Unit 12 - Exercise 12.2 Solutions

Question 1 Ungrouped Mean

Find the arithmetic mean in each of the following:

(i) 4, 6, 10, 12, 15, 20, 25, 28, 30
(ii) 12, 18, 19, 0, −19, −18, −12
(iii) 6.5, 11, 12.3, 9, 8.1, 16, 18, 20.5, 25
(iv) 8, 10, 12, 14, 16, 20, 22

Solution

(i) Values: 4, 6, 10, 12, 15, 20, 25, 28, 30

Number of observations (n) = 9

Mean (X̄) = Σxn = 4 + 6 + 10 + 12 + 15 + 20 + 25 + 28 + 309 = 1509 = 16.67

(ii) Values: 12, 18, 19, 0, −19, −18, −12

Number of observations (n) = 7

Mean (X̄) = 12 + 18 + 19 + 0 + (−19) + (−18) + (−12)7 = 49 − 497 = 07 = 0

(iii) Values: 6.5, 11, 12.3, 9, 8.1, 16, 18, 20.5, 25

Number of observations (n) = 9

Mean (X̄) = 6.5 + 11 + 12.3 + 9 + 8.1 + 16 + 18 + 20.5 + 259 = 126.49 = 14.04

(iv) Values: 8, 10, 12, 14, 16, 20, 22

Number of observations (n) = 7

Mean (X̄) = 8 + 10 + 12 + 14 + 16 + 20 + 227 = 1027 = 14.57

Question 2 Median Heights

Following are the heights in (inches) of 12 students. Find the median height:

55, 53, 54, 58, 60, 61, 62, 56, 57, 52, 51, 63.

Solution

Step 1: Arrange data in ascending order:

51, 52, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63

Step 2: Apply median formula for even observations:

Number of observations (n) = 12 (even).

Median (X̃) = 12 [ (n2)th observation + (n + 22)th observation ]

Median (X̃) = 12 [ 6th observation + 7th observation ]

Median (X̃) = 12 [ 56 + 57 ] = 1132 = 56.5 inches

Question 3 Workers' Earnings

Following are the earnings (in Rs.) of ten workers:

88, 70, 72, 125, 115, 95, 81, 90, 95, 90.

Calculate: (i) Arithmetic Mean, (ii) Median, (iii) Mode.

Solution

(i) Arithmetic Mean:

Sum (Σx) = 88 + 70 + 72 + 125 + 115 + 95 + 81 + 90 + 95 + 90 = 921

Mean (X̄) = 92110 = Rs. 92.1

(ii) Median:

Arrange data: 70, 72, 81, 88, 90, 90, 95, 95, 115, 125. (observations n = 10, even)

Median (X̃) = 12 [ 5th observation + 6th observation ] = 90 + 902 = Rs. 90

(iii) Mode:

The numbers 90 and 95 both appear 2 times (maximum frequency). Thus, the dataset is bimodal: Mode = 90 and 95.

Question 4 Grouped Mean & Median

The marks obtained by the students in the subject of English are given below:

Marks obtained	15–19	20–24	25–29	30–34	35–39
Frequency (f)	9	18	35	17	5

Find: (i) Arithmetic mean of their marks by direct and short formula. (ii) Median of their marks.

Solution

(i) Arithmetic Mean by Direct & Short Methods:

Let assumed mean (A) = 27. Class size (h) = 5.

Marks	Frequency (f)	Midpoint (x)	fx	Deviation D = x − 27	fD
15–19	9	17	153	−10	−90
20–24	18	22	396	−5	−90
25–29	35	27	945	0	0
30–34	17	32	544	5	85
35–39	5	37	185	10	50
Total	Σf = 84	—	Σfx = 2223	—	ΣfD = −45

Direct Method: X̄ = ΣfxΣf = 222384 = 26.46

Short Formula Method: X̄ = A + ΣfDΣf = 27 + −4584 = 27 − 0.54 = 26.46

(ii) Median of Grouped Data:

Marks	Frequency (f)	Class Boundaries	Cumulative Frequency (C.F.)
15–19	9	14.5–19.5	9
20–24	18	19.5–24.5	27
25–29	35	24.5–29.5	62 (Median class contains value 42)
30–34	17	29.5–34.5	79
35–39	5	34.5–39.5	84

Median Class = n2th observation = 842 = 42nd observation, which falls in **25–29** class.

Median formula: X̃ = l + hf ( n2 − C )

Where: l = 24.5, h = 5, f = 35, C = 27 (C.F. of preceding class), n = 84.

