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Every question from the end of every unit, with hints and answers to open. After you check an answer, mark how you did, and come back to the ones you didn’t get.

62 questions

Unit 13 · Statistics

Exercise 13.1

Question 1

i.

35, 36, 42, 45, 48, 52, 58, 60

Answer

Sum = 376, N = 8.
X‾=3768=47 \overline{X} = \frac{376}{8} = 47
The mean is 47.

ii.

13.5, 14.2, 15.8, 15.2, 16.9, 16.5, 17.4, 19.3, 15.3, 15.9

Answer

Sum = 160, N = 10.
X‾=16010=16 \overline{X} = \frac{160}{10} = 16
The mean is 16.

iii.
x 5 8 10 12 14 16
f 4 5 8 10 2 2
Answer

fx: 20, 40, 80, 120, 28, 32, so ∑fx=320 \sum fx = 320 and N = 31.
X‾=32031≈10.32 \overline{X} = \frac{320}{31} \approx 10.32
The mean is about 10.32.

iv.

The goals scored by players in a national football league:

Goals 12 13 14 15 16 17
Players 2 4 6 12 10 6
Answer

fx: 24, 52, 84, 180, 160, 102, so ∑fx=602 \sum fx = 602 and N = 40.
X‾=60240=15.05 \overline{X} = \frac{602}{40} = 15.05
The mean is 15.05 goals.

Question 2

a.

The ages of the passengers on a bus:

Age (years) 0–10 10–20 20–30 30–40 40–50
Passengers 5 9 15 7 4
Answer
Age f m fm d = m − 25 fd
0–10 5 5 25 −20 −100
10–20 9 15 135 −10 −90
20–30 15 25 375 0 0
30–40 7 35 245 10 70
40–50 4 45 180 20 80
40 960 −40

Direct: X‾=96040=24 \overline{X} = \frac{960}{40} = 24 . Short-cut: X‾=25+−4040=24 \overline{X} = 25 + \frac{-40}{40} = 24 .
The mean age is 24 years.

b.

The marks of Grade 10 students in Science:

Marks 10–20 20–30 30–40 40–50 50–60 60–70
Students 1 4 10 8 7 5
Answer
Marks f m fm d = m − 45 fd
10–20 1 15 15 −30 −30
20–30 4 25 100 −20 −80
30–40 10 35 350 −10 −100
40–50 8 45 360 0 0
50–60 7 55 385 10 70
60–70 5 65 325 20 100
35 1535 −40

Direct: X‾=153535≈43.86 \overline{X} = \frac{1535}{35} \approx 43.86 . Short-cut: X‾=45+−4035=45−1.14=43.86 \overline{X} = 45 + \frac{-40}{35} = 45 - 1.14 = 43.86 .
The mean mark is about 43.86.

c.

The daily wages of workers:

Wage (Rs) 200–400 400–600 600–800 800–1000 1000–1200
Workers 3 7 10 6 4
Answer
Wage (Rs) f m fm d = m − 700 fd
200–400 3 300 900 −400 −1200
400–600 7 500 3500 −200 −1400
600–800 10 700 7000 0 0
800–1000 6 900 5400 200 1200
1000–1200 4 1100 4400 400 1600
30 21200 200

Direct: X‾=2120030≈706.67 \overline{X} = \frac{21200}{30} \approx 706.67 . Short-cut: X‾=700+20030=706.67 \overline{X} = 700 + \frac{200}{30} = 706.67 .
The mean daily wage is about Rs 706.67.

d.

The marks of Grade 10 students in Mathematics:

Marks 0–10 10–20 20–30 30–40 40–50 50–60
Students 7 5 6 12 8 2
Answer
Marks f m fm d = m − 35 fd
0–10 7 5 35 −30 −210
10–20 5 15 75 −20 −100
20–30 6 25 150 −10 −60
30–40 12 35 420 0 0
40–50 8 45 360 10 80
50–60 2 55 110 20 40
40 1150 −250

Direct: X‾=115040=28.75 \overline{X} = \frac{1150}{40} = 28.75 . Short-cut: X‾=35+−25040=35−6.25=28.75 \overline{X} = 35 + \frac{-250}{40} = 35 - 6.25 = 28.75 .
The mean mark is 28.75.

Question 3

a.

X = 49, ∑fm=980 \sum fm = 980 , N=? N = ?

Answer

49=980NN=98049=20 \begin{aligned} 49 &= \frac{980}{N} \\ N &= \frac{980}{49} = 20 \end{aligned}
N = 20.

b.

