9th Class Mathematics Unit 12 Information Handling Numericals with Solutions

12.2.1.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.

Formula

Mean bar{X} = Sigma Xn

Working

(i) bar{X} = 4+6+10+12+15+20+25+28+309 = 1509

Result

= 16.67

Working

(ii) bar{X} = 12+18+19+0-19-18-127 = 07

Result

= 0

Working

(iii) bar{X} = 6.5+11+12.3+9+8.1+16+18+20.5+259 = 126.49

Result

= 14.04

Working

(iv) bar{X} = 8+10+12+14+16+20+227 = 1027

Result

= 14.57

12.2.2.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.

Given

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

Formula

Median (n even) Median = 12(6^th term + 7^th term)

Working

= 12(56+57) = 1132

Result

Median height = 56.5

12.2.3.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.

Given

Sorted data 70,72,81,88,90,90,95,95,115,125

Working

(i) Mean bar{X} = 70+72+81+88+90+90+95+95+115+12510 = 92110

Result

= 92.1

Working

(ii) Median Median = 12(5^th+6^th) = 12(90+90) = 1802

Result

= 90

(iii) Mode 90, 95 (most repeated values)

12.2.4.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 with Frequency: 9, 18, 35, 17, 5. Find: (i) Arithmetic mean of their marks by direct and short formula. (ii) Median of their marks.

12.2.5.Given below is a frequency distribution. Class Interval: 5–9, 10–14, 15–19, 20–24, 25–29 with Frequency: 1, 8, 18, 11, 2. Find the mode of the frequency distribution.

Given

Modal class l=14.5, f_m=18 (modal class 15-19), f₁=8, f₂=11, h=5

Formula

Mode Mode = l + f_m-f₁(f_m-f₁)+(f_m-f₂) × h

Working

= 14.5 + 18-8(18-8)+(18-11) × 5

Result

= 17.44

12.2.6.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.

Given

Sorted data 4250,4250,4300,4350,4350,4390,4400,4410,4500,4500

Short formula, A=4350 Y=X-A: -100,-100,-50,0,0,40,50,60,150,150

Working

bar{Y} = -100-100-50+0+0+40+50+60+150+15010 = 20010 = 20

Result

Mean bar{X} = bar{Y}+A = 20+4350 = 4370

Working

Median Median = 12(5^th+6^th) = 12(4350+4390) = 87402

Result

= 4370

Mode 4250, 4350, 4500 (most repeated values)

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

Formula

Mean bar{X} = Sigma Xn

Working

80 = Sigma X45 ⇒ Sigma X = 80 × 45

Result

Sum = 3600

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

Given

Sorted data 0,1,4,7,9

Working

Mean bar{X} = 0+1+4+7+95 = 215

Result

= 4.2

Median 4 (middle term out of 5)

Mode no mode (no entry is repeated)

12.2.9.A set of data contains the values as 148, 145, 160, 157, 156, 160. Show that Mode > Median > Mean.

Given

Sorted data 145,148,156,157,160,160

Working

Mean bar{X} = 145+148+156+157+160+1606 = 9266

Result

= 154.33

Working

Median Median = 12(3^rd+4^th) = 12(156+157) = 3132

Result

= 156.5

Mode 160

Comparison 160 > 156.5 > 154.33 ⇒ Mode > Median > Mean

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

12.2.11.On a prize distribution day, 50 students brought pocket money as under: Rupees: 5–10, 10–15, 15–20, 20–25, 25–30 with 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.

Cumulative frequency 12, 21, 39, 46, 50

Formula

Median Median = l + hfleft(n2-cright)

Working

= 15 + 518(25-21)

Result

= 16.11

Formula

Mode Mode = l + f_m-f₁(f_m-f₁)+(f_m-f₂) × h

Working

= 15 + 99+11 × 5

Result

= 17.25

Coding method, A=17.5 y=x-17.5: -10,-5,0,5,10; fy: -120,-45,0,35,40; Sigma fy = -90

Working

bar{Y} = -9050 = -1.8

Result

Mean bar{X} = bar{Y}+A = -1.8+17.5 = 15.70

12.2.12.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 age of the remaining boys?

