviation
σ12=n1.(σ12+d12)+n2.(σ22+d22)n1+n2{\sigma _{12}} = \sqrt {\dfrac{{{n_1}.\left( {\sigma _1^2 + d_1^2} \right) + {n_2}.\left( {\sigma _2^2 + d_2^2} \right)}}{{{n_1} + {n_2}}}}
Where,d1=x1x12{d_1} = {\overline x _1} - {\overline x _{12}}, d2=x2x12{d_2} = {\overline x _2} - {\overline x _{12}}
Putting the given mean and S.D. and number of observations in this formula we will get the required S.D.

Complete step-by-step answer:
It is given that a group of 1010 observations has mean 55 and S.D. 262\sqrt 6 . Another group of 2020 observations has mean 55 and S.D. 323\sqrt 2 .
We need to find out the S.D. of a combined group of 3030 observations.
If n1{n_1} and n2{n_2} are number of observations, x1{\overline x _1} and x2{\overline x _2} are the mean, σ1{\sigma _1} and σ2{\sigma _2} are the S.D. of both the groups then,
Hence the given of a group of 1010 observations,
n1=10,x1=5,σ1=26{n_1} = 10,{\overline x _1} = 5,{\sigma _1} = 2\sqrt 6
Hence the given of a group of 2020 observations,

n2=20,x2=5,σ2=32{n_2} = 20,{\overline x _2} = 5,{\sigma _2} = 3\sqrt 2
Combined mean is,
x12=n1.x1+n2.x2n1+n2\Rightarrow {\overline x _{12}} = \dfrac{{{n_1}.{{\overline x }_1} + {n_2}.{{\overline x }_2}}}{{{n_1} + {n_2}}}
Substituting the given values to find the mean of the 3030 observations,
x12=10×5+20×510+20\Rightarrow {\overline x _{12}} = \dfrac{{10 \times 5 + 20 \times 5}}{{10 + 20}}
Simplifying we get,
x12=50+10030\Rightarrow {\overline x _{12}} = \dfrac{{50 + 100}}{{30}}
x12=15030=5\Rightarrow {\overline x _{12}} = \dfrac{{150}}{{30}} = 5
Putting the combined mean and mean of the observations in the formula we get,

d1=x1x12=55=0 d2=x2x12=55=0  {d_1} = {\overline x _1} - {\overline x _{12}} = 5 - 5 = 0 \\\ {d_2} = {\overline x _2} - {\overline x _{12}} = 5 - 5 = 0 \\\

Now the formula for Standard deviation of combined group of 3030 observations is
σ12=n1.(σ12+d12)+n2.(σ22+d22)n1+n2\Rightarrow {\sigma _{12}} = \sqrt {\dfrac{{{n_1}.\left( {\sigma _1^2 + d_1^2} \right) + {n_2}.\left( {\sigma _2^2 + d_2^2} \right)}}{{{n_1} + {n_2}}}}
Substituting the given values to find the standard deviation of the 3030 observations,
\Rightarrow {\sigma _{12}} = \sqrt {\dfrac{{10.\left\\{ {{{\left( {2\sqrt 6 } \right)}^2} + 0} \right\\} + 20\left\\{ {{{\left( {3\sqrt 2 } \right)}^2} + 0} \right\\}}}{{10 + 20}}}
Simplifying the terms,
σ12=10×24+20×1830\Rightarrow {\sigma _{12}} = \sqrt {\dfrac{{10 \times 24 + 20 \times 18}}{{30}}}
Multiplying the terms we get,
σ12=240+36030\Rightarrow {\sigma _{12}} = \sqrt {\dfrac{{240 + 360}}{{30}}}
Adding the terms we get,
σ12=60030\Rightarrow {\sigma _{12}} = \sqrt {\dfrac{{600}}{{30}}}
Dividing the terms,
σ12=20\Rightarrow {\sigma _{12}} = \sqrt {20}
Taking square root,
σ12=25\Rightarrow {\sigma _{12}} = 2\sqrt 5
The S.D. of combined group of 3030 observations is σ12=25{\sigma _{12}} = 2\sqrt 5

So, the correct answer is “Option B”.

Note: Standard deviation:
In statistics, the standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.
σ=(xiμ)2N\sigma = \sqrt {\dfrac{{\sum {{{\left( {{x_i} - \mu } \right)}^2}} }}{N}}
Where,
σ\sigma =population standard deviation
NN= the size of the population
xi{x_i}=each value from the population
μ\mu = the population mean.