CMV6120Mathematics

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:

Reference

Suen, S.N. “Mathematics for Hong Kong 5A”; rev. ed.; Chapter 5; Canotta

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 and standard deviation ,

The standard score z is

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 / class average / class s.d. / John’s mark
Mathematics / 75 / 5 / 70
English / 56 / 2 / 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.

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 –) /

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

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.

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.

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

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.

Percentage of students with standard score greater than –1 is 34% + 50% = ______

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.

Solution

a)700 = 1000 – 2 (150)

=  2

Light bulbs with life hours lie greater than 2

=

The no. of light bulbs with life hours more than 700 hours

=

b)1150 = 1000 + 1 (150)

= 

The number of light bulbs with life hours less than 1150 hours =

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 

Solution

The possible highest resistance = 10 k (1 + 20%) = 12 k

12 k = 10 k + 3

 = (12 – 10) / 3 = 0.6667 k

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)

Practice

  1. Find Peter’s standard scores in English and Chinese

Subject / Peter’s mark / class mean / standard deviation
English / 70 / 55 / 10
Chinese / 66 / 50 / 8
  1. Keeping the standard score unchanged, find the adjusted marked for the original mark 55.

mean / standard deviation
original / 40 / 10
adjusted / 50 / 12
  1. Keeping the standard score unchanged, find the mean x of the adjusted marks.

Peter’s mark / mean / standard deviation
original / 38 / 42 / 8
adjusted / 44 / x / 12
  1. Find the mean m and standard deviation  of marks. It is given that:

raw mark / standard score
72 / -0.6
90 / 1.2
  1. 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)Find the number of students who score less than 68.

b)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?

  1. 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)number of students who are heavier than 74 kg.

b)number of students whose weight lie between 62 kg and 71 kg.

[Answer 1).1.5, 2;2) 683) 504) s = 10, m = 78

5a) 12600, 5b) 846a) 20, 6b) 820]

Unit 21: Applications of standard deviationPage 1 of 8