Statistics — NCERT Solutions, AI Tutor & Practice

Count how many times each value appears using tally marks (bundle every fifth stroke).

| Number of children | Tally marks | Frequency | |:---:|:---:|:---:| | 0 | ll | 2 | | 1 | llll | 4 | | 2 | llll lll | 8 | | 3 | llll | 4 | | 4 | ll | 2 | | Total | | 20 |

All frequencies add up to $2+4+8+4+2 = 20$, which equals the number of families, confirming the table is complete.

Use the exclusive (continuous) method: a value equal to the upper limit goes into the next class. So $30$ belongs to $30$–$40$, not $20$–$30$.

| Class interval | Tally | Frequency | |:---:|:---:|:---:| | 0–10 | lll | 3 | | 10–20 | llll l | 6 | | 20–30 | llll lll | 8 | | 30–40 | llll ll | 7 | | 40–50 | llll l | 6 | | Total | | 30 |

Check: $3+6+8+7+6 = 30$ students. ✓

Tally each blood group:

| Blood group | Tally | Frequency | |:---:|:---:|:---:| | A | llll ll | 7 | | B | llll l | 6 | | O | llll llll | 9 | | AB | lll | 3 | | Total | | 25 |

Check: $7+6+9+3 = 25$. ✓

The most common blood group is O, occurring 9 times.

Frequency table:

| Wage (₹) | Frequency | |:---:|:---:| | 200–250 | 8 | | 250–300 | 14 | | 300–350 | 12 | | 350–400 | 6 |

Bar graph: On the x-axis mark the wage classes and on the y-axis mark frequency (scale: 1 unit = 2 workers). Draw bars of heights 8, 14, 12 and 6 of equal width with equal gaps between them.

Workers earning less than ₹300 = classes 200–250 and 250–300: $8 + 14 = 22 \text{ workers}.$

Frequency table:

| Hours | Frequency | |:---:|:---:| | 1 | 4 | | 2 | 5 | | 3 | 4 | | 4 | 2 |

Check: $4+5+4+2 = 15$. ✓

Bar graph: x-axis = number of hours (1, 2, 3, 4), y-axis = frequency (scale 1 unit = 1 student). Bars of heights 4, 5, 4, 2 of equal width with equal spacing.

Range = maximum value − minimum value $= 4 - 1 = 3$ hours.

Bar graph: x-axis = sports, y-axis = number of students (scale 1 unit = 2 students). Draw bars of equal width and equal gaps with heights 18, 12, 8, 7, 5.

Total students $= 18+12+8+7+5 = 50$. ✓

Fraction preferring Cricket: $\frac{18}{50} = \frac{9}{25}.$

The first ten prime numbers are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29.$

Sum $= 2+3+5+7+11+13+17+19+23+29 = 129.$

Number of observations $n = 10$.

$\bar{x} = \frac{\text{Sum of observations}}{n} = \frac{129}{10} = 12.9.$

The mean of the first ten prime numbers is $12.9$.

Mean $= \dfrac{\text{sum}}{n}$, so sum of 8 observations: $\text{Sum} = 15 \times 8 = 120.$

Remove the observation 20: $\text{New sum} = 120 - 20 = 100,$ with $n = 7$ remaining observations.

$\text{New mean} = \frac{100}{7} \approx 14.29.$

Original sum $= 27 \times 5 = 135.$

When each of the 5 numbers is increased by 4, the total increases by $4 \times 5 = 20$: $\text{New sum} = 135 + 20 = 155.$

$\text{New mean} = \frac{155}{5} = 31.$

Shortcut: adding a constant $c$ to every observation increases the mean by $c$: $27 + 4 = 31$.

Use $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$.

| Goals $x_i$ | Frequency $f_i$ | $f_i x_i$ | |:---:|:---:|:---:| | 0 | 3 | 0 | | 1 | 6 | 6 | | 2 | 5 | 10 | | 3 | 4 | 12 | | 4 | 2 | 8 | | Total | 20 | 36 |

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{36}{20} = 1.8 \text{ goals per match}.$

Mean $= \dfrac{\text{sum}}{6} = 52$, so the total must be: $\text{Sum} = 52 \times 6 = 312.$

Add the known marks: $45 + 50 + 55 + x + 60 + 40 = 312$ $250 + x = 312$ $x = 312 - 250 = 62.$

So the missing mark is $x = 62$.

The class mark is $x_i = \dfrac{\text{lower limit} + \text{upper limit}}{2}$.

| Class | Class mark $x_i$ | $f_i$ | $f_i x_i$ | |:---:|:---:|:---:|:---:| | 150–155 | 152.5 | 4 | 610 | | 155–160 | 157.5 | 6 | 945 | | 160–165 | 162.5 | 7 | 1137.5 | | 165–170 | 167.5 | 3 | 502.5 | | Total | | 20 | 3195 |

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3195}{20} = 159.75 \text{ cm}.$