
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 $ 0 $ $ 8 $ $ 6 $ $ 4 $ C $ 8 $ $ 11 $ Did Not played $ 13 $
Who is the best performer?
Player | Game $ 1 $ | Game $ 2 $ | Game $ 3 $ | Game $ 4 $ |
A | $ 14 $ | $ 16 $ | $ 10 $ | $ 10 $ |
B | $ 0 $ | $ 8 $ | $ 6 $ | $ 4 $ |
C | $ 8 $ | $ 11 $ | Did Not played | $ 13 $ |
Answer
432.6k+ views
Hint: To know the best performer in the game we will find the average. Average is also known as the mean which is the sum of observations upon the total number of the observations. Find the mean for each player, and a player with a larger mean is the best performer.
Complete step-by-step answer:
To find the best performer, first of all find the average of the games played by each player.
Average is also known as the mean which is calculated by using the formula the total score of all the four games upon the total number of games.
For player A:
Mean, $ \overline {{x_A}} = \dfrac{{14 + 16 + 10 + 10}}{4} $
Simplify the above expression finding the sum of the terms in the numerator –
$ \overline {{x_A}} = \dfrac{{50}}{4} $
Find the division,
$ \overline {{x_A}} = 12.5 $ ….. (A)
Similarly find for Player B:
Mean, $ \overline {{x_B}} = \dfrac{{0 + 8 + 6 + 4}}{4} $
Simplify the above expression finding the sum of the terms in the numerator –
$ \overline {{x_B}} = \dfrac{{18}}{4} $
Find the division,
$ \overline {{x_B}} = 4.5 $ ….. (B)
Now for player C:
Mean, $ \overline {{x_C}} = \dfrac{{8 + 11 + 13}}{3} $
Simplify the above expression finding the sum of the terms in the numerator –
$ \overline {{x_C}} = \dfrac{{32}}{3} $
Find the division,
$ \overline {{x_C}} = 10.5 $ ….. (C)
From equation A, B and C - By observation and comparing we can say the A is the best performer.
Note: Remember the difference between mean and median. Mean can be defined as the sum of all the numbers upon the total number of the numbers. Mean is affected by the extreme values of the sequences or the observations in the range while in the median extreme values do not affect the extreme values
Complete step-by-step answer:
To find the best performer, first of all find the average of the games played by each player.
Average is also known as the mean which is calculated by using the formula the total score of all the four games upon the total number of games.
For player A:
Mean, $ \overline {{x_A}} = \dfrac{{14 + 16 + 10 + 10}}{4} $
Simplify the above expression finding the sum of the terms in the numerator –
$ \overline {{x_A}} = \dfrac{{50}}{4} $
Find the division,
$ \overline {{x_A}} = 12.5 $ ….. (A)
Similarly find for Player B:
Mean, $ \overline {{x_B}} = \dfrac{{0 + 8 + 6 + 4}}{4} $
Simplify the above expression finding the sum of the terms in the numerator –
$ \overline {{x_B}} = \dfrac{{18}}{4} $
Find the division,
$ \overline {{x_B}} = 4.5 $ ….. (B)
Now for player C:
Mean, $ \overline {{x_C}} = \dfrac{{8 + 11 + 13}}{3} $
Simplify the above expression finding the sum of the terms in the numerator –
$ \overline {{x_C}} = \dfrac{{32}}{3} $
Find the division,
$ \overline {{x_C}} = 10.5 $ ….. (C)
From equation A, B and C - By observation and comparing we can say the A is the best performer.
Note: Remember the difference between mean and median. Mean can be defined as the sum of all the numbers upon the total number of the numbers. Mean is affected by the extreme values of the sequences or the observations in the range while in the median extreme values do not affect the extreme values
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