X = 102.25, N = 8, ∑fm=? \sum fm = ?

Answer

∑fm=X‾×N=102.25×8=818 \sum fm = \overline{X} \times N = 102.25 \times 8 = 818
∑fm=818 \sum fm = 818 .

c.

A = 100, X = 90, N = 10, ∑fd=? \sum fd = ?

Answer

90=100+∑fd10∑fd=(90−100)×10=−100 \begin{aligned} 90 &= 100 + \frac{\sum fd}{10} \\ \sum fd &= (90 - 100) \times 10 = -100 \end{aligned}
∑fd=−100 \sum fd = -100 . It is negative because the assumed mean is above the mean.

d.

X = 41.75, ∑fd=270 \sum fd = 270 , N = 40, A=? A = ?

Answer

41.75=A+27040A=41.75−6.75=35 \begin{aligned} 41.75 &= A + \frac{270}{40} \\ A &= 41.75 - 6.75 = 35 \end{aligned}
A = 35.

Question 4

a.

The mean of these marks is 32.5. Find k.

Marks 0–10 10–20 20–30 30–40 40–50 50–60
Students 5 10 k 35 15 10
Answer

Mid-values 5, 15, 25, 35, 45, 55: N = 75 + k, ∑fm=25+150+25k+1225+675+550=2625+25k \sum fm = 25 + 150 + 25k + 1225 + 675 + 550 = 2625 + 25k .
32.5(75+k)=2625+25k2437.5+32.5k=2625+25k7.5k=187.5k=25 \begin{aligned} 32.5(75 + k) &= 2625 + 25k \\ 2437.5 + 32.5k &= 2625 + 25k \\ 7.5k &= 187.5 \\ k &= 25 \end{aligned}
k = 25.

b.

The mean of this table is 46.2. Find p.

x 0–20 20–40 40–60 60–80 80–100
f 35 400 350 p 65
Answer

Mid-values 10, 30, 50, 70, 90: N = 850 + p, ∑fm=350+12000+17500+70p+5850=35700+70p \sum fm = 350 + 12000 + 17500 + 70p + 5850 = 35700 + 70p .
46.2(850+p)=35700+70p39270+46.2p=35700+70p3570=23.8pp=150 \begin{aligned} 46.2(850 + p) &= 35700 + 70p \\ 39270 + 46.2p &= 35700 + 70p \\ 3570 &= 23.8p \\ p &= 150 \end{aligned}
p = 150.

c.

The mean of these marks is 36.4. Find y.

Marks 16–24 24–32 32–40 40–48 48–56 56–64
Students 6 8 y 8 4 2
Answer

Mid-values 20, 28, 36, 44, 52, 60: N = 28 + y, ∑fm=120+224+36y+352+208+120=1024+36y \sum fm = 120 + 224 + 36y + 352 + 208 + 120 = 1024 + 36y .
36.4(28+y)=1024+36y1019.2+36.4y=1024+36y0.4y=4.8y=12 \begin{aligned} 36.4(28 + y) &= 1024 + 36y \\ 1019.2 + 36.4y &= 1024 + 36y \\ 0.4y &= 4.8 \\ y &= 12 \end{aligned}
y = 12.

d.

The mean daily expenditure of these students is Rs 264.67 (264⅔). Find the missing frequency.

Daily expenditure (Rs) 0–100 100–200 200–300 300–400 400–500 500–600
Students 20 30 ? 20 18 12
Hint

Call the missing frequency x. The mean 264.67 is a rounded 26423=7943 264\tfrac{2}{3} = \frac{794}{3} ; using the exact value gives a whole number.

Answer

Mid-values 50, 150, …, 550: N = 100 + x, ∑fm=1000+4500+250x+7000+8100+6600=27200+250x \sum fm = 1000 + 4500 + 250x + 7000 + 8100 + 6600 = 27200 + 250x .
7943(100+x)=27200+250x79400+794x=81600+750x44x=2200x=50 \begin{aligned} \tfrac{794}{3}(100 + x) &= 27200 + 250x \\ 79400 + 794x &= 81600 + 750x \\ 44x &= 2200 \\ x &= 50 \end{aligned}
50 students. (With 264.67 itself the working gives 49.97, which rounds to the same 50.)

Question 5

a.