Median (X̃) = 24.5 + 535 ( 42 − 27 ) = 24.5 + 5 × 1535 = 24.5 + 7535 = 24.5 + 2.14 = 26.64 marks

Question 5 Grouped Mode

Given below is a frequency distribution. Find the mode of the frequency distribution:

Class Interval	5–9	10–14	15–19	20–24	25–29
Frequency	1	8	18	11	2

Solution

Modal class is the class interval having the **maximum frequency** (f_m = 18). So, the modal class is **15–19**.

Lower boundary of modal class (l) = 14.5
Frequency of modal class (f_m) = 18
Frequency of preceding class (f₁) = 8
Frequency of succeeding class (f₂) = 11
Class size (h) = 5

Mode formula: X̂ = l + [ f_m − f₁(f_m − f₁) + (f_m − f₂) ] × h

Mode (X̂) = 14.5 + [ 18 − 8(18 − 8) + (18 − 11) ] × 5 = 14.5 + [ 1010 + 7 ] × 5 = 14.5 + 5017 = 14.5 + 2.94 = 17.44

Question 6 Weekly Wages

Ten boys work on a petrol pump station. They get weekly wages as follows:

Wages (in Rs.): 4250, 4350, 4400, 4250, 4350, 4410, 4500, 4300, 4500, 4390.

Find the arithmetic mean by short formula, median and mode of their wages.

Solution

1. Arithmetic Mean by Short Formula:

Let Assumed Mean (A) = 4350, number of observations (n) = 10.

Wages (x)	Deviation D = x − 4350
4250	−100
4250	−100
4300	−50
4350	0
4350	0
4390	40
4400	50
4410	60
4500	150
4500	150
Total	ΣD = 200

Mean (X̄) = A + ΣDn = 4350 + 20010 = 4350 + 20 = Rs. 4370

2. Median:

Arrange wages in ascending order: 4250, 4250, 4300, 4350, 4350, 4390, 4400, 4410, 4500, 4500.

Median (X̃) = 12 [ 5th observation + 6th observation ] = 4350 + 43902 = 87402 = Rs. 4370

3. Mode:

The values 4250, 4350, and 4500 all occur 2 times (maximum). Hence, the distribution is trimodal: Mode = Rs. 4250, Rs. 4350, and Rs. 4500.

Questions 7–10 Properties & Deviations

Q7. The arithmetic mean of 45 numbers is 80. Find their sum.

Q8. Five numbers are 1, 4, 0, 7, 9. Find their mean, median and mode.

Q9. A set of data contains: 148, 145, 160, 157, 156, 160. Show that Mode > Median > Mean.

Q10. The monthly attendance of 10 students for lunch in hostel is: 21, 15, 16, 18, 14, 17, 15, 12, 13, 11. Find the median and mode of attendance. Also find mean if D = x − 20.

Solutions

Q7. Sum of Numbers

Given: Mean (X̄) = 80, n = 45.

We know: X̄ = Σxn ⇒ Σx = n × X̄ = 45 × 80 = 3600.

Q8. Analysis of (1, 4, 0, 7, 9)

Mean: X̄ = 1 + 4 + 0 + 7 + 95 = 215 = 4.2

Median: Arranged data: 0, 1, 4, 7, 9. Since n=5 (odd), Median = 3rd observation = 4.

Mode: Since no number is repeated, there is No mode.

Q9. Skewness inequality proof

Data: 148, 145, 160, 157, 156, 160. n = 6.

Mean: X̄ = 148 + 145 + 160 + 157 + 156 + 1606 = 9266 = 154.33

Median: Sorted: 145, 148, 156, 157, 160, 160. Median (X̃) = 156 + 1572 = 156.5

Mode: 160 (occurs twice). Mode (X̂) = 160.

Comparing values: 160 > 156.5 > 154.33 ⇒ Mode > Median > Mean (Verified).

Q10. Attendance analysis

Mean by short-cut formula: Let Assumed Mean A = 20.

Deviations (D = x − 20): −9, −8, −7, −6, −5, −5, −4, −3, −2, 1. Sum (ΣD) = −48.

Mean (X̄) = A + ΣDn = 20 + −4810 = 20 − 4.8 = 15.2 days.

Median: Arranged: 11, 12, 13, 14, 15, 15, 16, 17, 18, 21. Median = 15 + 152 = 15.

Mode: 15 (appears 2 times). Mode = 15.

Question 11 Pocket Money Table

On a prize distribution day, 50 students brought pocket money as under:

Rupees	5–10	10–15	15–20	20–25	25–30
Frequency (f)	12	9	18	7	4

(i) Find the median and mode of the above data.
(ii) Find the arithmetic mean of the data given above using coding method.

