Statistics Questions and Answers (41-45)

Question 41

The height h(cm) of the 270 students in a school are measured and the results are shown in the table below

h(cm)	Frequency
120 < h ≤ 130	15
130 < h ≤ 140	24
140 < h ≤ 150	36
150 < h ≤ 160	45
160 < h ≤ 170	50
170 < h ≤ 180	43
180 < h ≤ 190	37
190 < h ≤ 200	20

(a) Write down the modal group

(b) Calculate median height and use it to interpret the data

(c) Calculate an estimate of the mean height

(d) Explain why the answer in part (c) is an estimate?

Answer 41

(a) Modal group: 160 < h ≤ 170 cm (highest frequency 50)

(b) Median height:

Median position = (270+1)/2 = 135.5th term

Cumulative frequencies:

• Up to 130: 15
• Up to 140: 39
• Up to 150: 75
• Up to 160: 120
• Up to 170: 170

Median class: 160-170 cm

Median = 160 + (135.5-120)/50 × 10 = 163.1 cm

Interpretation: Half of the students are shorter than 163.1 cm and half are taller.

(c) Estimated mean height:

Class	Midpoint	f	fx
120-130	125	15	1,875
130-140	135	24	3,240
140-150	145	36	5,220
150-160	155	45	6,975
160-170	165	50	8,250
170-180	175	43	7,525
180-190	185	37	6,845
190-200	195	20	3,900
Total			43,830

Mean = 43,830/270 ≈ 162.33 cm

(d) The mean is an estimate because we use class midpoints to represent all values in each class, which may not reflect the actual distribution of heights within each class.

Question 42

The following table shows the cumulative frequency for the height of students

Height, h(cm)	Cumulative frequency
h ≤ 100	0
h ≤ 120	15
h ≤ 140	39
h ≤ 160	75
h ≤ 180	120
h ≤ 200	170
h ≤ 220	213
h ≤ 240	250
h ≤ 260	270
h ≤ 280	282
h ≤ 300	285

Use the given table to find:

(a) The median height

(b) Mode height

(c) All the players in the school's basketball team are chosen from the 30 tallest students.

(d) The least possible height of any player in the basketball team

Answer 42

First create frequency distribution:

Class	Frequency
100-120	15
120-140	24
140-160	36
160-180	45
180-200	50
200-220	43
220-240	37
240-260	20
260-280	12
280-300	3

(a) Median height:

Median position = 285/2 = 142.5th term

Median class: 180-200 cm

Median = 180 + (142.5-120)/50 × 20 ≈ 189 cm

(b) Mode height:

Modal class: 180-200 cm (highest frequency 50)

Mode ≈ 190 cm (midpoint of modal class)

(c) Basketball team selection:

30 tallest students come from the upper end of the distribution

(d) Least possible height:

285 total students - 30 = 255th student marks the cutoff

255th student falls in 240-260 cm class

Least possible height = 240 cm

Question 43

100 students are given a question to answer. The time taken t (seconds) by each student is recorded and the results are shown in following the table

t (sec)	Frequency
0 < t ≤ 20	10
20 < t ≤ 40	10
40 < t ≤ 60	15
60 < t ≤ 80	28
80 < t ≤ 100	22
100 < t ≤ 120	7
120 < t ≤ 140	8

(a) Calculate:

(i) Mean time taken

(ii) Mode time taken

(iii) Median time taken

(b) Two students are picked at random. What is the probability that they both took more than 50 seconds

Answer 43

(a) Calculations:

(i) Mean time:

Class	Midpoint	f	fx
0-20	10	10	100
20-40	30	10	300
40-60	50	15	750
60-80	70	28	1,960
80-100	90	22	1,980
100-120	110	7	770
120-140	130	8	1,040
Total			6,900

Mean = 6,900/100 = 69 seconds

(ii) Mode time:

Modal class: 60-80 seconds (highest frequency 28)

Mode ≈ 70 seconds (midpoint of modal class)

(iii) Median time:

Median position = 50th term

Median class: 60-80 seconds

Median = 60 + (50-35)/28 × 20 ≈ 70.71 seconds

(b) Probability both took >50 seconds:

Number who took >50 seconds = 15(40-60) + 28 + 22 + 7 + 8 - (15×10/20) = 70

Probability first student >50 sec = 70/100

Probability second student >50 sec = 69/99

Combined probability = (70/100) × (69/99) ≈ 0.4879 or 48.79%

Question 44

Calculate the median and mode from the following data

Marks	Number of students
Below 10	6
Below 20	15
Below 30	29
Below 40	41
Below 50	60
Below 60	70

Answer 44

First create frequency distribution:

Class	Frequency
0-10	6
10-20	9
20-30	14
30-40	12
40-50	19
50-60	10

Median:

Median position = 70/2 = 35th term

Median class: 30-40

Median = 30 + (35-29)/12 × 10 = 35

Mode:

Modal class: 40-50 (highest frequency 19)

Mode = 40 + (19-12)/(2×19-12-10) × 10 ≈ 44.38

Question 45

Given the cumulative frequency distribution table

Monthly income (Tsh)	Number of families
More than 100,000	15
More than 200,000	37
More than 300,000	50
More than 400,000	69
More than 500,000	85
More than 600,000	100

(a) Calculate the mean monthly income and use it to interpret the results

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

Answer 45

First convert to standard frequency distribution:

Class	Frequency
100,000-200,000	100-85 = 15
200,000-300,000	85-69 = 16
300,000-400,000	69-50 = 19
400,000-500,000	50-37 = 13
500,000-600,000	37-15 = 22
600,000+	15

(a) Mean monthly income:

Class	Midpoint	f	fx
100,000-200,000	150,000	15	2,250,000
200,000-300,000	250,000	16	4,000,000
300,000-400,000	350,000	19	6,650,000
400,000-500,000	450,000	13	5,850,000
500,000-600,000	550,000	22	12,100,000
600,000+	650,000*	15	9,750,000
Total			40,600,000

*Assuming upper limit of 700,000 for calculation

Mean = 40,600,000/100 = 406,000 Tsh

Interpretation: The average family earns about 406,000 Tsh per month.

(b) Cumulative frequency curve (ogive):

Would plot upper class boundaries against cumulative frequencies

Median estimate ≈ 380,000 Tsh (value at 50 on cumulative frequency)