CBSE Class 7 Mathematics Chapter 3: Data Handling NCERT Solutions

To find the range of heights, we first need to collect the heights of ten students from your class. Let's assume the following heights (in feet) for ten students:

Gunjan: 4.2, Aditi: 4.5, Nikhil: 5.1, Akhil: 5.2, Riya: 5.3, Akshat: 5.1, Abhishek: 4.7, Mayank: 4.9, Rahul: 4.5, Ayush: 4.5

The range of a dataset is the difference between the highest value and the lowest value.

Highest height = 5.3 feet

Lowest height = 4.2 feet

Range = Highest height – Lowest height

Range = $5.3\; -\; 4.2\; =\; 1.1$ feet.

Therefore, the range of the heights of these ten students is 1.1 feet.

First, we organize the given marks in a tabular form using tally marks and frequency.

The given marks are: 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7.

Let's create a frequency table:

Total number of students = 1 + 2 + 1 + 3 + 5 + 4 + 2 + 1 + 1 = 20.

(i) The highest number is 9.

(ii) The lowest number is 1.

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

Range = Highest number – Lowest number = $9\; -\; 1\; =\; 8$.

(iv) The arithmetic mean is calculated as the sum of all observations divided by the total number of observations.

Sum of scores = 4+6+7+5+3+5+4+5+2+6+2+5+1+9+6+5+8+4+6+7 = 100

Total number of observations = 20

Arithmetic Mean = $\backslash frac\{\backslash text\{Sum\; of\; scores\}\}\{\backslash text\{Total\; number\; of\; observations\}\}\; =\; \backslash frac\{100\}\{20\}\; =\; 5$.

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 = $0\; +\; 1\; +\; 2\; +\; 3\; +\; 4\; =\; 10$.

Mean = $\backslash frac\{\backslash text\{Sum\; of\; numbers\}\}\{\backslash text\{Total\; number\; of\; numbers\}\}\; =\; \backslash frac\{10\}\{5\}\; =\; 2$.

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 scores and divide by the number of innings.

Sum of scores = $58\; +\; 76\; +\; 40\; +\; 35\; +\; 46\; +\; 45\; +\; 0\; +\; 100\; =\; 400$.

Mean score = $\backslash frac\{\backslash text\{Sum\; of\; scores\}\}\{\backslash text\{Number\; of\; innings\}\}\; =\; \backslash frac\{400\}\{8\}\; =\; 50$.

Thus, the mean score of the cricketer is 50.

We are given the scores of three players A, B, and C in four games. We need to calculate their average scores and determine the best performer.

(i) To find the mean number of points scored per game by Player A:

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

Number of games played by A = 4.

Mean of Player A = $\backslash frac\{\backslash text\{Sum\; of\; scores\; by\; A\}\}\{\backslash text\{No.\; of\; games\; played\; by\; A\}\}\; =\; \backslash frac\{14\; +\; 16\; +\; 10\; +\; 10\}\{4\}\; =\; \backslash frac\{50\}\{4\}\; =\; 12.5$.

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

(ii) To find the mean number of points per game for Player C:

Player C's scores are 8, 11, and 13. Player C 'Did not play' in Game 3.

We should divide the total points by 3 because Player C only played in 3 games. The 'Did not play' entry does not contribute to the number of games played or the score.

Mean of Player C = $\backslash frac\{8\; +\; 11\; +\; 13\}\{3\}\; =\; \backslash frac\{32\}\{3\}\; \backslash approx\; 10.67$.

(iii) To find the mean score for Player B:

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

Number of games played by B = 4.

Mean of Player B = $\backslash frac\{\backslash text\{Sum\; of\; scores\; by\; B\}\}\{\backslash text\{No.\; of\; games\; played\; by\; B\}\}\; =\; \backslash frac\{8\; +\; 4\; +\; 0\; +\; 6\}\{4\}\; =\; \backslash frac\{18\}\{4\}\; =\; 4.5$.

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

Mean of Player A = 12.5

Mean of Player B = 4.5

Mean of Player C = 10.67 (approximately)

Comparing these means, Player A has the highest average score (12.5 points per game).

Therefore, Player A is the best performer.