Tung Wah College

GEN1008/ GED1008 Introduction to Statistics

Assignment 1
[Deadline: 8 February 2025 Saturday, 11:59pm]

Instructions
1. Answer all questions. Write your answers on standard A4 papers. You may use electronic devices
to write on “virtual papers”. Do not type.
2. Submit the softcopy as one PDF file in the blackboard. You can scan your assignment using any
Mobile Phone Apps, such as Adobe Scan, FastScanner, which can save your file as a single PDF.
3. Write your steps clearly. Answers without any steps will have no marks.
4. Make reasonable rounding unless the rounding rule is explicitly stated in the question.
5. Make sure that all your answers are written in proper English. Marks could be deducted for any
English or grammatical mistakes.
6. No plagiarism is tolerated. All plagiarized work will score 0.
7. Using ChatGPT to generate answers directly may not work as I observed that wrong answers
could be produced. You cannot learn reasoning on similar questions in the midterm or
examination if you rely on ChatGPT to complete the assignment.

Question 1 (16 marks)

Suppose you work in a physiotherapy team. The team performed a study about the flexibility of patients
recovering from lower back pain. The forward bending flexibility (measured in centimeters from the
floor) were recorded as follow.

26 21 48 31 37 36 38
27 25 28 27 26 40 29
20 43 27 33 31 22 28
33 34 21 34 24 41 37

a) What is the level of measurement of the variable in the study? Is it discrete or continuous? [2]
b) Construct a frequency distribution with title, class limits, class boundaries and frequency using 6
classes. Use 20 as the lower limit of the first class, and a convenient number for the class interval
width. [4]
c) Suppose you have to report the mean and the standard deviation of data above. Find the answers
using the Excel software.

To show your work, please paste a screenshot of the Excel software with the data values, the

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 mean value and the standard deviation value, and write down the Excel functions with
parameters you have used to obtain the mean and the standard deviation. Your screenshot
should also include the row / column headings, similar to that below. [3]

column headings

row headings

[Screenshot: 1 mark ; Excel function and numerical answer for mean: 1 mark ; Excel function and
numerical answer for standard deviation: 1 mark]

d) Choose a correct option for (iii), and fill in the blanks for (i), (ii), (iv) to (vi) to complete the
following descriptions about the data set. [4]

In this study, the sample is chosen from a population that consists of (i) _____________. The
sample size is (ii) __________. From the collected sample data, there are (iii) __________% of
patients having forwarding bending flexibility less than 35 centimeters. The sample data shows
that the forwarding bending flexibility is (iv) ( positively-skewed / symmetric / negatively-
skewed ). In the dataset, the sample mean equals (v) ____________ centimeters. If the population
data follows the same distribution, the population mean is expected to be (vi) ( larger than / the
same as / less than ) the median.

e) Suppose two more values are added into the dataset. Both are larger than 50. How will (i) the
mean, (ii) the standard deviation, and (iii) the degree of skewness of the sample data change? [3]

[Solution]

a) The variable is measured at ratio level. [1]

The variable is continuous. [1]
Reason: The flexibility is measured in centimeters, which is a distance from the patient’s fingertip
and the floor. The distance has a true zero and the measure is continuous.

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b) The frequency distribution is [3]

Class Limits Class Boundaries Frequency

20 - 24 19.5 - 24.5 5
25 - 29 24.5 - 29.5 9
30 - 34 29.5 - 34.5 6
35 - 39 34.5 - 39.5 4
40 - 44 39.5 - 44.5 3
45 - 49 44.5 - 49.5 1
Total 28
Frequency distribution of forward bending flexibility (measured in
centimeters from the floor) of a sample of patients recovering from lower back
pain [1]

c) Screenshot [1]:

The mean is 30.96 cm (found by =AVERAGE(A1:G4)) [1]

The standard deviation is 7.183 cm (found by =STDEV.S(A1:G4)) [1]

d) (i) all patients recovering from lower back pain [1]

(ii) 28 [0.5]
(iii) 71.4% [0.5]
(iv) positively-skewed [0.5]
(v) 30.96 [0.5]
(vi) larger than [1]

e) The mean will be larger. [1]

The standard deviation will be larger. [1]
The distribution is more positively-skewed. [1]

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Question 2 (9 marks)

Your research team has collected data from a sample of 24 high school students. The aptitude test
scores (from 0 to 100) and the stress scores (from 0 to 60) of students are obtained. They are organized
as the frequency distributions below (with some information hidden).

Your assistant has obtained the following summary statistics based on the distribution for you, but
some information is found missing.

Sum of 𝑋 Sum of 𝑋 2 Mean Standard

values values Deviation
Aptitude test score ? 174,936 85.083 ?
Stress level ? ? ? 6.753

a) Find the mean stress level of students. [2]

b) Find the standard deviation of the aptitude test scores of students. [3]
c) Find the coefficient of variation of the aptitude test scores and the stress levels of students.
respectively. [2]
d) Which of the two variables, aptitude test score or stress level, is more dispersed? Explain your
answer by using a value for comparing the dispersion. [2]

[Solution]

a) The mean stress level of students is found by [2]:

Class Limit Frequency (𝑓) Mid-point (𝑋𝑚 ) 𝑓 × 𝑋𝑚

21 - 25 2 23 46
26 - 30 4 28 112
31 - 35 4 33 132
36 - 40 8 38 304
41 - 45 5 43 215
4
 46 - 50 1 48 48
Total 24 857

857
𝑥̅ = = 35.71
24

b) The sum of aptitude test scores of students is [1]

∑ 𝑥 = 85.083 × 24 = 2042

The sample standard deviation of aptitude test scores of students is [2]

𝑛
1 𝑛 ∑ 𝑥 2 − (∑ 𝑥)2 24(174936) − (2042)2
2
𝑠 = ̅ )
∑(𝑋 − 𝑋 = 2
= = 51.993
𝑛−1 𝑛(𝑛 − 1) 24(23)
𝑖=1

𝑠 = √𝑠 2 = 7.211 𝑚𝑎𝑟𝑘𝑠

c) For aptitude test score, the coefficient of variation is [1]

7.211
𝐶𝑉𝑎𝑟 = × 100% = 8.475%
85.083
For stress level, the coefficient of variation is [1]
6.753
𝐶𝑉𝑎𝑟 = × 100% = 18.91%
35.71

d) Since the coefficient of variation of stress level (18.91%) is higher than that of the aptitude test
score (8.475%), the stress levels of students are more dispersed. [2]

Remarks:
- 0.5 marks if the student uses the standard deviation to do the comparison.