Variance and standard deviation

MBM 106
VARIANCE AND STANDARD
DEVIATION
PRESENTED BY:
Amrit Swaroop

OBJECTIVES
 To find the variance of a data set.
 To find the standard deviation of a data set.

VARIANCE
 Variance is the average squared deviation
from the mean of a set of data.
 It is used to find the standard deviation.

VARIANCE
 Find the mean of the data.
 Mean is the average so add up the values and divide
by the number of items.
 Subtract the mean from each value – the result
is called the deviation from the mean.
 Square each deviation of the mean.
 Find the sum of the squares.
 Divide the total by the number of items.

VARIANCE FORMULA
2
( )x
n

The variance formula includes the Sigma
Notation, , which represents the sum of all
the items to the right of Sigma.

Mean is represented by and n is the number
of items.


STANDARD DEVIATION
 Standard Deviation shows the variation in
data.
 If the data is close together, the standard
deviation will be small.
 If the data is spread out, the standard
deviation will be large.
 Standard Deviation is often denoted
by the lowercase Greek letter sigma, .

STANDARD DEVIATION
 Find the variance.
a) Find the mean of the data.
b) Subtract the mean from each value.
c) Square each deviation of the mean.
d) Find the sum of the squares.
e) Divide the total by the number of items.
 Take the square root of the variance.

STANDARD DEVIATION FORMULA
The standard deviation formula can be represented
using Sigma Notation:
2
( )x
n



 
The standard deviation formula is the
square root of the variance.

FIND THE VARIANCE AND STANDARD
DEVIATION
The math test scores of five students are:
92,88,80,68 and 52.

2) Find the deviation from the mean:
92-76=16
88-76=12
80-76=4
68-76= -8
52-76= -24
1) Find the mean:
(92+88+80+68+52)/5 = 76.

3) Square the deviation from the
mean:
(16)² = 256
(12)² = 144
(4)² = 16
(-8)² = 64
(-24)² = 576

4) Find the sum of the squares of the deviation
from the mean:
256+144+16+64+576= 1056
5) Divide by the number of data items to find
the variance:
1056/5 = 211.2

6) Find the square root of the variance:
211.2 14.53
Thus the standard deviation of the test
scores is 14.53.

 As we have seen, standard deviation
measures the dispersion of data.
 The greater the value of the standard
deviation, the further the data tend to be
dispersed from the mean.
Conclusion

Variance and standard deviation

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