Question

Two sets each of $20$ observations, have the same standard deviation $5$. The first set has a mean $17$ and second a mean $22$. Then the standard deviation of the set obtained by combining the given two sets is
${\text{(A) 5}}$
${\text{(B) 4}}{\text{.5}}$
${\text{(C) 5}}{\text{.59}}$
${\text{(D) 4}}$

Answer 1

Hint: Here in this question as known the values of mean, standard deviation and the number of terms in both the distribution, we will substitute all the values in the combined standard deviation formula to get the required answer.

Formula used: ${\text{Combined S}}{\text{.D = }}\sqrt {\dfrac{{{{\text{n}}_{\text{1}}}{{{\sigma}}_{\text{1}}}^{\text{2}}{\text{+}}{{\text{n}}_{\text{2}}}{{{\sigma }}_{\text{2}}}^{\text{2}}}}{{{{\text{n}}_{\text{1}}}{\text{ + }}{{\text{n}}_{\text{2}}}}}{\text{+}}\dfrac{{{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}{{{\text{(}}{{{{\bar x}}}_{\text{1}}}{\text{-}}{{{{\bar x}}}_{\text{2}}}{\text{)}}}^{\text{2}}}}}{{{{{\text{(}}{{\text{n}}_{\text{1}}}{\text{+}}{{\text{n}}_{\text{2}}}{\text{)}}}^{\text{2}}}}}} $
Where ${\text{S}}{\text{.D}}$ stands for the standard deviation

Complete step-by-step solution:
Let the number of terms in both the distribution be ${n_1}$ and ${n_2}$, since the total number of observations are same in both the sets,
${n_1} = 20$ and ${n_2} = 20$
Let ${\sigma _1}$ and ${\sigma _2}$ be the standard deviation of both the sets, since the standard deviation is same for both the sets, it can be written as:
${\sigma _1} = 17$ and ${\sigma _2} = 22$
Let ${{{\bar x}}_{\text{1}}}$ and ${{{\bar x}}_2}$ be the mean of both the distributions therefore,
${{{\bar x}}_{\text{1}}} = 17$ and ${{{\bar x}}_2} = 22$
On substituting all the values in the formula, we get:
${\text{Combined S}}{\text{.D = }}\sqrt {\dfrac{{{\text{20}} \times {5^{\text{2}}}{\text{ + 20}} \times {5^{\text{2}}}}}{{20 + 20}}{\text{ + }}\dfrac{{{\text{20}} \times 20 \times {{{\text{(17 - 22)}}}^{\text{2}}}}}{{{{{\text{(20 + 20)}}}^{\text{2}}}}}} $
On squaring the terms we get:
$ \Rightarrow \sqrt {\dfrac{{{\text{20}} \times 25{\text{ + 20}} \times 25}}{{20 + 20}}{\text{ + }}\dfrac{{{\text{20}} \times 20 \times {{{\text{( - 5)}}}^{\text{2}}}}}{{{{{\text{(20 + 20)}}}^{\text{2}}}}}} $
Let us add the denominator term and we get
$ \Rightarrow \sqrt {\dfrac{{{\text{20}} \times 25{\text{ + 20}} \times 25}}{{40}}{\text{ + }}\dfrac{{{\text{20}} \times 20 \times 25}}{{{\text{4}}{{\text{0}}^{\text{2}}}}}} $
Let us multiply the numerator term and we can write it as,
$ \Rightarrow \sqrt {\dfrac{{500 + 500}}{{40}}{\text{ + }}\dfrac{{{\text{25}} \times 400}}{{{\text{4}}{{\text{0}}^{\text{2}}}}}} $
On adding the numerator term and we get,
\[ \Rightarrow \sqrt {\dfrac{{1000}}{{40}}{\text{ + }}\dfrac{{10000}}{{1600}}} \]
Let us divide the term and we get
$ \Rightarrow \sqrt {{\text{25 + }}\dfrac{{25}}{{\text{4}}}} $
On taking the L.C.M we get:
$ \Rightarrow \sqrt {\dfrac{{25 \times 4 + 25}}{4}} $
This can be simplified as:
$ \Rightarrow \sqrt {\dfrac{{125}}{4}} $
Since the square root of $4$ is $2$ we take it out of the root part.
$ \Rightarrow \dfrac{1}{2}\sqrt {125} $
Now the root value of $\sqrt {125} $ is $11.18$ therefore,
$ \Rightarrow \dfrac{{11.18}}{2}$
$ \Rightarrow 5.59$
${\text{Combined S}}{\text{.D = 5}}{\text{.59}}$, which is the required answer.

Therefore, the correct option is ${\text{(C)}}$ which is $5.59$.

Note: The combined Standard deviation of two distributions would always be a very close answer to the original standard deviations of the two sets.
Also, in statistics there is a relation between the variance and standard deviation of a distribution. The standard deviation is the square root of the variance, it can be expressed as:
${\text{Standard deviation = }}\sqrt {{\text{variance}}} $