Statistics — NCERT Solutions, AI Tutor & Practice

Direct method: Mean $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$, where $x_i$ is the class mark.

| Class | $f_i$ | $x_i$ | $f_i x_i$ | |---|---|---|---| | 10–20 | 4 | 15 | 60 | | 20–30 | 6 | 25 | 150 | | 30–40 | 10 | 35 | 350 | | 40–50 | 6 | 45 | 270 | | 50–60 | 4 | 55 | 220 | | Total | 30 | | 1050 |

$\bar{x} = \frac{1050}{30} = 35$

The mean pocket money is ₹35.

Assumed-mean method: $\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}$, where $d_i = x_i - a$.

Take $a = 50$.

| Class | $f_i$ | $x_i$ | $d_i = x_i-50$ | $f_i d_i$ | |---|---|---|---|---| | 0–20 | 8 | 10 | -40 | -320 | | 20–40 | 12 | 30 | -20 | -240 | | 40–60 | 15 | 50 | 0 | 0 | | 60–80 | 10 | 70 | 20 | 200 | | 80–100 | 5 | 90 | 40 | 200 | | Total | 50 | | | -160 |

$\bar{x} = 50 + \frac{-160}{50} = 50 - 3.2 = 46.8$

The mean mark is 46.8.

Step-deviation method: $\bar{x} = a + h\left(\dfrac{\sum f_i u_i}{\sum f_i}\right)$, where $u_i = \dfrac{x_i - a}{h}$.

Take $a = 145$, $h = 10$.

| Class | $f_i$ | $x_i$ | $u_i$ | $f_i u_i$ | |---|---|---|---|---| | 120–130 | 5 | 125 | -2 | -10 | | 130–140 | 8 | 135 | -1 | -8 | | 140–150 | 12 | 145 | 0 | 0 | | 150–160 | 9 | 155 | 1 | 9 | | 160–170 | 6 | 165 | 2 | 12 | | Total | 40 | | | 3 |

$\bar{x} = 145 + 10\left(\frac{3}{40}\right) = 145 + 0.75 = 145.75$

The mean height is 145.75 cm.

Using class marks $x_i$: 12, 14, 16, 18, 20, 22, 24.

| $x_i$ | $f_i$ | $f_i x_i$ | |---|---|---| | 12 | 3 | 36 | | 14 | 6 | 84 | | 16 | 9 | 144 | | 18 | 13 | 234 | | 20 | $f$ | $20f$ | | 22 | 5 | 110 | | 24 | 4 | 96 | | Total | $40+f$ | $704 + 20f$ |

Mean $= 18$: $\frac{704 + 20f}{40 + f} = 18$ $704 + 20f = 720 + 18f$ $2f = 16 \implies f = 8$

The missing frequency is 8.

Mode formula: $\text{Mode} = l + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)\times h$.

The modal class is the one with maximum frequency = 20, i.e. 40–50.

• $l = 40$ (lower boundary of modal class)
• $f_1 = 20$ (frequency of modal class)
• $f_0 = 12$ (frequency of preceding class)
• $f_2 = 11$ (frequency of succeeding class)
• $h = 10$

$\text{Mode} = 40 + \left(\frac{20-12}{2(20)-12-11}\right)\times 10 = 40 + \frac{8}{17}\times 10$ $= 40 + 4.71 = 44.71$

The mode is approximately 44.71 goals.

The maximum frequency is 23, so the modal class is 35–45.

• $l = 35$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$

$\text{Mode} = 35 + \left(\frac{23-21}{2(23)-21-14}\right)\times 10$ $= 35 + \frac{2}{46-35}\times 10 = 35 + \frac{2}{11}\times 10$ $= 35 + 1.82 = 36.82$

The modal age is approximately 36.82 years.

Empirical relationship: $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$

Substitute Mean $= 53.5$ and Median $= 52.4$: $\text{Mode} = 3(52.4) - 2(53.5)$ $= 157.2 - 107 = 50.2$

The mode is approximately 50.2.

Median formula: $\text{Median} = l + \left(\dfrac{\frac{n}{2} - cf}{f}\right)\times h$.

Build the cumulative frequency (cf) table:

| Class | $f$ | cf | |---|---|---| | 100–120 | 12 | 12 | | 120–140 | 14 | 26 | | 140–160 | 8 | 34 | | 160–180 | 6 | 40 | | 180–200 | 10 | 50 |

$n = 50$, so $\frac{n}{2} = 25$.

The cf just reaching 25 is 26, so the median class is 120–140.

• $l = 120$, $cf = 12$ (cf of class before), $f = 14$, $h = 20$

$\text{Median} = 120 + \left(\frac{25 - 12}{14}\right)\times 20 = 120 + \frac{13}{14}\times 20$ $= 120 + 18.57 = 138.57$

The median wage is approximately ₹138.57.

Equation 1 (total frequency = 60): $5 + x + 20 + 15 + y + 5 = 60 \implies x + y = 15$

Since median $= 28.5$ lies in 20–30, that is the median class.

cf before median class $= 5 + x$, $f = 20$, $l = 20$, $h = 10$, $\frac{n}{2} = 30$.

$28.5 = 20 + \left(\frac{30 - (5+x)}{20}\right)\times 10$ $8.5 = \frac{25 - x}{2}$ $17 = 25 - x \implies x = 8$

From $x + y = 15$: $y = 15 - 8 = 7$.

So $x = 8$, $y = 7$.

'Less than' cumulative frequency table (accumulate up to upper class boundaries):

| Marks less than | Cumulative frequency | |---|---| | 10 | 10 | | 20 | 25 | | 30 | 50 | | 40 | 80 | | 50 | 100 |

Here $n = 100$, so $\frac{n}{2} = 50$.

The cumulative frequency first reaches 50 at 'less than 30', meaning entries beyond cf = 50 lie in 30–40. Since $\frac{n}{2}=50$ is reached exactly at cf of class 20–30, the median class is 30–40 (the class where cf first exceeds 50 for the next value).

Computing the median with $l = 30$, $cf = 50$, $f = 30$, $h = 10$: $\text{Median} = 30 + \left(\frac{50 - 50}{30}\right)\times 10 = 30$

So the median is 30 marks, and the 'less than' table shows how cf helps locate it.

Cumulative frequency table:

| Class | $f$ | cf | |---|---|---| | 200–300 | 4 | 4 | | 300–400 | 8 | 12 | | 400–500 | 12 | 24 | | 500–600 | 10 | 34 | | 600–700 | 6 | 40 |

$n = 40$, $\frac{n}{2} = 20$. The cf first reaching 20 is 24, so the median class is 400–500.

• $l = 400$, $cf = 12$, $f = 12$, $h = 100$

$\text{Median} = 400 + \left(\frac{20 - 12}{12}\right)\times 100 = 400 + \frac{8}{12}\times 100$ $= 400 + 66.67 = 466.67$

The median lifetime is approximately 466.67 hours.