CMV6120
Mathematics
Unit 21 : Applications of standard deviation
Learning Objectives
The Students should be able to:

Calculate standard score from a given set of data.

Determine, in the case of normal distribution, the percentages of data lying within a certain
number of standard deviations from the mean.
Activities
Teacher demonstration and student hand-on exercise.
Web Reference:
http://www.mste.uiuc.edu/hill/dstate/dstate.html
http://frey.newcastle.edu.au/Stat/stat101.html
Reference
Suen, S.N. “Mathematics for Hong Kong 5A”; rev. ed.; Chapter 5; Canotta
Unit 21: Applications of standard deviation
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CMV6120
Mathematics
Applications of standard deviation
1. Standard score
Standard scores are used to compare students’ performances in different tests.
For a distribution of marks with mean x and standard deviation ,
The standard score z is
z
xx

Example 1
In the final examination, John obtained 70 in Mathematics and 60 in Vocational English.
His results are compared with those for the whole class.
Subject
Mathematics
English
class average
75
56
class s.d.
5
2
John’s mark
70
60
a) Find John’s standard scores in Mathematics and Vocational English.
b) State, for each subject, whether his performance is above or below average.
Solution
a) standard score in Mathematics = (70 – 75)  5 =
standard score in Vocational English =
b) John is below average in Mathematics. (70<75)
John is above average in Vocational English.
(60>56)
Although John’s raw mark in Mathematics is higher than that in English (70>60),
he is better in English than Mathematics when compared to other students.
This is reflected by the standard scores (2 > 1).
Remark:
1. Negative standard score means below average.
2. Zero standard score means at the average.
3. Positive standard score means above average.
Unit 21: Applications of standard deviation
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CMV6120
Mathematics
Example 2
The statistical data of an examination are tabulated below. A student obtained 43 marks and 48
marks in Information Technology Application and Chinese respectively. Calculate the standard
scores that the student obtained in the examination.
Subject
mean Standard deviation
Information Technology Application 51.5
13.1
Chinese
63.3
14.5
Solution
Standard score Z = (x – x ) /
Information Technology Application: Z =
Chinese:
Z=
Example 3
A student obtained 65 marks in Engineering Science. The mark was equal to a standard score of 3.2.
Calculate the standard deviation of the marks if the mean mark was 45.
Solution
Standard deviation  =
Example 4
The mean mark and the standard deviation of the marks in an examination are 58.9 and 19.4
respectively. The standard score of a student is 1.5. What is the actual mark that the student
obtained? .
Solution
Unit 21: Applications of standard deviation
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CMV6120
Mathematics
2. Percentages of Normal data lying within a certain number of standard
deviations from the mean
A distribution curve for a set of data is basically a frequency or relative frequency curve of the data.
It is found that the distribution curves for a lot of commonly occurring data sets follow a certain
pattern that came to be known as normal distributions.
A normal distribution has a bell-shaped curve as shown.
y
50%
50%
99.7%
95%
68%
34%
13.5%
13.5% 2.35%
2.35%
m3 m2 m
34%
m
m+
m+2 m+3
x
A normal curve has the following characteristics:
1. It is symmetrical about the mean.
2. Mean = mode = median. They all lie at the centre of the curve.
3. There are fewer data for values further away from the mean.
a) about 68% of the data lie within 1 standard deviation from the mean.
b) about 95% of the data lie within 2 standard deviations from the mean.
c) about 99.7% of the data lie within 3 standard deviations from the mean.
Unit 21: Applications of standard deviation
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CMV6120
Mathematics
Example 5
John got 70 in a Mathematics examination.
The marks of the examination are normally distributed with mean = 75 and standard deviation = 5.
If there are 100 students, how many students do better than John?
Solution
Higher than 70
y
By symmetry, ______% students do better than 75 marks,
______ students do worse.
John’s mark is 70. He has a standard score of –1.
John is at 1 standard deviation below the mean.
There are ______% of students lying within one s.d.
below the mean.
34%
Percentage of students with standard score greater than –
1 is 34% + 50% = _________
34%
13.5%
13.5%
2.35%
Total no. of students doing better than John out of the
100 students =
Example 6
The life hours of 200 sample light bulbs are normally
distributed with a mean of 1000 hours and the standard
deviation is 150 hours. Find the number of light bulbs
having life hours:
a)
more than 700 hours;
b)
less than 1150 hours.
60
70 75
65
80
2.35%
85 90
x
y
34%
34%
Solution
a)
700 = 1000 – 2 (150)
= x  2
13.5%
13.5%
2.35%
550
700
850
1000
1150
2.35%
1300
1450
x
Light bulbs with life hours lie greater than x 2
=
y
The no. of light bulbs with life hours more than 700 hours
=
b)
1150 = 1000 + 1 (150)
34%
= x 
The number of light bulbs with life hours less than
1150 hours =
13.5%
2.35%
34%
13.5%
2.35%
x
Unit 21: Applications of standard deviation
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CMV6120
Mathematics
Example 7
The cheesecake made by a cake factory are normally distributed with a mean of 500 gm and a
standard deviation of 12 gm. How many per cent of cheesecake have weigh more than 524 gm?
Solution
Example 8
The length of a sample of 400 conduits are normally distributed with a mean of 2.92 m and a
standard deviation of 2 cm. How many conduits in the sample have length between 2.9 m and 2.96
m?
Solution
Example 9
The resistance of a sample of 100 resistors is normally distributed with a mean of 10k. If all the
resistors have a resistance of 10k  20%, lie within 3 standard deviations from the mean. Find the
standard deviation and the possible resistance lying within x  
Solution
The possible highest resistance = 10 k (1 + 20%) = 12 k
12 k = 10 k + 3
 = (12 – 10) / 3 = 0.6667 k
Unit 21: Applications of standard deviation
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CMV6120
Mathematics
Example 10
The movies have a mean playing time of 88 min. and the standard deviation is 5 min. If the playing
times are normally distributed, find the percentage of the movies with playing time
a)
less than 78 min;
b)
between 83 min. and 103 min..
Solution
a)
Unit 21: Applications of standard deviation
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CMV6120
Mathematics
Practice
1.
Find Peter’s standard scores in English and Chinese
Peter’s mark
70
66
Subject
English
Chinese
2.
mean
40
50
Peter’s mark
38
44
standard deviation
8
12
standard score
-0.6
1.2
In a Mathematics examination, the marks obtained by 15000 students are normally
distributed with a mean of 52 and a standard deviation of 16. The percentages of marks
lying within 1 and 2 standard deviations from the mean are 68% and 96% respectively.
a)
b)
6.
mean
42
x
Find the mean m and standard deviation  of marks. It is given that:
raw mark
72
90
5.
standard deviation
10
12
Keeping the standard score unchanged, find the mean x of the adjusted marks.
original
adjusted
4.
standard deviation
10
8
Keeping the standard score unchanged, find the adjusted marked for the original mark 55.
original
adjusted
3.
class mean
55
50
Find the number of students who score less than 68.
If the top 2% of students are awarded a distinction, what is the minimum mark a
student must get in order to receive a distinction?
The weights of 1000 students are normally distributed with mean 68 kg and standard
deviation 3 kg. If 68% of the students lie within one standard deviation of the mean and
96% lie within 2 standard deviations of the mean, find
a)
b)
number of students who are heavier than 74 kg.
number of students whose weight lie between 62 kg and 71 kg.
[Answer 1).1.5, 2;
2) 68
5a) 12600, 5b) 84
Unit 21: Applications of standard deviation
3) 50
4) s = 10, m = 78
6a) 20, 6b) 820]
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