discrete random variables variance © christine crisp “teach a level maths” statistics 1
TRANSCRIPT
Discrete Random Discrete Random VariablesVariablesVarianceVariance
© Christine Crisp
““Teach A Level Teach A Level Maths”Maths”
Statistics 1Statistics 1
Variance of a Discrete Random VariableThe variance of a discrete random variable is
found in a similar way to the one we used for the mean.
22
2 xn
fxs variance
For a frequency distribution, the formula is
22
22
1 xn
...xfxf
Replacing by etc.
gives n
f11p
22
221
21
2 ... xpxpxs
22
21
22 ... pxpx
22
221
21
2 ... xpxpxs
So,
222 )( xXPx
But we must replace by and we replace s by the letter ( which is the Greek lowercase s, pronounced sigma ).
x
The variance of X is also written as Var(X).
( Notice that this expression contains the Greek capital S, , and the lowercase s, . )
1910
100
Tip: With a bit of practice you’ll find you can simplify the fractions without a calculator. It’s quicker and more accurate. Try these before you see my answer.
Solution:
222 )( xXPx
e.g. 1 Find the variance of X for the following:
1 2x
P(X = x )
3 4
10
1
10
2
10
3
10
4
222 )( xXPx
10
44
10
33
10
22
10
11
We first need to find the mean,
310
30
)( xXxP
222222 )3(10
44
10
33
10
22
10
11
910
64
10
27
10
8
10
1
SUMMARY
222 )()(Var xXPxX
• The variance, , of a discrete random variable is given by
2
• The mean of a discrete random variable is given by
)()(E xXxPX
For probability distributions ( models ) use and ( the Greek alphabet ).
2
N.B. For frequency distributions use and for the mean and variance ( the “English” alphabet ).
x 2s
Exercise1. Find the variance of X for each of the
following:(a) (b)
P 6
16
21 2 3x
(X = )x P 30 60
0 1 2x
(X = )x6
310
Solution:
(a)
222 )( xXPx2
222
3
7
6
33
6
22
6
11
)( xXxP
6
33
6
22
6
11
6
14
7
3 3
7
9
49
6
27
6
8
6
1
9
496
9
5
Exercise
(b)
P 30 60
0 1 2x
(X = )x 10
Solution:
222 )( xXPx
222 31602101
)( xXxP
602101 31
69152 810