15, 51, 32, 12, 32, 33, 23, 43, 35, 46, 57, 19, 59, 25, 20, 38, 16, 45, 39, 40 (classes 10 wide)

Answer
Class Tally f m fm
10–20 IIII 4 15 60
20–30 III 3 25 75
30–40 ⫽⫽⫽⫽ I 6 35 210
40–50 IIII 4 45 180
50–60 III 3 55 165
20 690

X‾=69020=34.5 \overline{X} = \frac{690}{20} = 34.5
The mean is 34.5. (The 20 numbers themselves average 34.)

b.

25, 15, 24, 42, 22, 35, 34, 41, 33, 38, 54, 50, 36, 40, 27, 18, 35, 16, 51, 31, 23, 9, 16, 23, 31, 51, 7, 30, 17, 40, 60, 32, 50, 10, 23, 12, 21, 28, 37, 20, 58, 39, 10, 41, 13 (classes 5 wide)

Answer

The smallest is 7 and the largest 60, so classes 5–10, 10–15, …, 60–65.

Class f m fm
5–10 2 7.5 15
10–15 4 12.5 50
15–20 5 17.5 87.5
20–25 7 22.5 157.5
25–30 3 27.5 82.5
30–35 6 32.5 195
35–40 6 37.5 225
40–45 5 42.5 212.5
45–50 0 47.5 0
50–55 5 52.5 262.5
55–60 1 57.5 57.5
60–65 1 62.5 62.5
45 1407.5

X‾=1407.545≈31.28 \overline{X} = \frac{1407.5}{45} \approx 31.28
The mean is about 31.28.

Question 6

a.
Marks 0–10 10–20 20–30 30–40 40–50 50–60
Frequency 8 10 14 10 8 10
Answer

With A = 25: fd = −160, −100, 0, 100, 160, 300, so ∑fd=300 \sum fd = 300 , N = 60.
X‾=25+30060=30 \overline{X} = 25 + \frac{300}{60} = 30
The mean is 30. (Directly: 180060=30 \frac{1800}{60} = 30 .)

b.
Daily wage (Rs) 0–50 50–100 100–150 150–200 200–250 250–300
Workers 1 2 3 4 1 2
Answer

With A = 175: fd = −150, −200, −150, 0, 50, 200, so ∑fd=−250 \sum fd = -250 , N = 13.
X‾=175+−25013=175−19.23=155.77 \overline{X} = 175 + \frac{-250}{13} = 175 - 19.23 = 155.77
The mean daily wage is about Rs 155.77.

Exercise 13.2

Question 1

a.

2.5, 4.5, 3.6, 4.9, 5.4, 2.9, 3.1, 4.2, 4.6, 2.2, 1.5

Answer

In order: 1.5, 2.2, 2.5, 2.9, 3.1, 3.6, 4.2, 4.5, 4.6, 4.9, 5.4. N = 11, the 11+12= \frac{11 + 1}{2} = 6th item.
The median is 3.6.

b.

100, 105, 104, 197, 97, 108, 120, 148, 144, 190, 148, 22, 169, 171, 92, 100

Answer

In order: 22, 92, 97, 100, 100, 104, 105, 108, 120, 144, 148, 148, 169, 171, 190, 197. N = 16, so the 16+12= \frac{16 + 1}{2} = 8.5th item, halfway between the 8th and 9th.
Md=108+1202=114 M_d = \frac{108 + 120}{2} = 114
The median is 114.

c.
Marks 18 25 28 29 34 40 44 46
Students 3 6 5 7 8 12 5 4
Answer

cf: 3, 9, 14, 21, 29, 41, 46, 50. N = 50, so the 25.5th item: the 25th and 26th are both 34 (items 22 to 29).
The median is 34.

d.
x 102 105 125 140 170 190 200
f 10 18 22 25 15 12 8
Answer

cf: 10, 28, 50, 75, 90, 102, 110. N = 110, so the 55.5th item: items 51 to 75 are 140.
The median is 140.

