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Question: The first two samples have 100 items with mean 15 and S.D. 3. If the whole group has 250 items with ...

The first two samples have 100 items with mean 15 and S.D. 3. If the whole group has 250 items with mean 15.6 and s.d.=13.44s.d. = \sqrt {13.44} , find the standard deviation of the second group.
A) 5
B) 4
C) 6
D) 3.52

Explanation

Solution

First of all, find the number of items in the second group by subtracting the number of items in the first group from the total number of items. Then, find the mean of the second group using the direct formula of total mean, when more than one group is present. Hence, find the standard deviation of the second group by substituting the values in the direct formula.

Complete step by step solution:
We are given that in group one there are 100 items with mean as 15 and the standard deviation is 3.
There are a total 250 items with mean as 15.6 and standard deviation as s.d.=13.44s.d. = \sqrt {13.44} .
Hence, in the group, there are 150 items left.
We have to calculate the mean of second group using the formula,
a=n1m1+n2m2n1+n2a = \dfrac{{{n_1}{m_1} + {n_2}{m_2}}}{{{n_1} + {n_2}}}, where aa is the total mean, n1{n_1} is the number of items in group one and n2{n_2} is the number of items in group 2.
We have m1=15{m_1} = 15, a=15.6a = 15.6, n1=100{n_1} = 100 and n2=150{n_2} = 150.
On substituting the values, we get,
15.6=100(15)+(150)m2250 m2=250(15.6)100(15)150 m2=240015 m2=16  15.6 = \dfrac{{100\left( {15} \right) + \left( {150} \right){m_2}}}{{250}} \\\ \Rightarrow {m_2} = \dfrac{{250\left( {15.6} \right) - 100\left( {15} \right)}}{{150}} \\\ \Rightarrow {m_2} = \dfrac{{2400}}{{15}} \\\ \Rightarrow {m_2} = 16 \\\
Thus, the mean of the second group is 16.
Also, standard deviation of whole group is given by,
σ2=n1(σ12+d12)+n2(σ22+d22)n1+n2{\sigma ^2} = \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, σ\sigma is the standard deviation of whole group, σ1{\sigma _1} is the standard deviation of group one and σ2{\sigma _2} is the standard deviation of group 2, d1=m1a{d_1} = {m_1} - a and d2=m2a{d_2} = {m_2} - a
(13.44)2=100(9+(1515.6)2)+150(σ22+(1615.6)2)100+150 13.44=100(9+0.36)+150(σ22+0.16)250 13.44(250)=936+150σ22+24 3360=936+150σ22+24 σ22=2400150 σ22=16 σ=4  {\left( {\sqrt {13.44} } \right)^2} = \dfrac{{100\left( {9 + {{\left( {15 - 15.6} \right)}^2}} \right) + 150\left( {\sigma _2^2 + {{\left( {16 - 15.6} \right)}^2}} \right)}}{{100 + 150}} \\\ \Rightarrow 13.44 = \dfrac{{100\left( {9 + 0.36} \right) + 150\left( {\sigma _2^2 + 0.16} \right)}}{{250}} \\\ \Rightarrow 13.44\left( {250} \right) = 936 + 150\sigma _2^2 + 24 \\\ \Rightarrow 3360 = 936 + 150\sigma _2^2 + 24 \\\ \Rightarrow \sigma _2^2 = \dfrac{{2400}}{{150}} \\\ \Rightarrow \sigma _2^2 = 16 \\\ \Rightarrow \sigma = 4 \\\
Therefore, the standard deviation of the second group is 4.

Hence, option B is correct.

Note:
We use standard deviation when our data is the sample of the whole population. Suppose in a state there are 50 cities. But, we have a population of 19 cities only and now we can estimate the population of the state by using only this data. Standard deviation is defined as the value by which members of a group differ from the mean value of the group. Also, one must remember the formulas for such questions.