Solution

(i) Median and Mode:

Rupees	Frequency (f)	Class Boundaries	Cumulative Frequency (C.F.)
5–10	12	4.5–9.5	12
10–15	9	9.5–14.5	21
15–20	18	14.5–19.5	39 (Median & Modal class)
20–25	7	19.5–24.5	46
25–30	4	24.5–29.5	50
Total	Σf = 50	—

Median Class: n2 = 502 = 25th observation, which lies in 15–20.

Median (X̃) = l + hf ( n2 − C ) = 14.5 + 518 ( 25 − 21 ) = 14.5 + 2018 = 14.5 + 1.11 = Rs. 15.61

Mode: Modal class is 15–20 (frequency = 18).

Mode (X̂) = l + f_m − f₁(f_m − f₁) + (f_m − f₂) × h = 14.5 + 18 − 9(18 − 9) + (18 − 7) × 5 = 14.5 + 4520 = 14.5 + 2.25 = Rs. 16.75

(ii) Arithmetic Mean using Coding Method:

Assumed Mean (A) = 17.5. Class width (h) = 5. Coding variable U = x − Ah.

Rupees	Frequency (f)	Midpoint (x)	Coding U = x − 17.55	fU
5–10	12	7.5	−2	−24
10–15	9	12.5	−1	−9
15–20	18	17.5	0	0
20–25	7	22.5	1	7
25–30	4	27.5	2	8
Total	Σf = 50	—	—	ΣfU = −18

Mean (X̄) = A + [ ΣfUΣf ] × h = 17.5 + [ −1850 ] × 5 = 17.5 − 1.8 = Rs. 15.7

Question 12 Time Conversion

The arithmetic mean of the ages of 20 boys is 13 years, 4 months and 5 days. Find the sum of their ages. If one of the boys is of age exactly 15 years, what is the average of the remaining boys?

Solution

Step 1: Calculate the Sum of Ages of 20 Boys

Mean Age = 13 years, 4 months, 5 days

Sum of Ages of 20 boys = 20 × (13y + 4m + 5d)

Sum = (20 × 13)y + (20 × 4)m + (20 × 5)d = 260y + 80m + 100d

Convert units (1 month = 30 days, 1 year = 12 months):

100 days = 3 months + 10 days ⇒ Sum = 260y + 83m + 10d

83 months = 6 years + 11 months ⇒ Sum = 266 years, 11 months, 10 days.

Step 2: Calculate Average Age of Remaining 19 Boys

Subtract 15 years: Sum of 19 boys = 251y + 11m + 10d

Convert the sum entirely to days (1 year = 365 days):

Sum in days = (251 × 365)d + (11 × 30)d + 10d = 91615 + 330 + 10 = 91955 days.

Average of 19 boys = 9195519 = 4839.74 days.

Convert back to years: 4839.74 ÷ 365 = 13.26 years ⇒ 13 years + 0.26 year.

0.26 year × 12 = 3.12 months ⇒ 3 months + 0.12 month.

0.12 month × 30 = 3.6 days ≈ 4 days.

Thus, average age of remaining 19 boys is **13 years, 3 months, 4 days** approximately.

Question 13 Summary Mean Formulas

Calculate the arithmetic mean from the following information:

(i) If D = X − 140, ΣD = 500 and n = 10
(ii) If U = x − 1306, ΣU = −150 and n = 15
(iii) If D = x − 25, ΣfD = 300 and Σf = 20
(iv) If U = x − 1205, ΣfU = 60 and Σf = 100

Solution

(i) Ungrouped Deviation Method: Here A = 140, ΣD = 500, n = 10.

X̄ = A + ΣDn = 140 + 50010 = 140 + 50 = 190

(ii) Ungrouped Coding Method: Here A = 130, h = 6, ΣU = −150, n = 15.

X̄ = A + [ ΣUn ] × h = 130 + [ −15015 ] × 6 = 130 − 60 = 70

(iii) Grouped Deviation Method: Here A = 25, ΣfD = 300, Σf = 20.

X̄ = A + ΣfDΣf = 25 + 30020 = 25 + 15 = 40

(iv) Grouped Coding Method: Here A = 120, h = 5, ΣfU = 60, Σf = 100.

X̄ = A + [ ΣfUΣf ] × h = 120 + [ 60100 ] × 5 = 120 + 3 = 123

Question 14 Game Scores Award

The three children Haris, Maham and Minal made the following scores in a game conducted by a group of teachers in the school:

Haris	50	55	70	85	90
Maham	75	60	60	45	53
Minal	80	77	66	42	48

It is decided that the candidate who gets the highest average score will be awarded rupees 1000. Who will get the awarded amount?

