Exercise 13.2 answers: Statistics

Open NCERT p. 186Matches NCERT’s answer

1. Age (years)	Patients (f)	Class mark (x)	f × x
   5 − 15	6	10	60
   15 − 25	11	20	220
   25 − 35	21	30	630
   35 − 45	23	40	920
   45 − 55	14	50	700
   55 − 65	5	60	300

2. The largest frequency is 23, in the class 35 − 45 — that is the modal class. Its neighbours have frequencies 21 (below) and 14 (above).
3. Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h = 35 + [(23−21) ÷ (46−21−14)] × 10 = 35 + 20/11 ≈ 36.82 years.
4. For the mean, total of f×x = 60+220+630+920+700+300 = 2830, and the frequencies total 80.
5. Mean = 2830 ÷ 80 = 35.375 years.
6. The mode (≈36.82) is a little higher than the mean (35.375): the age at which the most patients arrive is close to, but not the same as, the average age of a patient.

AnswerMode = 405/11 ≈ 36.82 years; mean = 35.375 years. The maximum number of patients admitted are of age about 36.82 years, while on average a patient's age is 35.375 years — the two are close but not equal.

Watch this explained “Mode against mean, in both directions”, 12:23 into Finding the busiest class, then placing the mode inside it

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1. Lifetime (hours)	Components (f)
   0 − 20	10
   20 − 40	35
   40 − 60	52
   60 − 80	61
   80 − 100	38
   100 − 120	29

2. The largest frequency is 61, in the class 60 − 80 — the modal class. Its neighbours have frequencies 52 (below) and 38 (above).
3. Mode = 60 + [(61−52) ÷ (122−52−38)] × 20 = 60 + (9/32) × 20 = 60 + 5.625 = 65.625 hours.

AnswerModal lifetime = 65.625 hours.

Watch this explained “A ratio of two excesses”, 5:15 into Finding the busiest class, then placing the mode inside it

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1. Expenditure (₹)	Families (f)	Class mark (x)	u = (x−2750)/500	f × u
   1000 − 1500	24	1250	−3	−72
   1500 − 2000	40	1750	−2	−80
   2000 − 2500	33	2250	−1	−33
   2500 − 3000	28	2750	0	0
   3000 − 3500	30	3250	1	30
   3500 − 4000	22	3750	2	44
   4000 − 4500	16	4250	3	48
   4500 − 5000	7	4750	4	28

2. The largest frequency is 40, in the class 1500 − 2000 — the modal class. Its neighbours have frequencies 24 (below) and 33 (above).
3. Mode = 1500 + [(40−24) ÷ (80−24−33)] × 500 = 1500 + (16/23) × 500 ≈ 1847.83.
4. For the mean, use the step-deviation method with assumed mean a = 2750 and h = 500: f×u totals −72−80−33+0+30+44+48+28 = −35.
5. Mean = a + h × (Σf u ÷ Σf) = 2750 + 500 × (−35/200) = 2750 − 87.5 = 2662.5.

AnswerModal monthly expenditure = ₹42500/23 ≈ ₹1847.83; mean monthly expenditure = ₹2662.50.

Watch this explained “A ratio of two excesses”, 5:15 into Finding the busiest class, then placing the mode inside it

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1. Students per teacher	States/U.T.s (f)	Class mark (x)	f × x
   15 − 20	3	17.5	52.5
   20 − 25	8	22.5	180
   25 − 30	9	27.5	247.5
   30 − 35	10	32.5	325
   35 − 40	3	37.5	112.5
   40 − 45	0	42.5	0
   45 − 50	0	47.5	0
   50 − 55	2	52.5	105

2. The largest frequency is 10, in the class 30 − 35, so that is the modal class: l = 30, h = 5, f₁ = 10, and its neighbours have f₀ = 9 (below) and f₂ = 3 (above).
3. Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h = 30 + [(10 − 9) ÷ (20 − 9 − 3)] × 5 = 30 + 5/8 = 30.625.
4. For the mean, total of f × x = 52.5 + 180 + 247.5 + 325 + 112.5 + 0 + 0 + 105 = 1022.5, and the frequencies total 35.
5. Mean = 1022.5 ÷ 35 = 409/14 ≈ 29.21.
6. Interpreting them: the mode says the largest number of states/U.T.s have about 30.6 students per teacher. The mean says that, taken over all 35 states/U.T.s, the average is about 29.2 students per teacher.
7. The mean is a little below the mode because far more states/U.T.s lie in the classes below 30 − 35 (3 + 8 + 9 = 20) than above it (3 + 0 + 0 + 2 = 5).

AnswerMode = 30.625; mean = 409/14 ≈ 29.21. Most states/U.T.s have about 30.6 students per teacher, while on average the ratio is about 29.2 students per teacher.

Watch this explained “The mode is local, the mean is not”, 11:01 into Finding the busiest class, then placing the mode inside it

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1. Runs scored	Batsmen (f)
   3000 − 4000	4
   4000 − 5000	18
   5000 − 6000	9
   6000 − 7000	7
   7000 − 8000	6
   8000 − 9000	3
   9000 − 10000	1
   10000 − 11000	1

2. The largest frequency is 18, in the class 4000 − 5000 — the modal class. Its neighbours have frequencies 4 (below) and 9 (above).
3. Mode = 4000 + [(18−4) ÷ (36−4−9)] × 1000 = 4000 + (14/23) × 1000 ≈ 4608.70 runs.

AnswerMode of runs scored = 106000/23 ≈ 4608.70.

Watch this explained “A ratio of two excesses”, 5:15 into Finding the busiest class, then placing the mode inside it

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1. Number of cars	Periods (f)
   0 − 10	7
   10 − 20	14
   20 − 30	13
   30 − 40	12
   40 − 50	20
   50 − 60	11
   60 − 70	15
   70 − 80	8

2. The largest frequency is 20, in the class 40 − 50 — the modal class. Its neighbours have frequencies 12 (below) and 11 (above).
3. Mode = 40 + [(20−12) ÷ (40−12−11)] × 10 = 40 + (8/17) × 10 ≈ 44.71 cars.

AnswerMode of the number of cars = 760/17 ≈ 44.71.

Watch this explained “A ratio of two excesses”, 5:15 into Finding the busiest class, then placing the mode inside it