Question 2

a.
Weight (kg) 30–40 40–50 50–60 60–70 70–80 80–90 90–100
Students 3 5 7 11 10 3 1
Answer

cf: 3, 8, 15, 26, 36, 39, 40. N = 40, the 20th item, in 60–70.
Md=60+20−1511×10=60+4.55=64.55 M_d = 60 + \frac{20 - 15}{11} \times 10 = 60 + 4.55 = 64.55
The median weight is about 64.55 kg.

b.
Height (cm) 140–145 145–150 150–155 155–160 160–165 165–170 170–175
Frequency 2 5 8 10 7 5 3
Answer

cf: 2, 7, 15, 25, 32, 37, 40. N = 40, the 20th item, in 155–160.
Md=155+20−1510×5=155+2.5=157.5 M_d = 155 + \frac{20 - 15}{10} \times 5 = 155 + 2.5 = 157.5
The median height is 157.5 cm.

c.
Expenditure per day (Rs) less than 100 100–200 200–300 300–400 400–500 more than 500
Frequency 22 34 52 20 19 13
Hint

The end classes are open. Take them as 0–100 and 500–600, as the classes are 100 wide; the median isn’t in either, so it doesn’t matter.

Answer

cf: 22, 56, 108, 128, 147, 160. N = 160, the 80th item, in 200–300.
Md=200+80−5652×100=200+46.15=246.15 M_d = 200 + \frac{80 - 56}{52} \times 100 = 200 + 46.15 = 246.15
The median expenditure is about Rs 246.15 a day.

d.
Marks less than 20 less than 40 less than 60 less than 80 less than 100
Students 21 44 66 79 90
Hint

These are already cumulative frequencies. The classes are 0–20, 20–40, …, with frequencies 21, 44 − 21, …

Answer
Marks f cf
0–20 21 21
20–40 23 44
40–60 22 66
60–80 13 79
80–100 11 90

N = 90, the 45th item, in 40–60.
Md=40+45−4422×20=40+0.91=40.91 M_d = 40 + \frac{45 - 44}{22} \times 20 = 40 + 0.91 = 40.91
The median mark is about 40.91.

Question 3

a.

The median is 35.

Marks 20–25 25–30 30–35 35–40 40–45 45–50
Students 2 5 8 k 4 5
Hint

35 is where the class 35–40 starts. When is the median exactly a class’s lower limit?

Answer

cf: 2, 7, 15, 15 + k, 19 + k, 24 + k; N = 24 + k. The median 35 is in 35–40 with L = 35, cf = 15, f = k, h = 5.
35=35+24+k2−15k×50=24+k2−15k=6 \begin{aligned} 35 &= 35 + \frac{\dfrac{24 + k}{2} - 15}{k} \times 5 \\ 0 &= \frac{24 + k}{2} - 15 \\ k &= 6 \end{aligned}
k = 6. The median is a lower limit when exactly half the values are below it: 15 of 30.

b.

The median is 132.5.

Wages (Rs) 100–110 110–120 120–130 130–140 140–150 150–160
Workers 5 6 p 4 7 5
Answer

N = 27 + p. The median 132.5 is in 130–140: L = 130, cf = 11 + p, f = 4, h = 10.
132.5=130+27+p2−(11+p)4×102.5=27+p−22−2p8×102=5−pp=3 \begin{aligned} 132.5 &= 130 + \frac{\dfrac{27 + p}{2} - (11 + p)}{4} \times 10 \\ 2.5 &= \frac{27 + p - 22 - 2p}{8} \times 10 \\ 2 &= 5 - p \\ p &= 3 \end{aligned}
p = 3.

c.

The median is 3956 39\tfrac{5}{6} years (about 39.83).

Age (years) 20–25 25–30 30–35 35–40 40–45 45–50 50–55 55–60
People 50 70 100 300 ? 220 70 60
Answer

Call the missing frequency x: N = 870 + x. The median is in 35–40: L = 35, cf = 220, f = 300, h = 5.
3956=35+870+x2−220300×5296=435+x2−22060290=215+x2x=150 \begin{aligned} 39\tfrac{5}{6} &= 35 + \frac{\dfrac{870 + x}{2} - 220}{300} \times 5 \\ \tfrac{29}{6} &= \frac{435 + \frac{x}{2} - 220}{60} \\ 290 &= 215 + \tfrac{x}{2} \\ x &= 150 \end{aligned}
150 people.

