Mathematical Methods for Physics and Engineering : A Comprehensive Guide

31.2 SAMPLE STATISTICS

and thesample standard deviationis the positive square root of the sample

variance, i.e.

s=

√

√

√

√^1

N

∑N

i=1

(xi− ̄x)^2. (31.7)

Find the sample variance and sample standarddeviation of the data given in table 31.1.

We have already found that the sample mean is 185.0 to one decimal place. However,
when the mean is to be used in the subsequent calculation of the sample variance it is
better to use the most accurate value available. In this case the exact value is 184.975, and
so using (31.6),

s^2 =

1

8

[

(188. 7 − 184 .975)^2 +···+ (200. 0 − 184 .975)^2

]

=

1608. 36

8

= 201. 0 ,

where once again we have quoted the result to one decimal place. The sample standard
deviation is then given bys=

√

201 .0=14.2. As it happens, in this case the difference
between the true mean and the rounded value is very small compared with the variation
of the individual readings about the mean and using the rounded value has a negligible
effect; however, this would not be so if the difference were comparable to the sample
standard deviation.

Using the definition (31.7), it is clear that in order to calculate the standard

deviation of a sample we must first calculate the sample mean. This requirement

can be avoided, however, by using an alternative form fors^2. From (31.6), we see

that

s^2 =

1

N

∑N

i=1

(xi− ̄x)^2

=

1

N

∑N

i=1

x^2 i−

1

N

∑N

i=1

2 xi ̄x+

1

N

∑N

i=1

̄x^2

=x^2 − 2 ̄x^2 + ̄x^2 =x^2 − ̄x^2

We may therefore write the sample variances^2 as

s^2 =x^2 − ̄x^2 =

1

N

∑N

i=1

x^2 i−

(

1

N

∑N

i=1

xi

) 2

, (31.8)

from which the sample standard deviation is found by taking the positive square

root. Thus, by evaluating the quantities

∑N

i=1xiand

∑N

i=1x

2
ifor our sample, we
can calculate the sample mean and sample standard deviation at the same time.