Data Handling — NCERT Solutions, AI Tutor & Practice

Arrange the data in a frequency table:

| Class Interval | Frequency | |---|---| | 0-10 | 2 | | 10-20 | 5 | | 20-30 | 7 | | 30-40 | 4 | | 40-50 | 2 |

Highest frequency: The class interval $20\text{-}30$ has the highest frequency, which is $7$.

Range of marks: The data spans from the lower limit of the first class to the upper limit of the last class. $\text{Range} = 50 - 0 = 50$

Total students $= 2+5+7+4+2 = 20$ (check).

Total number of balls $= 5 + 3 + 2 = 10$.

(a) P(red): $P(\text{red}) = \frac{\text{red balls}}{\text{total}} = \frac{5}{10} = \frac{1}{2}$

(b) P(not blue): Balls that are not blue $= 5 + 3 = 8$. $P(\text{not blue}) = \frac{8}{10} = \frac{4}{5}$

(c) P(green or red): Favourable $= 3 + 5 = 8$. $P(\text{green or red}) = \frac{8}{10} = \frac{4}{5}$

A die has outcomes $\{1, 2, 3, 4, 5, 6\}$, so total outcomes $= 6$.

(a) Even number: favourable outcomes $\{2, 4, 6\}$, count $= 3$. $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$

(b) Number greater than 4: favourable outcomes $\{5, 6\}$, count $= 2$. $P(>4) = \frac{2}{6} = \frac{1}{3}$

(c) Prime number: primes are $\{2, 3, 5\}$, count $= 3$. $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$

The central angle for each item $= \dfrac{\text{percentage}}{100} \times 360^\circ$.

| Item | Percentage | Central Angle | |---|---|---| | Food | 40% | $\frac{40}{100}\times 360^\circ = 144^\circ$ | | Rent | 25% | $\frac{25}{100}\times 360^\circ = 90^\circ$ | | Education | 15% | $\frac{15}{100}\times 360^\circ = 54^\circ$ | | Savings | 10% | $\frac{10}{100}\times 360^\circ = 36^\circ$ | | Others | 10% | $\frac{10}{100}\times 360^\circ = 36^\circ$ |

Check: $144^\circ + 90^\circ + 54^\circ + 36^\circ + 36^\circ = 360^\circ$. ✓

Draw a circle and mark sectors with these angles using a protractor.

The fraction of the circle for Sports is: $\frac{72^\circ}{360^\circ} = \frac{1}{5}$

Number of students who chose Sports: $= \frac{72}{360} \times 200 = \frac{1}{5} \times 200 = 40$

So 40 students chose Sports.

Count the occurrences of each value using tally marks:

| Hours | Tally | Frequency | |---|---|---| | 1 | |||| | | 6 | | 2 | |||| |||| | 10 | | 3 | |||| || | 8 | | 4 | |||| | 4 | | 5 | || | 2 |

Check total: $6 + 10 + 8 + 4 + 2 = 30$ students. ✓

The most common value is 2 hours (frequency 10).

When two coins are tossed, the sample space is: $\{HH, HT, TH, TT\}$ Total outcomes $= 4$.

(a) Two heads: favourable $\{HH\}$, count $= 1$. $P(\text{two heads}) = \frac{1}{4}$

(b) Exactly one head: favourable $\{HT, TH\}$, count $= 2$. $P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}$

(c) At least one tail: favourable $\{HT, TH, TT\}$, count $= 3$. $P(\text{at least one tail}) = \frac{3}{4}$

Total frequency: $4 + 9 + 6 + 3 = 22$

Difference between a histogram and a bar graph:

1. A histogram represents grouped (continuous) data using class intervals, so the bars are drawn adjacent to each other with no gaps.
2. A bar graph represents discrete or categorical data, so the bars have equal gaps between them.
3. In a histogram, the width of each bar corresponds to the class interval; in a bar graph, the width has no special meaning.

Here, since the class intervals (100-150, etc.) are continuous, a histogram is appropriate and the bars will touch each other.

Total outcomes $= 25$ (numbers $1$ to $25$).

(a) Multiple of 5: $\{5, 10, 15, 20, 25\}$, count $= 5$. $P = \frac{5}{25} = \frac{1}{5}$

(b) Perfect square: $\{1, 4, 9, 16, 25\}$, count $= 5$. $P = \frac{5}{25} = \frac{1}{5}$

(c) Neither prime nor 1: Primes up to 25 are $\{2,3,5,7,11,13,17,19,23\}$ (9 numbers), plus the number $1$. So numbers that ARE prime or 1 $= 9 + 1 = 10$. Therefore numbers that are neither $= 25 - 10 = 15$. $P = \frac{15}{25} = \frac{3}{5}$

Central angle $= \dfrac{\text{value of item}}{\text{total}} \times 360^\circ$. Total $= 720$.

| Fruit | Number | Central Angle | |---|---|---| | Mango | 240 | $\frac{240}{720}\times 360^\circ = 120^\circ$ | | Apple | 180 | $\frac{180}{720}\times 360^\circ = 90^\circ$ | | Banana | 120 | $\frac{120}{720}\times 360^\circ = 60^\circ$ | | Orange | 180 | $\frac{180}{720}\times 360^\circ = 90^\circ$ |

Check: $120^\circ + 90^\circ + 60^\circ + 90^\circ = 360^\circ$. ✓

Total equally likely outcomes $= 8$: $\{1,2,3,4,5,6,7,8\}$.

(a) Odd number: $\{1,3,5,7\}$, count $= 4$. $P(\text{odd}) = \frac{4}{8} = \frac{1}{2}$

(b) Number less than 3: $\{1,2\}$, count $= 2$. $P(<3) = \frac{2}{8} = \frac{1}{4}$

(c) Divisible by 3: $\{3,6\}$, count $= 2$. $P(\text{divisible by 3}) = \frac{2}{8} = \frac{1}{4}$

Frequency table:

| Wages (₹) | Workers | |---|---| | 200-250 | 5 | | 250-300 | 12 | | 300-350 | 15 | | 350-400 | 8 |

Tallest bar: The class interval $300\text{-}350$ has the highest frequency ($15$), so its bar will be the tallest.

Workers earning ₹300 or more: These fall in intervals $300\text{-}350$ and $350\text{-}400$. $= 15 + 8 = 23 \text{ workers}$

Fraction of total workers: $\frac{23}{40}$

Total check: $5 + 12 + 15 + 8 = 40$. ✓