Question 4

a.
Marks 50–60 60–70 70–80 80–90 90–100
Students 2 3 6 5 4
Answer

cf: 2, 5, 11, 16, 20. N = 20, the 10th item, in 70–80.
Md=70+10−56×10=70+8.33=78.33 M_d = 70 + \frac{10 - 5}{6} \times 10 = 70 + 8.33 = 78.33
The median mark is about 78.33.

b.
Marks less than 20 less than 40 less than 60 less than 80 less than 100
Students 9 23 43 55 60
Answer

Frequencies: 9, 14, 20, 12, 5 for 0–20, 20–40, 40–60, 60–80, 80–100. N = 60, the 30th item, in 40–60 (23 below 40).
Md=40+30−2320×20=40+7=47 M_d = 40 + \frac{30 - 23}{20} \times 20 = 40 + 7 = 47
The median mark is 47.

c.
Income (Rs) less than 600 less than 700 less than 800 less than 900 less than 1000
Workers 30 98 152 177 200
Answer

The classes are 100 wide, so the first is 500–600. Frequencies: 30, 68, 54, 25, 23. N = 200, the 100th item, in 700–800 (98 below 700).
Md=700+100−9854×100=700+3.70=703.70 M_d = 700 + \frac{100 - 98}{54} \times 100 = 700 + 3.70 = 703.70
The median income is about Rs 703.70.

d.
Temperature (°C) 0–9 10–19 20–29 30–39 40–49
Days 8 10 20 15 7
Answer

The classes have gaps of 1, so the correction is 0.5: −0.5–9.5, 9.5–19.5, 19.5–29.5, 29.5–39.5, 39.5–49.5. cf: 8, 18, 38, 53, 60. N = 60, the 30th item, in 19.5–29.5.
Md=19.5+30−1820×10=19.5+6=25.5 M_d = 19.5 + \frac{30 - 18}{20} \times 10 = 19.5 + 6 = 25.5
The median temperature is 25.5 °C.

Question 5

a.

The marks of 30 students in a class test: 22, 56, 62, 37, 48, 30, 58, 42, 29, 39, 37, 50, 38, 41, 32, 20, 28, 16, 43, 18, 40, 52, 44, 27, 35, 45, 36, 49, 55, 40. Make a frequency table with classes 10 wide and find the mean and the median.

Answer
Marks f m fm cf
10–20 2 15 30 2
20–30 5 25 125 7
30–40 8 35 280 15
40–50 9 45 405 24
50–60 5 55 275 29
60–70 1 65 65 30
30 1180

Mean: 118030≈39.33 \frac{1180}{30} \approx 39.33 . Median: the 15th item, in 30–40: 30+15−78×10=40 30 + \frac{15 - 7}{8} \times 10 = 40 .
The mean is about 39.33 and the median 40. (The 30 marks themselves average 38.97.)

b.

The heights in cm of 40 Grade 10 students: 142, 145, 151, 157, 159, 160, 165, 162, 156, 158, 155, 141, 147, 149, 148, 159, 154, 155, 166, 168, 169, 172, 174, 173, 176, 161, 164, 163, 149, 150, 154, 153, 152, 164, 158, 159, 162, 157, 156, 155. Make a frequency table with classes 5 wide and find the mean and the median.

Answer
Height (cm) f m fm cf
140–145 2 142.5 285 2
145–150 5 147.5 737.5 7
150–155 6 152.5 915 13
155–160 12 157.5 1890 25
160–165 7 162.5 1137.5 32
165–170 4 167.5 670 36
170–175 3 172.5 517.5 39
175–180 1 177.5 177.5 40
40 6330

Mean: 633040=158.25 \frac{6330}{40} = 158.25 . Median: the 20th item, in 155–160: 155+20−1312×5=155+2.92=157.92 155 + \frac{20 - 13}{12} \times 5 = 155 + 2.92 = 157.92 .
The mean height is 158.25 cm and the median about 157.92 cm.

Exercise 13.3

Question 1

a.

29 cm, 34 cm, 29 cm, 26 cm, 55 cm, 34 cm, 35 cm, 40 cm, 34 cm, 56 cm

Answer

34 cm occurs 3 times (29 cm twice, the rest once).
The mode is 34 cm.

b.

99 kg, 135 kg, 182 kg, 49 kg, 189 kg, 196 kg, 78 kg, 192 kg, 182 kg

Answer

182 kg occurs twice, every other weight once.
The mode is 182 kg.

Question 2

a.
Marks 5 10 15 20 25 30 35 40 45
Students 2 6 7 9 11 5 15 2 3
Answer

The highest frequency, 15, is at 35.
The mode is 35 marks.

b.
Wage (Rs) 50 75 100 125 150 175 200 225
Workers 8 12 17 29 30 27 20 11
Answer

The highest frequency, 30, is at Rs 150.
The mode is Rs 150.

