Answer:
The mean of the remaining 3 boys is 51.
Step-by-step explanation:
The mean mark of the entire group ([tex]p_{10}[/tex]), dimensionless, is:
[tex]p_{10} = \frac{1}{10}\cdot \Sigma_{i = 1}^{10} x_{i}[/tex] (Eq. 1)
([tex]p_{10} = 58[/tex])
[tex]\frac{1}{10}\cdot \Sigma_{i=1}^{10} x_{i} = 58[/tex] (Eq. 1b)
Where [tex]x_{i}[/tex] is the mark of the i-th boy, dimensionless.
In addition, we know the following mean marks from statement:
[tex]p_{7} = \frac{1}{7} \cdot \Sigma_{i = 1}^{7} x_{i}[/tex] (Eq. 2)
([tex]p_{7} = 61[/tex])
[tex]\frac{1}{7}\cdot \Sigma_{i=1}^{7} x_{i} = 61[/tex] (Eq. 2b)
[tex]p_{3} = \frac{1}{3}\cdot \Sigma_{i=8}^{10}x_{i}[/tex] (Eq. 3)
Where:
[tex]p_{7}[/tex] - Mean mark of the first 7 boys, dimensionless.
[tex]p_{3}[/tex] - Mean mark of the remaining 3 boys, dimensionless.
By applying sum properties in (Eq. 1b) and using (Eq. 2b) and (Eq. 3), we obtain the mean of the remaining 3 boys:
[tex]\frac{1}{10}\cdot [\Sigma_{i = 1}^{7}x_{i}+\Sigma_{i=8}^{10}x_{i}] = 58[/tex]
[tex]\frac{1}{10}\cdot [7\cdot (61)+3\cdot p_{3}] = 58[/tex]
[tex]7\cdot (61) + 3\cdot p_{3} = 580[/tex]
[tex]3\cdot p_{3} = 153[/tex]
[tex]p_{3} = \frac{153}{3}[/tex]
[tex]p_{3} = 51[/tex]
The mean of the remaining 3 boys is 51.
