Let the heights (in cm) of 10 students of our class be

125, 127, 132, 133, 134, 136,138, 141, 144, 146

Highest value among these observations = 146

Lowest value among these observations = 125

Range = Highest value − Lowest value

= (146 − 125) cm

= 21 cm

First five whole numbers are 0, 1, 2, 3, and 4.

Mean =$\frac{0+1+2+3+4}{5} = \frac{10}{5} = 2$

Therefore, the mean of first five whole numbers is 2.

Runs scored by the cricketer are 58, 76, 40, 35, 46, 45, 0, and 100.

$\mathrm{Mean} \mathrm{score} = \frac{\mathrm{Total} \mathrm{runs} \mathrm{scored} \mathrm{in} \mathrm{all} \mathrm{the} \mathrm{innings}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{the} \mathrm{innings}}$

Mean score = $\frac{58+76+40+35+46+45+0+100}{8} = \frac{400}{8} = 50$

Therefore, mean score is 50.

(i) A’s average number of points =$\frac{14+16+10+10}{4} = \frac{50}{4}=12.5$

(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played 3 games.

(iii) Mean of B’s score =$\frac{0+8+6+4}{4}=\frac{18}{4}=4.5$

(iv) The best performer will have the greatest average among all. Now we can observe that the average of A is 12.5 which is more than that of B and C. Therefore, A is the best performer among these three.