Question 3

a.
Marks 20–25 25–30 30–35 35–40 40–45 45–50
Students 2 5 8 6 4 5
Answer

Modal class 30–35: L = 30, f₁ = 8, f₀ = 5, f₂ = 6, h = 5.
Mo=30+8−516−5−6×5=30+35×5=33 M_o = 30 + \frac{8 - 5}{16 - 5 - 6} \times 5 = 30 + \frac{3}{5} \times 5 = 33
The mode is 33 marks.

b.
Wage (Rs) 100–110 110–120 120–130 130–140 140–150 150–160
Workers 5 6 4 7 5 4
Answer

Modal class 130–140: L = 130, f₁ = 7, f₀ = 4, f₂ = 5, h = 10.
Mo=130+7−414−4−5×10=130+6=136 M_o = 130 + \frac{7 - 4}{14 - 4 - 5} \times 10 = 130 + 6 = 136
The mode is Rs 136.

c.
Age (years) 20–25 25–30 30–35 35–40 40–45 45–50 50–55 55–60
People 50 70 100 300 220 150 70 60
Answer

Modal class 35–40: L = 35, f₁ = 300, f₀ = 100, f₂ = 220, h = 5.
Mo=35+300−100600−100−220×5=35+200280×5=35+3.57=38.57 M_o = 35 + \frac{300 - 100}{600 - 100 - 220} \times 5 = 35 + \frac{200}{280} \times 5 = 35 + 3.57 = 38.57
The mode is about 38.57 years.

Exercise 13.4

Question 1

a.

10, 12, 14, 11, 22, 15, 27, 14, 16, 13, 25

Answer

In order: 10, 11, 12, 13, 14, 14, 15, 16, 22, 25, 27. N = 11: Q₁ is the 124= \frac{12}{4} = 3rd item and Q₃ the 364= \frac{36}{4} = 9th.
Q₁ = 12, Q₃ = 22.

b.
Marks 42 48 49 53 56 59 60 65 68 70
Students 2 3 5 8 9 11 7 8 6 4
Answer

cf: 2, 5, 10, 18, 27, 38, 45, 53, 59, 63. N = 63.
Q₁: the 644= \frac{64}{4} = 16th item; items 11 to 18 are 53.
Q₃: the 3×644= \frac{3 \times 64}{4} = 48th item; items 46 to 53 are 65.
Q₁ = 53, Q₃ = 65.

Question 2

a.
Age (years) 2–4 4–6 6–8 8–10 10–12 12–14 14–16 16–18
Students 5 12 25 26 24 28 20 15
Answer

cf: 5, 17, 42, 68, 92, 120, 140, 155. N = 155.
Q₁: the 38.75th item, in 6–8: 6+38.75−1725×2=6+1.74=7.74 6 + \frac{38.75 - 17}{25} \times 2 = 6 + 1.74 = 7.74 .
Q₃: the 116.25th item, in 12–14: 12+116.25−9228×2=12+1.73=13.73 12 + \frac{116.25 - 92}{28} \times 2 = 12 + 1.73 = 13.73 .
Q1≈7.74 Q_1 \approx 7.74 years, Q3≈13.73 Q_3 \approx 13.73 years.

b.
Marks 10–20 20–30 30–40 40–50 50–60 60–70 70–80
Students 2 3 6 12 13 11 7
Answer

cf: 2, 5, 11, 23, 36, 47, 54. N = 54.
Q₁: the 13.5th item, in 40–50: 40+13.5−1112×10=42.08 40 + \frac{13.5 - 11}{12} \times 10 = 42.08 .
Q₃: the 40.5th item, in 60–70: 60+40.5−3611×10=64.09 60 + \frac{40.5 - 36}{11} \times 10 = 64.09 .
Q1≈42.08 Q_1 \approx 42.08 , Q3≈64.09 Q_3 \approx 64.09 .

c.
Height (cm) 100–110 110–120 120–130 130–140 140–150 150–160 160–170
Students 3 4 9 15 20 14 7
Answer

cf: 3, 7, 16, 31, 51, 65, 72. N = 72.
Q₁: the 18th item, in 130–140: 130+18−1615×10=131.33 130 + \frac{18 - 16}{15} \times 10 = 131.33 .
Q₃: the 54th item, in 150–160: 150+54−5114×10=152.14 150 + \frac{54 - 51}{14} \times 10 = 152.14 .
Q1≈131.33 Q_1 \approx 131.33 cm, Q3≈152.14 Q_3 \approx 152.14 cm.

