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Expectation and variance of the sample mean!
Starter
1. (Review of last lesson) Independent random variables and are such that
, , and . Find:
" (a)" the possible values of !
" (b)" "!
Notes
In the A2 mathematics course, we looked at the distribution of the sample mean.
When a number of random samples of size are taken from a normal distribution with mean and
variance such that , then the distribution of the sample means of the samples will
be normally distributed such that . The standard deviation of this distribution,
, is called the standard error of the mean.
This can be generalised for all random variables with random samples of size :
and .
Binomial: Mean:
Variance:
Geometric: Mean:
Variance:
Poisson: Mean:
Variance:
N.B. With the binomial distribution, the above refers to the number of trials, whereas the in
the formula for refers to the size of the sample.
There is no need to copy this part.
Proof
Let the mean and variance of the population of random variable be and
respectively. A random sample of values is taken from the population. The sample mean, , is
given by:
This is an estimate for the population mean, .
Each of the sample values can be thought of as a value from the
independent variables . These variables have the same distribution as the
population so and .
So the sample mean is a value of the of the random variable given by:
X
Y
E(X
2
) = 14
E(Y
2
) = 20
Var(X ) = 10
Var(Y ) = 11
Var(5X − 2Y )
n
μ
σ
2
X ∼ N(μ, σ
2
)
¯
X
n
∼ N
(
μ,
σ
2
n
)
σ
n
n
E(
¯
X ) = E(X )
Var(
¯
X ) =
Var(X )
n
X ∼ B(n, p)
μ = E(X ) = np
σ
2
= Var(X ) = np(1 − p)
X ∼ Geo( p)
E(X ) =
1
p
Var!(X ) =
1 − p
p
2
X ∼ Po(λ)
μ = E(X ) = λ
σ
2
= Var(X ) = λ
n
n
Var(
¯
X )
X
μ = E(X )
σ
2
= Var(X )
n
¯x
¯x =
x
1
+ x
2
+ x
3
+ . . . + x
n
n
E(X )
x
1
+ x
2
+ x
3
+ . . . + x
n
X
1
, X
2
, X
3
, . . . , X
n
E(X
i
) = E(X )
Var(X
i
) = Var(X )
¯
X
¯
X =
X
1
+ X
2
+ X
3
+ . . . + X
n
n
=
1
n
X
1
+
1
n
X
2
+
1
n
X
3
+ . . . +
1
n
X
n
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Therefore,
i.e. the population mean
Given the fact that are independent,
Please start copying again.
The distribution of the sample means is called the sampling distribution of the means or just
sampling distribution.
As increases, the variance of the sample decreases. In other words, the value of is more
reliable when it is calculated from a large sample which is logical.
E.g. 1 Find the expected value and the variance of the sample mean:
(a) , ,
(b) ,
(c) ,
(d) ,
(e) ,
Working: (a)
E.g. 2 A machine fills cans of drink with a mean liquid content of ml and standard deviation
ml. A sample of cans is taken. Calculate the expectation and variance of the sample
mean of the cans.
E.g. 3 Bananas are sold in bags of with the mass of the bag being exactly g. The mass of
one banana has mean mass g and standard deviation g. Find the mean and
standard deviation of a bag of bananas.
Comparing this question to E.g. 4 from the previous lesson and may wonder what is the difference.
Here is the question from the previous lesson:
“A crane is lifting a crate with large boxes and small boxes. The large boxes have mean mass
kg and standard deviation kg while the small boxes have mean mass kg and standard
deviation kg. Given that the crate has mass kg, calculate the expectation and standard
deviation of the total mass of the crate with large boxes and small boxes loaded on it.”
E(
¯
X ) =
1
n
E(X
1
) +
1
n
E(X
2
) +
1
n
E(X
3
) + . . . +
1
n
E(X
n
)
E(
¯
X ) =
1
n
E(X ) +
1
n
E(X ) +
1
n
E(X ) + . . . +
1
n
E(X )
= E(X )
X
1
, X
2
, X
3
, . . . , X
n
Var(
¯
X ) = Var
(
X
1
n
)
+ Var
(
X
2
n
)
+ Var
(
X
3
n
)
+ . . . + Var
(
X
n
n
)
=
1
n
2
Var(X
1
) +
1
n
2
Var(X
2
) +
1
n
2
Var(X
3
) + . . . +
1
n
2
Var(X
n
)
=
1
n
2
Var(X ) +
1
n
2
Var(X ) +
1
n
2
Var(X ) + . . . +
1
n
2
Var(X )
=
1
n
Var(X )
n
¯x
E(X ) = 10
Var(X ) = 1.6
n = 20
X ∼ N(120, 5
2
)
n = 8
X ∼ B(9, 0.4)
n = 15
X ∼ Po(8.5)
n = 30
X ∼ Geo(0.25)
n = 32
E(
¯
X ) = E(X ) = 10
Var(
¯
X ) =
Var(X )
n
=
1.6
20
=
2
25
= 0.08
355
17
30
30
5
37
180
16.4
4
5
18
3
12
1.5
25
4
5
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The answer is that there is no real difference in the question and similar working for one will work
for the other.
Here is sample mean working for the crane question:
and
and
Standard deviation is .
The mean and standard deviation are g and g (3 s.f.)
Here is the independent observations working for the bananas question:
and
" !
" " " " !
" " " " !
" " " " !
" !
" " " " !
" " " " !
" " " " !
" " " " !
" " " " !
Video: Deriving the mean and variance of the sample mean
Video: Expectation and variance of the sample mean
Solutions to Starter and E.g.s
Exercise
p155 8B Qu 1i, 2-5, (7-8 red)
Summary
E(
¯
L) = E(L) = 18
Var(
¯
L) =
Var(L )
n
=
3
2
4
= 2.25
E(
¯
S ) = E(S ) = 12
Var(
¯
S ) =
Var(S )
n
=
1.5
2
4
= 0.45
E(4
¯
L + 5
¯
S + 25) = 4E(
¯
L) + 5E(
¯
S ) + 25
= 4E(L) + 5E(S ) + 25
= 4 × 18 + 5 × 12 + 25
= 157
Var(4
¯
L + 5
¯
S + 25) = 4
2
Var(
¯
L) + 5
2
Var(
¯
S )
= 16 × 2.25 + 25 × 0.45
= 47.25
82 5
5
≈ 36.7
937
36.7
E(B) = 180
Var(B) = 16.4
2
E(5#bananas#plus#bag) = E(B
1
+ B
2
+ . . . B
5
+ 37)
= E(B
1
) + . . . + E(B
5
) + 37
= 5 × 180 + 37
= 937
Var(5#bananas#plus#bag) = Var(B
1
+ B
2
+ . . . + B
5
)
= Var(B
1
) + Var(B
2
) + . . . + Var(B
5
)
= Var(B) + Var(B) + Var(B) + Var(B) + Var(B)
= 5Var(B)
= 5 × 16.4
2
= 1344.8
E(
¯
X ) = E(X )
Var(
¯
X ) =
Var(X )
n
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