Working

Sum of all ages 20 × (13 years) + 20 × (4 months) + 20 × (5 days)

= 260 years + 80 months + 100 days = 260y + (6y 8m) + (3m 10d)

Result

= 266 years 11 months 10 days

Excluding the 15-year-old boy 266y11m10d - 15y = 251 years 11 months 10 days

Working

Convert to days 251 × 360 + 11 × 30 + 10 = 90700 days

Average of remaining 19 boys 9070019 ≈ 4774 days

Result

= 13 years 3 months 4 days

12.2.13.Calculate the arithmetic mean from the following information: (i) If D = X − 140, ΣD = 500 and n = 10. (ii) If U = (x−130)/6, ΣU = −150 and n = 15. (iii) If D = x − 25, ΣfD = 300 and Σf = 20. (iv) If U = (x−120)/5, ΣfU = 60 and Σf = 100.

Working

(i) bar{D} = 50010 = 50 ⇒ bar{X} = bar{D}+140 = 190

(ii) bar{U} = Sigma Un = -15015 = -10

The book prints the denominator as 50 instead of the given n=15 ("-150/50=-10"), though -150/50 actually equals -3, not -10; only n=15 correctly gives -10. This is a printing slip that doesn't affect the downstream value used.

Result

bar{X} = 6bar{U}+130 = -60+130 = 70

The book writes the formula as "6\bar{U}+30" but then computes -60+130 (not -60+30); the label "+30" is a typo for "+130". The final numeric answer 70 is correct.

Working

(iii) bar{D} = 30020 = 15 ⇒ bar{X} = bar{D}+25 = 40

(iv) bar{U} = 60100 = 0.6 ⇒ bar{X} = 5bar{U}+120 = 3+120 = 123

12.2.14.The three children Haris, Maham and Minal made the following scores in a game conducted by a group of teachers in the school. Haris scores: 50, 55, 70, 85, 90. Maham scores: 75, 60, 60, 45, 53. Minal scores: 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?

Working

Average of Haris 50+55+70+85+905 = 3505

Result

= 70 (winner with highest average)

Working

Average of Maham 75+60+60+45+535 = 2935

Result

= 58.6

Working

Average of Minal 80+77+66+42+485 = 3135

Result

= 62.6

12.2.15.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 with f: 1, 3, 6, 20, 26, 12, 2.

fD table fD: -6,-12,-12,0,52,48,12

Working

Sigma f = 70, Sigma fD = 82

bar{D} = 8270 = 1.17

Result

Mean bar{X} = bar{D}+20 = 1.17+20 = 21.17

12.2.16.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.

D = X - 58 -13,-7,0,3,16,-10,-12,-8

Working

bar{D} = -13-7+0+3+16-10-12-88 = -318

Result

= -3.87

Mean bar{X} = bar{D}+58 = -3.87+58 = 54.13

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

Wx table 10(96)=960, 12(48)=576, 4(190)=760, 3(49)=147, 2(650)=1300

Working

Sigma w = 31, Sigma wx = 3743

Formula

Weighted mean bar{X} = Sigma wXSigma w = 374331

Result

= 120.74

12.2.18.For the following data, find the weighted mean. Item: Washing Machine (Quantity 5, Cost 35), Heater (Quantity 3, Cost 5), Stove (Quantity 2, Cost 13), Dispenser (Quantity 6, Cost 18), cost of item in thousands.

wx table 5(35)=175, 3(5)=15, 2(13)=26, 6(18)=108

Working

Sigma w = 16, Sigma wx = 324

Formula

Weighted mean bar{X} = Sigma wXSigma w = 32416

Result

= 20.25 thousands

12.2.19.A company is planning its next year marketing budget across five years: yearly budgets (in million) are: 5, 7, 8, 6, 7. Find the average budget for the next year.

Formula

Average Average Budget = Sigma Xn

Working

= 5+6+6+7+85 = 335

The addition shown (5+6+6+7+8) actually equals 32, not 33, and also does not match the stem values (5,7,8,6,7); the correct addition is 5+7+8+6+7=33. The numerator 33 used afterward is coincidentally correct.

Result

= 6.6 millions

12.2.20.Ahmad obtained the following marks in a certain examination. Find the weighted mean if weights 5, 4, 2, 3, 2, 4 respectively are allotted to the subjects. Urdu 78, English 65, Science 80, Math 90, Islamiyat 85, Computer 72.

wx table 78(5)=390, 65(4)=260, 80(2)=160, 90(3)=270, 85(2)=170, 72(4)=288

Working

Sigma w = 20, Sigma wx = 1538

Formula

Weighted mean bar{X} = Sigma wXSigma w = 153820

Result

= 76.9

Unit 12: Information Handling — Exercise 12 2