d.
Wage (Rs) 100–150 150–200 200–250 250–300 300–350 350–400
Workers 6 11 21 34 25 22
Answer

cf: 6, 17, 38, 72, 97, 119. N = 119.
Q₁: the 29.75th item, in 200–250: 200+29.75−1721×50=230.36 200 + \frac{29.75 - 17}{21} \times 50 = 230.36 .
Q₃: the 89.25th item, in 300–350: 300+89.25−7225×50=334.5 300 + \frac{89.25 - 72}{25} \times 50 = 334.5 .
Q1≈ Q_1 \approx Rs 230.36, Q₃ = Rs 334.50.

e.
Marks 0–20 20–40 40–60 60–80 80–100 100–120 120–140
Frequency 8 12 15 14 12 9 10
Answer

cf: 8, 20, 35, 49, 61, 70, 80. N = 80.
Q₁: the 20th item, in 20–40: 20+20−812×20=40 20 + \frac{20 - 8}{12} \times 20 = 40 .
Q₃: the 60th item, in 80–100: 80+60−4912×20=98.33 80 + \frac{60 - 49}{12} \times 20 = 98.33 .
Q₁ = 40, Q3≈98.33 Q_3 \approx 98.33 .

f.
Time (minutes) 0–10 10–20 20–30 30–40 40–50 50–60
Frequency 5 3 10 6 4 2
Answer

cf: 5, 8, 18, 24, 28, 30. N = 30.
Q₁: the 7.5th item, in 10–20: 10+7.5−53×10=18.33 10 + \frac{7.5 - 5}{3} \times 10 = 18.33 .
Q₃: the 22.5th item, in 30–40: 30+22.5−186×10=37.5 30 + \frac{22.5 - 18}{6} \times 10 = 37.5 .
Q1≈18.33 Q_1 \approx 18.33 minutes, Q₃ = 37.5 minutes.

g.
Time (minutes) 20–25 25–30 30–35 35–40 40–45 45–50
Frequency 2 5 8 6 4 5
Answer

cf: 2, 7, 15, 21, 25, 30. N = 30.
Q₁: the 7.5th item, in 30–35: 30+7.5−78×5=30.31 30 + \frac{7.5 - 7}{8} \times 5 = 30.31 .
Q₃: the 22.5th item, in 40–45: 40+22.5−214×5=41.88 40 + \frac{22.5 - 21}{4} \times 5 = 41.88 .
Q1≈30.31 Q_1 \approx 30.31 minutes, Q3≈41.88 Q_3 \approx 41.88 minutes.

Question 3

a.

Q₁ = 8. Find k.

Age (years) 0–6 6–12 12–18 18–24 24–30 30–36
Pupils 9 6 5 k 7 9
Answer

N = 36 + k. Q₁ = 8 is in 6–12: L = 6, cf = 9, f = 6, h = 6.
8=6+36+k4−96×62=36+k4−944=36+kk=8 \begin{aligned} 8 &= 6 + \frac{\dfrac{36 + k}{4} - 9}{6} \times 6 \\ 2 &= \frac{36 + k}{4} - 9 \\ 44 &= 36 + k \\ k &= 8 \end{aligned}
k = 8.

b.

Q₁ = 31. Find the missing frequency.

Class 10–20 20–30 30–40 40–50 50–60 60–70
Frequency 4 5 ? 8 7 6
Answer

Call it x: N = 30 + x. Q₁ = 31 is in 30–40: L = 30, cf = 9, f = x, h = 10.
31=30+30+x4−9x×10x=(30+x−36)×1044x=10x−60x=10 \begin{aligned} 31 &= 30 + \frac{\dfrac{30 + x}{4} - 9}{x} \times 10 \\ x &= \frac{(30 + x - 36) \times 10}{4} \\ 4x &= 10x - 60 \\ x &= 10 \end{aligned}
The missing frequency is 10.

c.

Q₃ = 53.75. Find q.

Weight (kg) 40–44 44–48 48–52 52–56 56–60 60–64
Frequency 8 10 14 q 3 1
Answer

N = 36 + q. Q₃ = 53.75 is in 52–56: L = 52, cf = 32, f = q, h = 4.
53.75=52+3(36+q)4−32q×41.75q=108+3q−1281.25q=20q=16 \begin{aligned} 53.75 &= 52 + \frac{\dfrac{3(36 + q)}{4} - 32}{q} \times 4 \\ 1.75q &= 108 + 3q - 128 \\ 1.25q &= 20 \\ q &= 16 \end{aligned}
q = 16.

