CBSE Class 7 Mathematics Chapter 3: Data Handling NCERT Solutions

First, we organize the given marks in a tabular form using tally marks to count the frequency of each mark.

(i) The highest number in the data is 9.

(ii) The lowest number in the data is 1.

(iii) The range of the data is the difference between the highest and lowest numbers.

Range = Highest number – Lowest number

Range = $9\; -\; 1$

Range = $8$.

(iv) To find the arithmetic mean, we sum all the marks and divide by the total number of students.

Sum of marks = $4+6+7+5+3+5+4+5+2+6+2+5+1+9+6+5+8+4+6+7$

Total number of students = 20

Arithmetic mean = $\backslash frac\{\backslash text\{Sum\; of\; marks\}\}\{\backslash text\{Total\; number\; of\; students\}\}$

Arithmetic mean = $\backslash frac\{100\}\{20\}$

Arithmetic mean = $5$.

Thus, the arithmetic mean of the marks is 5.

The first five whole numbers are 0, 1, 2, 3, and 4. Whole numbers start from 0.

To find the mean, we sum these numbers and divide by the count of numbers, which is 5.

Sum of the first five whole numbers = $0\; +\; 1\; +\; 2\; +\; 3\; +\; 4$

Sum = $10$

Mean = $\backslash frac\{\backslash text\{Sum\; of\; numbers\}\}\{\backslash text\{Total\; number\; of\; numbers\}\}$

Mean = $\backslash frac\{10\}\{5\}$

Mean = $2$.

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

The runs scored by the cricketer in eight innings are 58, 76, 40, 35, 46, 45, 0, and 100.

The number of innings is $8$.

To find the mean score, we sum all the runs scored and divide by the number of innings.

Sum of scores = $58\; +\; 76\; +\; 40\; +\; 35\; +\; 46\; +\; 45\; +\; 0\; +\; 100$

Sum of scores = $400$

Mean score = $\backslash frac\{\backslash text\{Sum\; of\; scores\}\}\{\backslash text\{Number\; of\; innings\}\}$

Mean score = $\backslash frac\{400\}\{8\}$

Mean score = $50$.

Thus, the mean score of the cricketer is 50.

(i) To find the mean points scored by Player A per game, we sum A's scores and divide by the number of games played.

Player A's scores: 14, 16, 10, 10.

Number of games played by A = 4.

Mean of player A = $\backslash frac\{\backslash text\{Sum\; of\; scores\; by\; A\}\}\{\backslash text\{No.\; of\; games\; played\; by\; A\}\}$

Mean of player A = $\backslash frac\{14\; +\; 16\; +\; 10\; +\; 10\}\{4\}$

Mean of player A = $\backslash frac\{50\}\{4\}$

Mean of player A = $12.5$.

Player A's average score per game is 12.5 points.

(ii) Player C played in Game 1, Game 2, and Game 4. Player C did not play in Game 3. Therefore, Player C played only 3 games.

To find the mean number of points per game for C, we should divide the total points scored by Player C by the number of games Player C actually played, which is 3.

(iii) Player B played in all four games. To find the mean score for Player B, we sum B's scores and divide by the total number of games played, which is 4.

Player B's scores: 8, 4, 0, 6.

Number of games played by B = 4.

Mean of player B = $\backslash frac\{\backslash text\{Sum\; of\; scores\; by\; B\}\}\{\backslash text\{No.\; of\; games\; played\; by\; B\}\}$

Mean of player B = $\backslash frac\{8\; +\; 4\; +\; 0\; +\; 6\}\{4\}$

Mean of player B = $\backslash frac\{18\}\{4\}$

Mean of player B = $4.5$.

(iv) To determine the best performer, we compare the mean scores of all players.

Mean of player A = 12.5 points per game.

Mean of player B = 4.5 points per game.

For Player C, the scores are 8, 11, and 13 (since 'Did not play' means 0 points for that game, but for calculating the average per game played, we exclude the game not played or consider only the games played). Player C played 3 games.

Mean of player C = $\backslash frac\{\backslash text\{Sum\; of\; scores\; by\; C\}\}\{\backslash text\{No.\; of\; games\; played\; by\; C\}\}$

Mean of player C = $\backslash frac\{8\; +\; 11\; +\; 13\}\{3\}$

Mean of player C = $\backslash frac\{32\}\{3\}$

Mean of player C ≈ $10.67$ points per game.

Comparing the means: Player A (12.5), Player B (4.5), Player C (10.67).

Since Player A has the highest average score per game, Player A is the best performer.