Question

The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of the remaining 55% is

Answer 1

Hint: We first assume the total number of students for the group as $100x$. We then take the percentage values for the two divided groups and find their total number. We then find the total number allotted for the remaining students and find the mean.

Complete step-by-step solution:
Let us assume the number of students in the group is $100x$.
The arithmetic mean of the scores of a group of students in a test was 52.
Therefore, the total numbers obtained by the students is $100x\times 52=5200x$.
We need to find the 20% and 25% of the total number of students.
We know for any arbitrary percentage value of a%, we can write it as $\dfrac{a}{100}$. The percentage is to find the respective value out of 100.
Therefore, 20% and 25% can be written as $\dfrac{20}{100}$ and $\dfrac{25}{100}$ respectively.
The number of students will be $100x\times \dfrac{20}{100}=20x$ and $100x\times \dfrac{25}{100}=25x$ respectively.
The respective total scores will be $20x\times 80=1600x$ and $25x\times 31=775x$.
Therefore, the remaining scores $5200x-1600x-775x=2825x$ is allotted for the rest of the students.
The number of those students is $100x\times \dfrac{55}{100}=55x$.
The mean for those students will be $\dfrac{2825x}{55x}=51.36$.
The mean score of the remaining 55% is $51.36$.

Note: These percentages can be considered as the weighted means. We can use them as frequencies to find the mean of the whole observation. The formula will be the same for the mean that we used earlier.