Question 4

a.
Height (cm) less than 125 less than 130 less than 135 less than 140 less than 145 less than 150 less than 155
Students 0 5 11 24 45 60 72
Answer

Frequencies for 125–130, 130–135, …, 150–155: 5, 6, 13, 21, 15, 12. N = 72.
Q₁: the 18th item, in 135–140: 135+18−1113×5=137.69 135 + \frac{18 - 11}{13} \times 5 = 137.69 .
Q₃: the 54th item, in 145–150: 145+54−4515×5=148 145 + \frac{54 - 45}{15} \times 5 = 148 .
Q1≈137.69 Q_1 \approx 137.69 cm, Q₃ = 148 cm.

b.
Weight (lb) 110–119 120–129 130–139 140–149 150–159 160–169 170–179 180–189
Frequency 5 7 12 20 16 10 7 3
Answer

Correction 0.5: 109.5–119.5, 119.5–129.5, … cf: 5, 12, 24, 44, 60, 70, 77, 80. N = 80.
Q₁: the 20th item, in 129.5–139.5: 129.5+20−1212×10=136.17 129.5 + \frac{20 - 12}{12} \times 10 = 136.17 .
Q₃: the 60th item, in 149.5–159.5: 149.5+60−4416×10=159.5 149.5 + \frac{60 - 44}{16} \times 10 = 159.5 .
Q1≈136.17 Q_1 \approx 136.17 lb, Q₃ = 159.5 lb.

c.
Expenditure per day (Rs) less than 100 100–200 200–300 300–400 400–500 more than 500
Frequency 22 34 52 20 19 13
Answer

cf: 22, 56, 108, 128, 147, 160. N = 160.
Q₁: the 40th item, in 100–200: 100+40−2234×100=152.94 100 + \frac{40 - 22}{34} \times 100 = 152.94 .
Q₃: the 120th item, in 300–400: 300+120−10820×100=360 300 + \frac{120 - 108}{20} \times 100 = 360 .
Q1≈ Q_1 \approx Rs 152.94, Q₃ = Rs 360. (Neither is in an open class, so the open ends don’t matter.)

d.
Marks less than 20 less than 40 less than 60 less than 80 less than 100
Students 21 44 66 79 90
Answer

Frequencies: 21, 23, 22, 13, 11. N = 90.
Q₁: the 22.5th item, in 20–40: 20+22.5−2123×20=21.30 20 + \frac{22.5 - 21}{23} \times 20 = 21.30 .
Q₃: the 67.5th item, in 60–80: 60+67.5−6613×20=62.31 60 + \frac{67.5 - 66}{13} \times 20 = 62.31 .
Q1≈21.30 Q_1 \approx 21.30 , Q3≈62.31 Q_3 \approx 62.31 .

Question 5

a.

The marks of 30 students in an internal examination: 42, 65, 78, 70, 62, 50, 72, 34, 30, 40, 58, 53, 30, 34, 51, 54, 42, 59, 20, 40, 42, 60, 25, 35, 35, 28, 46, 60, 47, 52. Make a frequency table with classes 10 wide and find Q₁ and Q₃.

Answer
Marks 20–30 30–40 40–50 50–60 60–70 70–80
f 3 6 7 7 4 3
cf 3 9 16 23 27 30

Q₁: the 7.5th item, in 30–40: 30+7.5−36×10=37.5 30 + \frac{7.5 - 3}{6} \times 10 = 37.5 .
Q₃: the 22.5th item, in 50–60: 50+22.5−167×10=59.29 50 + \frac{22.5 - 16}{7} \times 10 = 59.29 .
Q₁ = 37.5, Q3≈59.29 Q_3 \approx 59.29 .

b.

The number of eggs a chicken farm produced on 26 days: 32, 87, 17, 51, 99, 79, 64, 39, 25, 95, 53, 49, 78, 32, 42, 48, 59, 86, 69, 57, 15, 27, 44, 66, 77, 92. Make a frequency table with classes 20 wide and find Q₁ and Q₃.

Answer
Eggs 0–20 20–40 40–60 60–80 80–100
f 2 5 8 6 5
cf 2 7 15 21 26

Q₁: the 6.5th item, in 20–40: 20+6.5−25×20=38 20 + \frac{6.5 - 2}{5} \times 20 = 38 .
Q₃: the 19.5th item, in 60–80: 60+19.5−156×20=75 60 + \frac{19.5 - 15}{6} \times 20 = 75 .
Q₁ = 38 eggs, Q₃ = 75 eggs.