Solution

We find the simple arithmetic mean score of each child (n = 5):

Haris's average score: X̄_H = 50 + 55 + 70 + 85 + 905 = 3505 = 70.0
Maham's average score: X̄_Ma = 75 + 60 + 60 + 45 + 535 = 2935 = 58.6
Minal's average score: X̄_Mi = 80 + 77 + 66 + 42 + 485 = 3135 = 62.6

Since Haris has the highest average score (70.0), Haris will get the award of Rs. 1000.

Questions 15–16 Deviation Calculations

Q15. Given below is a frequency distribution derived by making a substitution as D = X − 20. Calculate the arithmetic mean:

D	−6	−4	−2	0	2	4	6
f	1	3	6	20	26	12	2

Q16. Being partners Hafsa and Fatima took part in a quiz programme. They made the following number of points: 45, 51, 58, 61, 74, 48, 46 and 50. Compute the average number of points using deviation D = X − 58.

Solutions

Q15. Grouped Deviation Mean

Given: D = X − 20 ⇒ Assumed Mean A = 20.

ΣfD = [ (−6)×1 + (−4)×3 + (−2)×6 + 0×20 + 2×26 + 4×12 + 6×2 ] = −6 − 12 − 12 + 0 + 52 + 48 + 12 = 82.

Σf = 1 + 3 + 6 + 20 + 26 + 12 + 2 = 70.

Mean (X̄) = A + ΣfDΣf = 20 + 8270 = 20 + 1.17 = 21.17

Q16. Ungrouped Deviation Mean

Given points: 45, 51, 58, 61, 74, 48, 46, 50. n = 8. Assumed Mean A = 58.

Deviations (D = X − 58): −13, −7, 0, 3, 16, −10, −12, −8. Sum (ΣD) = −31.

Mean (X̄) = A + ΣDn = 58 + −318 = 58 − 3.875 = 54.125 ≈ 54.13 points.

Questions 17–20 Weighted Mean

Q17. A person purchased: Rice (10 kg @ Rs. 96/kg), Flour (12 kg @ Rs. 48/kg), Ghee (4 kg @ Rs. 190/kg), Sugar (3 kg @ Rs. 49/kg), Mutton (2 kg @ Rs. 650/kg). What is the weighted mean of cost of food items per kg?

Q18. For the following data, find the weighted mean of appliance costs: Washing Machine (5 @ Rs. 35k), Heater (3 @ Rs. 5k), Stove (2 @ Rs. 13k), Dispenser (6 @ Rs. 18k).

Q19. Company budgets across five years are: 5, 7, 8, 6, 7 (million). Find the average budget.

Q20. Ahmad obtained: Urdu (78, wt=5), English (65, wt=4), Science (80, wt=2), Math (90, wt=3), Islamyat (85, wt=2), Computer (72, wt=4). Find the weighted mean marks.

Solutions

Q17. Weighted Mean Food Cost

Food Item	Quantity (W)	Cost/kg (X)	WX
Rice	10	96	960
Flour	12	48	576
Ghee	4	190	760
Sugar	3	49	147
Mutton	2	650	1300
Total	ΣW = 31	—	ΣWX = 3743

Weighted Mean (X̄_w) = ΣWXΣW = 374331 = Rs. 120.74 per kg.

Q18. Weighted Mean Appliance Cost

ΣW = 5 + 3 + 2 + 6 = 16. Cost in thousands (X).

ΣWX = 5×35 + 3×5 + 2×13 + 6×18 = 175 + 15 + 26 + 108 = 324.

Weighted Mean (X̄_w) = 32416 = 20.25 (thousands Rs.) = Rs. 20,250.

Q19. Average Company Budget

Mean (X̄) = 5 + 7 + 8 + 6 + 75 = 335 = 6.6 million Rs.

Q20. Weighted Mean Marks

ΣW = 5 + 4 + 2 + 3 + 2 + 4 = 20.

ΣWX = 78×5 + 65×4 + 80×2 + 90×3 + 85×2 + 72×4 = 390 + 260 + 160 + 270 + 170 + 288 = 1538.

Weighted Mean (X̄_w) = 153820 = 76.9 marks.

9th Mathematics

Exercise 12.2 Solutions

Q7. Sum of Numbers

Q8. Analysis of (1, 4, 0, 7, 9)

Q9. Skewness inequality proof

Q10. Attendance analysis

Q15. Grouped Deviation Mean

Q16. Ungrouped Deviation Mean

Q17. Weighted Mean Food Cost

Q18. Weighted Mean Appliance Cost

Q19. Average Company Budget

Q20. Weighted Mean Marks

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