Ex.3.1 Q5 Data Handling - NCERT Maths Class 7
Question
Following table shows the point of each player scored in four games: -
Player |
Game 1 |
Game 2 |
Game 3 |
Game 4 |
\(A\) |
\(14\) |
\(16\) |
\(10\) |
\(10\) |
\(B\) |
\(10\) |
\(8\) |
\(6\) |
\(4\) |
C |
\(8\) |
\(11\) |
Didn’t play |
\(13\) |
Now answer the following questions: -
(i) Find the mean to detetextine \(A’s\) average number of points scored per game.
(ii) To find the mean number of points per game for \(C,\) would you divide the points by \(3\) or \(4\)? Why?
(iii) \(B\) played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Text Solution
Reasoning:
What is known?
Point of each player scored in four games.
What is unknown?
Average score, Mean
Reasoning: Mean \(=\)\( \begin{align}\rm\frac{Sum\text{ }of \text{ score }}{Number\text{ }of\text{ games}}\end{align}\)
Steps:
Total number of games played by \(A = 4\)
Scores obtained by \(A = 14, 16, 10, 10\)
\[\begin{align}{\text{Mean of player A}} &= \frac{{14 + 16 + 10 + 10}}{4}\\ &= \frac{{50}}{4}\\ &= 12.5\end{align}\]
ii) We should divide total points by \(3\) because player \(C\) played only three games.
iii) Total number of games played by \(B = 4\)
Scores obtained by \(B = 0, 8, 6, 4\)
\[\begin{align}{\text{Mean of player B}} &= \frac{{{\text{Sum of scores by B}}}}{{{\text{No}}{\text{. of games played by B}}}}\\ &= \frac{{0 + 8 + 6 + 4}}{4}\\ &= \frac{{18}}{4}\\ &= 4.5\end{align}\]
iv) To find the best performer,
We should find the mean of all players.
Mean of player \(A = 12.5\)
Mean of player \(B = 4.5\)
\[\begin{align}{\text{Mean of player C}} &= \frac{{{\text{sum of scores by C}}}}{{{\text{No}}{\text{. of games played by C}}}}\\ &=\frac{{8 + 11 + 13}}{3}\\ &= \frac{{32}}{3}\\ &= 10.67\end{align}\]
Therefore, on comparing means of all players, \(A\) is the best performer.