Statistics Questions and Answers (31-40)

Question 31

Find the missing frequencies f₁ and f₂ in the table given below, it being given that the mean of the given frequency distribution is 50 and the total frequencies is 120

Class	Frequency
0-20	17
20-40	f₁
40-60	32
60-80	f₂
80-100	19

(Given: f₁ = 28, f₂ = 24)

Answer 31

Finding f₁ and f₂:

Total frequency = 17 + f₁ + 32 + f₂ + 19 = 120 ⇒ f₁ + f₂ = 52

Given mean = 50

Class	Midpoint	Frequency	fx
0-20	10	17	170
20-40	30	f₁	30f₁
40-60	50	32	1600
60-80	70	f₂	70f₂
80-100	90	19	1710

Mean = Σfx/Σf ⇒ 50 = (170 + 30f₁ + 1600 + 70f₂ + 1710)/120

Solving the equations gives f₁ = 28, f₂ = 24

Question 32

The examination scores of 33 students is given on the following cumulative frequency

Class marks	Cumulative frequency
44.5	2
54.5	4
64.5	9
74.5	17
84.5	29
94.5	33

From the above data determine:

(a) Mean mark

(b) Modal class and mode mark

(c) Median class and median mark

Answer 32

(a) Mean mark:

Class Mark	Frequency	fx
44.5	2	89
54.5	2	109
64.5	5	322.5
74.5	8	596
84.5	12	1014
94.5	4	378
Total	33	2,508.5

Mean = 2,508.5/33 ≈ 76.02

(b) Modal class: 80-90 (highest frequency 12)

Mode ≈ 84.5 (midpoint of modal class)

(c) Median position = (33+1)/2 = 17th term

Median class: 70-80 (cumulative frequency reaches 17 here)

Median ≈ 74.5

Question 33

The data below shows a master's graduate, summarized his research findings by table as follows:

Class marks	Frequency
20	3
25	2
30	10
35	5
40	3
45	2

(a) Construct the frequency distribution table, showing intervals, frequencies and class marks.

(b) Calculate the mean

(c) Draw a cumulative frequency curve and use it to estimate median

Answer 33

(a) Frequency distribution table:

Class Interval	Class Mark	Frequency
17.5-22.5	20	3
22.5-27.5	25	2
27.5-32.5	30	10
32.5-37.5	35	5
37.5-42.5	40	3
42.5-47.5	45	2

(b) Mean calculation:

Σfx = (20×3)+(25×2)+(30×10)+(35×5)+(40×3)+(45×2) = 60+50+300+175+120+90 = 795

Mean = 795/25 = 31.8

(c) Cumulative frequency curve would show:

Upper Bound	Cumulative Freq
22.5	3
27.5	5
32.5	15
37.5	20
42.5	23
47.5	25

Median ≈ 31.5 (value at 12.5 on cumulative frequency)

Question 34

The mass in kilograms of 72 athletes were recorded as shown below.

Mass(kg)	Frequency
42	7
47	8
52	11
57	10
62	4
67	2
72	15
77	12
82	3

Find:

(a) The mode and the modal class

(b) The median and the median class

(c) The mean

Answer 34

(a) Mode: 72 kg (highest frequency 15)

Modal class: 70-74 kg

(b) Median position = (72+1)/2 = 36.5th term

Cumulative frequencies:

• Up to 42: 7
• Up to 47: 15
• Up to 52: 26
• Up to 57: 36

Median class: 55-59 kg

Median ≈ 57 kg

(c) Mean calculation:

Σfx = (42×7)+(47×8)+(52×11)+(57×10)+(62×4)+(67×2)+(72×15)+(77×12)+(82×3)

= 294 + 376 + 572 + 570 + 248 + 134 + 1,080 + 924 + 246 = 4,444

Mean = 4,444/72 ≈ 61.72 kg

Question 35

(NECTA CSEE 2024). In the terminal examination of a certain school, the scores of students in Geography subject were grouped as shown in the following table.

Scores	Cumulative frequency
65-69	10
70-74	22
75-79	43
80-84	49
85-89	58
90-94	62
95-99	66

Using the information given in the table:

(a) Find the mean score correct to 2 decimal places, given an assumed mean of 77

(b) Draw the cumulative frequency curve (ogive) of the score

(c) Calculate the mode of the scores, correct to 3 decimal places

Answer 35

(a) Mean using assumed mean (77):

Class	Midpoint	f	d=x-A	fd
65-69	67	10	-10	-100
70-74	72	12	-5	-60
75-79	77	21	0	0
80-84	82	6	5	30
85-89	87	9	10	90
90-94	92	4	15	60
95-99	97	4	20	80
Total				100

Mean = A + (Σfd/Σf) = 77 + (100/66) ≈ 78.52

(b) Ogive would plot upper class boundaries against cumulative frequencies.

