Exercise 13.3 answers: Statistics

Open NCERT p. 198Matches NCERT’s answer

1. Consumption (units)	Consumers (f)	Cumulative f	Class mark (x)	f × x
   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

2. Median: n = 68, so n/2 = 34. The first cumulative frequency to reach 34 is 42, so the median class is 125 − 145, with l = 125, cf = 22 (the class before it), f = 20 and h = 20.
3. Median = l + [(n/2 − cf) ÷ f] × h = 125 + [(34 − 22) ÷ 20] × 20 = 125 + 12 = 137.
4. Mean: total of f × x = 300 + 475 + 1495 + 2700 + 2170 + 1400 + 780 = 9320, over 68 consumers.
5. Mean = 9320 ÷ 68 = 2330/17 ≈ 137.06.
6. Mode: the largest frequency is 20, in the class 125 − 145, with neighbours f₀ = 13 and f₂ = 14.
7. Mode = 125 + [(20 − 13) ÷ (40 − 13 − 14)] × 20 = 125 + 140/13 ≈ 135.77.
8. Comparing: mode ≈ 135.77, median = 137 and mean ≈ 137.06 lie within about 1.3 units of each other, because the frequencies rise to the peak and fall away at about the same rate on both sides.

AnswerMedian = 137 units; mean = 2330/17 ≈ 137.06 units; mode = 1765/13 ≈ 135.77 units. The three measures are nearly equal, so the distribution is nearly symmetric. (NCERT's answers print 137.05 and 135.76: the same values cut off after two decimal places rather than rounded.)

Watch this explained “The easy case - a balanced table”, 7:21 into Which of the three averages a given question actually wants

Open NCERT p. 198Matches NCERT’s answer

1. Class	Frequency
   0 − 10	5
   10 − 20	x
   20 − 30	20
   30 − 40	15
   40 − 50	y
   50 − 60	5
   Total	60

2. The frequencies add to 60, so 5+x+20+15+y+5 = 60, which gives x + y = 15.
3. The median 28.5 lies in the class 20 − 30, so this is the median class: l = 20, f = 20, h = 10, and cf (before this class) = 5 + x.
4. Median formula: 28.5 = 20 + [(30 − (5+x)) ÷ 20] × 10, so 8.5 = (25 − x) ÷ 2, giving 17 = 25 − x, so x = 8.
5. Then y = 15 − x = 15 − 8 = 7.
6. Check: cumulative frequencies become 5, 13, 33, 48, 55, 60 — the running total does first reach 30 inside the class 20 − 30, confirming that class was correctly chosen.

Answerx = 8 and y = 7.

Watch this explained “The formula run backwards”, 10:02 into Locating the middle class and interpolating across it

Open NCERT p. 198Matches NCERT’s answer

1. Age (below)	Policy holders (cumulative)
   20	2
   25	6
   30	24
   35	45
   40	78
   45	89
   50	92
   55	98
   60	100

2. Difference the running totals to get each class's own frequency: 2, 4, 18, 21, 33, 11, 3, 6, 2 — these total 100, which checks the subtraction.
3. The table's own lowest row reads 'below 20', but policies start at 18, so the first class is really 18 − 20, not an open class from 0.

4. Age (years)	Policy holders (f)	Cumulative f
   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

5. n/2 = 50. The first cumulative frequency reaching 50 is 78, in the class 35 − 40 — the median class.
6. Median = 35 + [(50 − 45) ÷ 33] × 5 = 35 + 25/33 ≈ 35.76 years.

AnswerMedian age = 1180/33 ≈ 35.76 years.

Watch this explained “The limit the table does not print”, 10:37 into Running totals, and converting between the two cumulative tables

Open NCERT p. 199Matches NCERT’s answer

1. Length (mm)	Leaves (f)
   118 − 126	3
   127 − 135	5
   136 − 144	9
   145 − 153	12
   154 − 162	5
   163 − 171	4
   172 − 180	2

2. As the hint says, these classes have gaps of 1 between them, so convert to continuous classes: 117.5−126.5, 126.5−135.5, ..., 171.5−180.5, each 9 wide.
3. n = 40, so n/2 = 20. Cumulative frequencies: 3, 8, 17, 29, 34, 38, 40 — the first to reach 20 is 29, in the class 144.5 − 153.5.
4. Median = 144.5 + [(20 − 17) ÷ 12] × 9 = 144.5 + 2.25 = 146.75 mm.

AnswerMedian length of the leaves = 146.75 mm.

Watch this explained “What the formula demands: classes that meet”, 11:47 into Locating the middle class and interpolating across it

Open NCERT p. 200Matches NCERT’s answer

1. Life time (hours)	Lamps (f)	Cumulative f
   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

2. n = 400, so n/2 = 200. The first cumulative frequency reaching 200 is 216, in the class 3000 − 3500 — the median class.
3. Median = 3000 + [(200 − 130) ÷ 86] × 500 = 3000 + 35000/86 ≈ 3406.98 hours.

AnswerMedian life time = 146500/43 ≈ 3406.98 hours.

Watch this explained “The formula, clause by clause”, 6:26 into Locating the middle class and interpolating across it

Open NCERT p. 200Matches NCERT’s answer

1. Number of letters	Surnames (f)	Cumulative f
   1 − 4	6	6
   4 − 7	30	36
   7 − 10	40	76
   10 − 13	16	92
   13 − 16	4	96
   16 − 19	4	100

2. For the median: n/2 = 50, and the first cumulative frequency reaching 50 is 76, in the class 7 − 10.
3. Median = 7 + [(50 − 36) ÷ 40] × 3 = 7 + 1.05 = 8.05.
4. For the mean, use class marks 2.5, 5.5, 8.5, 11.5, 14.5, 17.5: f×x total = 15+165+340+184+58+70 = 832, over 100 surnames.
5. Mean = 832 ÷ 100 = 8.32.
6. For the mode, the largest frequency is 40, in the same class 7 − 10, with neighbours 30 and 16.
7. Mode = 7 + [(40−30) ÷ (80−30−16)] × 3 = 7 + 30/34 ≈ 7.88.

AnswerMedian = 8.05 letters; mean = 8.32 letters; mode = 134/17 ≈ 7.88 letters.

Watch this explained “The formula, clause by clause”, 6:26 into Locating the middle class and interpolating across it

Open NCERT p. 200Matches NCERT’s answer

1. Weight (kg)	Students (f)	Cumulative f
   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

2. n = 30, so n/2 = 15. The first cumulative frequency reaching 15 is 19, in the class 55 − 60 — the median class.
3. Median = 55 + [(15 − 13) ÷ 6] × 5 = 55 + 10/6 ≈ 56.67 kg.

AnswerMedian weight = 170/3 ≈ 56.67 kg.

Watch this explained “The worked case, and what the number claims”, 7:36 into Locating the middle class and interpolating across it