NCERT Solutions for Class 10 Maths Chapter 13 Statistics Exercise 13.3#

Monthly consumption (units)	$f_i$	$cf$	$x_i$	$f_i x_i$
65-85	4	4	75	300
85-105	5	9	95	475
105-125	13	22	115	1495
125-145	20	42	135	2700
145-165	14	56	155	2170
165-185	8	64	175	1400
185-205	4	68	195	780
Total	68			9320

Mean:

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{9320}{68} = 137.05 \text{ (about)}$$

Median: $n = 68$, $\dfrac{n}{2} = 34$

Median class: 125-145 (cf before = 22, cf after = 42 ≥ 34)

$l = 125$, $cf = 22$, $f = 20$, $h = 20$

Using Median $= l + \dfrac{\frac{n}{2} - cf}{f} \times h$, we find the median:

$$= 125 + \frac{34 - 22}{20} \times 20$$

$$= 125 + 12$$

$$= 137$$

Mode: Modal class = 125-145

$l = 125$, $f_1 = 20$, $f_0 = 13$, $f_2 = 14$, $h = 20$

Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:

$$= 125 + \frac{20 - 13}{2(20) - 13 - 14} \times 20$$

$$= 125 + \frac{7}{13} \times 20$$

$$= 125 + 10.76$$

$$= 135.76 \text{ (about)}$$

∴ Median = 137 units, mean ≈ 137.05 units and mode ≈ 135.76 units.

Comparison: The three measures are nearly equal (about 137 units), so the distribution is nearly symmetric.

Class interval	$f_i$	$cf$
0-10	5	5
10-20	x	5+x
20-30	20	25+x
30-40	15	40+x
40-50	y	40+x+y
50-60	5	45+x+y
Total	60

The total frequency is 60, so:

$$5 + x + 20 + 15 + y + 5 = 60$$

$$\Rightarrow x + y = 15 \quad \cdots (1)$$

The median is 28.5, which lies in the class 20-30, so this is the median class. Here:
$\dfrac{n}{2} = 30$
$l = 20$, $cf = 5 + x$, $f = 20$, $h = 10$

Using Median $= l + \dfrac{\frac{n}{2} - cf}{f} \times h$, the median is 28.5, so:

$$28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10$$

$$\Rightarrow 28.5 = 20 + \frac{25 - x}{2}$$

$$\Rightarrow 8.5 = \frac{25 - x}{2}$$

$$\Rightarrow 17 = 25 - x$$

$$\Rightarrow x = 8$$

Substituting $x = 8$ in equation (1):

$$x + y = 15$$

$$8 + y = 15$$

$$\Rightarrow y = 7$$

∴ $x = \mathbf{8}$ and $y = \mathbf{7}$

Age (years)	Number of policy holders	$cf$
18-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

$n = 100$, $\dfrac{n}{2} = 50$

Median class: 35-40 (cf = 45 < 50, next cf = 78 ≥ 50)

$l = 35$, $cf = 45$, $f = 33$, $h = 5$

Using Median $= l + \dfrac{\frac{n}{2} - cf}{f} \times h$, we find the median:

$$= 35 + \frac{50 - 45}{33} \times 5$$

$$= 35 + 0.76$$

$$= 35.76 \text{ (about)}$$

∴ The median age is about 35.76 years.

The classes are not continuous, so we use the class boundaries: 118-126 becomes 117.5-126.5, and so on.

Length (mm)	$f_i$	$cf$
118-126	3	3
127-135	5	8
136-144	9	17
145-153	12	29
154-162	5	34
163-171	4	38
172-180	2	40

Using actual boundaries (117.5-126.5, etc.), $h = 9$.

$n = 40$, $\dfrac{n}{2} = 20$

Median class: 145-153 (actual: 144.5-153.5), cf = 17 < 20, next = 29 ≥ 20.

$l = 144.5$, $cf = 17$, $f = 12$, $h = 9$

Using Median $= l + \dfrac{\frac{n}{2} - cf}{f} \times h$, we find the median:

$$= 144.5 + \frac{20 - 17}{12} \times 9$$

$$= 144.5 + 2.25$$

$$= 146.75$$

∴ The median length of the leaves is 146.75 mm.

Life time (hours)	$f_i$	$cf$
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

$n = 400$, $\dfrac{n}{2} = 200$

Median class: 3000-3500 (cf = 130 < 200, next = 216 ≥ 200)

$l = 3000$, $cf = 130$, $f = 86$, $h = 500$

Using Median $= l + \dfrac{\frac{n}{2} - cf}{f} \times h$, we find the median:

$$= 3000 + \frac{200 - 130}{86} \times 500$$

$$= 3000 + \frac{35000}{86}$$

$$= 3000 + 406.98$$

$$= 3406.98 \text{ (about)}$$

∴ The median life time of a lamp is about 3406.98 hours.

Number of letters	$f_i$	$cf$	$x_i$	$f_i x_i$
1-4	6	6	2.5	15
4-7	30	36	5.5	165
7-10	40	76	8.5	340
10-13	16	92	11.5	184
13-16	4	96	14.5	58
16-19	4	100	17.5	70
Total	100			832

Mean:

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{832}{100} = 8.32$$

Median: $\dfrac{n}{2} = 50$; Median class: 7-10 (cf = 36 < 50, next = 76 ≥ 50)

$l = 7$, $cf = 36$, $f = 40$, $h = 3$

Using Median $= l + \dfrac{\frac{n}{2} - cf}{f} \times h$, we find the median:

$$= 7 + \frac{50 - 36}{40} \times 3$$

$$= 7 + 1.05$$

$$= 8.05$$

Mode: Modal class = 7-10

$l = 7$, $f_1 = 40$, $f_0 = 30$, $f_2 = 16$, $h = 3$

Using Mode $= l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$, we find the mode:

$$= 7 + \frac{40 - 30}{2(40) - 30 - 16} \times 3$$

$$= 7 + \frac{10}{34} \times 3$$

$$= 7 + 0.88$$

$$= 7.88 \text{ (about)}$$

∴ Median = 8.05, mean = 8.32 and modal size ≈ 7.88 letters.

Weight (kg)	$f_i$	$cf$
40-45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

$n = 30$, $\dfrac{n}{2} = 15$

Median class: 55-60 (cf = 13 < 15, next = 19 ≥ 15)

$l = 55$, $cf = 13$, $f = 6$, $h = 5$

Using Median $= l + \dfrac{\frac{n}{2} - cf}{f} \times h$, we find the median:

$$= 55 + \frac{15 - 13}{6} \times 5$$

$$= 55 + 1.67$$

$$= 56.67 \text{ (about)}$$

∴ The median weight is about 56.67 kg.