(c) Mode calculation:

Modal class: 75-79 (highest frequency 21)

Mode = L + (f1-f0)/(2f1-f0-f2) × h

= 75 + (21-12)/(2×21-12-6) × 4 ≈ 76.500

Question 36

The ages of a sample of 100 families is given in the following cumulative frequency table

Age(in years)	Cumulative frequency
0-15	35
16-31	55
32-47	79
48-63	91
64-79	100

From the given data:

(a) Find the mean age of this sample of villagers

(b) Find the median age

(c) Find the mode age

Answer 36

(a) Mean age:

Class	Midpoint	Frequency	fx
0-15	7.5	35	262.5
16-31	23.5	20	470
32-47	39.5	24	948
48-63	55.5	12	666
64-79	71.5	9	643.5
Total			2,990

Mean = 2,990/100 = 29.9 years

(b) Median age:

Median position = 50th term

Median class: 16-31

Median = 15.5 + (50-35)/20 × 16 ≈ 27.5 years

(c) Mode age:

Modal class: 0-15 (highest frequency 35)

Mode ≈ 7.5 years (midpoint of modal class)

Question 37

The following table shows the marks obtained by 100 students of class X in a school during a particular academic session

Marks	Cumulative frequency
< 10	7
< 20	21
< 30	34
< 40	46
< 50	66
< 60	77
< 70	92
< 80	100

Calculate:

(a) Mode

(b) Median

(c) Mean

Answer 37

First create frequency distribution:

Class	Frequency
0-10	7
10-20	14
20-30	13
30-40	12
40-50	20
50-60	11
60-70	15
70-80	8

(a) Mode:

Modal class: 40-50 (highest frequency 20)

Mode = 40 + (20-12)/(2×20-12-11) × 10 ≈ 44.71

(b) Median:

Median position = 50th term

Median class: 40-50

Median = 40 + (50-46)/20 × 10 = 42

(c) Mean:

Class	Midpoint	f	fx
0-10	5	7	35
10-20	15	14	210
20-30	25	13	325
30-40	35	12	420
40-50	45	20	900
50-60	55	11	605
60-70	65	15	975
70-80	75	8	600
Total			4,070

Mean = 4,070/100 = 40.7

Question 38

Compute the arithmetic mean, mode and median for the following data

Marks	Number of students
Less than 10	14
Less than 20	22
Less than 30	37
Less than 40	58
Less than 50	67
Less than 60	75

Answer 38

First create frequency distribution:

Class	Frequency
0-10	14
10-20	8
20-30	15
30-40	21
40-50	9
50-60	8

Arithmetic mean:

Class	Midpoint	f	fx
0-10	5	14	70
10-20	15	8	120
20-30	25	15	375
30-40	35	21	735
40-50	45	9	405
50-60	55	8	440
Total			2,145

Mean = 2,145/75 = 28.6

Mode:

Modal class: 30-40 (highest frequency 21)

Mode = 30 + (21-15)/(2×21-15-9) × 10 ≈ 33.33

Median:

Median position = (75+1)/2 = 38th term

Median class: 30-40

Median = 30 + (38-37)/21 × 10 ≈ 30.48

Question 39

The cumulative frequency table below represents the scores of candidates in an examination

Marks	Cumulative frequency
≤ 10	0
≤ 20	2
≤ 30	5
≤ 40	10
≤ 50	22
≤ 60	39
≤ 70	59
≤ 80	75
≤ 90	80

(a) How many candidates:

(i) Sat for the examination?

(ii) Score over 50?

(iii) Score over 40 but not more than 70?

(b) What percentage of candidates scores more than 60?

(c) Calculate the frequency distribution table for these marks

(d) Calculate:

(i) Mean marks

(ii) Median marks

(iii) Mode marks

Answer 39

(a) (i) Total candidates: 80

(ii) Score >50: 80 - 22 = 58

(iii) Score >40 but ≤70: 59 - 10 = 49

(b) Percentage >60: (80-39)/80 × 100 = 51.25%

(c) Frequency distribution:

Class	Frequency
0-10	0
10-20	2
20-30	3
30-40	5
40-50	12
50-60	17
60-70	20
70-80	16
80-90	5

(d) Calculations:

(i) Mean:

Class	Midpoint	f	fx
10-20	15	2	30
20-30	25	3	75
30-40	35	5	175
40-50	45	12	540
50-60	55	17	935
60-70	65	20	1300
70-80	75	16	1200
80-90	85	5	425
Total			4,680

Mean = 4,680/80 = 58.5

(ii) Median:

Median position = 40th term

Median class: 60-70

Median = 60 + (40-39)/20 × 10 = 60.5

(iii) Mode:

Modal class: 60-70 (highest frequency 20)

Mode = 60 + (20-17)/(2×20-17-16) × 10 ≈ 62.31

Question 40

The following table shows the masses in grams of 100 apples

Mass, x(g)	Frequency
120 ≤ x < 130	25
130 ≤ x < 140	19
140 ≤ x < 150	23
150 ≤ x < 160	16
160 ≤ x < 170	12
170 ≤ x < 180	5

Find:

(a) Mean mass of the apple

(b) Mode mass of the apple

(c) Median mass of the apple

Answer 40

(a) Mean mass:

Class	Midpoint	f	fx
120-130	125	25	3,125
130-140	135	19	2,565
140-150	145	23	3,335
150-160	155	16	2,480
160-170	165	12	1,980
170-180	175	5	875
Total			14,360

Mean = 14,360/100 = 143.6g

(b) Mode:

Modal class: 120-130 (highest frequency 25)

Mode ≈ 125g (midpoint of modal class)

(c) Median:

Median position = 50th term

Median class: 140-150

Median = 140 + (50-44)/23 × 10 ≈ 142.61g