Limit Central Theorem
Notation. N (, 2) refers to normal with mean
and variance 2.
Have just seen that when X1, . . . , Xn} is a random sample from N (, 2) then sample means
X N (, 2/n). This follows from a more general result that:
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If Xi N (i, i ) are independent normals then
any linear combination
n
Y = a0 +
aiXi
i=1
is also normal. By rules of linear combinations
n
E[Y ] = a0 +
aii
i=1
n
Var[Y ] =
2
a2i
i
i=1
1
The previous example used the special case where all i are same , all
standard deviations i are the same ,
1
a0 = 0, all other ai are the same n .
Then Y becomes the sample mean
1n
Xi
X=
n i=1
E[X ] = a0 +
=0+
Var[X ] =
n
with
n
aii
i=1
n1
i=1 n
=
and
2
a2i
i
i=1
n
2
122
=
( ) =
n
i=1 n
2
That is
X N (, 2/n)
By standardizing, the random variable
X
Z=
/ n
is standard normal.
Another special
case when all i = and all i =
is the sample sum
n
T=
Xi
so
i=1
T N (n, n 2)
which standardizes using
T n
Z=
n
3
But what if {X1, . . . , Xn} is a random
sample from a non-normal population?
Then no general statement (except for
a few special cases) can be made about
the exact distribution of a linear com
bination, including a sample mean X or
sample sum T except when n is large.
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Central Limit Theorem. When n is
large there is a famous result known
as the Central Limit Theorem, which
states that if {X1, . . . , Xn} is a random
sample from any population with mean
and standard deviation then
X N (, 2/n)
n
T
Xi N (n, n 2)
i=1
In other words sums and averages are
approximately normal for large n.
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By standardizing, the last two approximations are equivalent to
X
N (0, 1)
/ n
n
i=1 Xi n N (0, 1)
n
Next plots show distribution for average of sums for randomly thrown dice.
See how uniform turns into normal as
number of tosses gets large
6
fig_ 0 7 _ 0 1
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For some populations, even small n gives
accurate approximation, as in the next
example, from your text.
Example 4.1-3 p 223 X1, X2 independent U (0, 1) and T = X1 + X2. Find
P (0.5 < T < 1.5) using central limit
theorem.
Recall for uniform U (a, b),
b+a 2
(b a)2
=
, =
.
2
12
Hence E[X1] = = (0 + 1)/2 = 1/2
and VarX1 = (1 0)2/12 = 1/12.
Note. These calculations were done in class in
response to a question.
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E[T ] = 2 = 2(1/2) = 1, Var T = 2 2 =
2(1/12) = 1/6. Trying the Central Limit
Approximation
P (0.5 < T < 1.5)
0. 5 2
1.5 2
P(
<Z<
)
2
2
0.5 2(1/2)
1.5 2(1/2)
)
= P(
<Z<
1/12 2
1/12 2
0 . 5
0.5
= P(
<Z<
)
1/6
1/6
= (1.22) (1.22) = 0.8888 0.1112 =
The exact value (see problem 4.1-10)
OR try simulating) is 0.75.
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Example. # accidents/week at hazardous intersection is r.v. X with = 2.2, = 1.4. so
not normal.
(a) Let X be average # accidents/week at intersection during a year (52 weeks). What
is approximate distribution of X ?
X integer values so not normal, n = 52.
X = = 2 . 2
1.4
X =
=
= 0.194
52
52
X N (2.2, 0.1942)
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(b) What is the approximate probability that
X is less than 2?
2 2 .2
P (X < 2) P (Z <
)
0.194
= P (Z < 1.03) = 0.1515
(c) What is the approximate probability that
there are fewer than 100 accidents at the
intersection in a year?
Let T = X1 + X2 + . . . + X52, then
T = n = 52(2.2) = 114.4
T = n = 52(1.4) = 10.0955
P (fewer than 100) = P (T < 100)
100 114.4
P (Z <
)
10.0955
= P (Z < 1.43) = 0.0764
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Example. Suppose # accidents/week at hazardous intersection is Poisson X with parameter = 2. What is the exact probability that
there are fewer 100 than accidents at the intersection in a year?
Fact: sums of independent Poissons are Poisson; parameter is sum of individual parameters.
Let T = X1 + X2 + . . . + X52. Then T is Poisson
with parameter = 2(52) = 104.
99
e104104i
P (T < 100) =
= 0.3343
i!
i=0
using software.
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What is the approximate probability using the
central limit theorem?
T = 104,
T = 104 = 10.198
P (T < 100) = P (T 99)
99 104
P (Z
)
10.198
= P (Z < 0.4903) = 0.3120
Better approximation uses continuity correction:
P (T < 100) = P (T 99.5)
99.5 104
P (Z
)
10.198
= P (Z < 0.4413) = 0.3295
Continuity correction used when approximating discrete random variable by continuous one.
Draw histogram, then smooth curve through
it. The curve will intersect each rectangle in
the histogram close to the midpoint. So take
boundary half way to next integer value. Continuity correction always adds more probability.
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Normal Approximation to Binomial
First known case of Central Limit Theorem was for binomial. Called normal
approximation to binomial.
Let X binomial with parameters n, p.
2
Know X = np, X = np(1 p). So
X N (np, np(1 p))
In terms of proportion P = X
n
p(1 p)
)
P N (p,
n
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Example. Multiple-choice tests. Suppose student has probability p = 0.75 of correctly answering question chosen at random from a universe of all possible questions. Correctness of
answers to dierent questions are independent.
(a) Use normal approximation to nd probability that student scores 80% or higher on a
20-question test.
n = 20
p = p = 0.75
p(1 p)
.75(.25)
p =
=
= 0.0968
n
20
p 0.75
0.80 0.75
P ( 0.80) = P (
p
)
0.0968
0.0968
= P (Z 0.5165)
= 1 (0.5165) = 0.3027
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(b) If test contains 100 questions, what is probability student scores 80% or higher?
p = p = 0.75, p =
.75(.25)
= 0.0433
100
p 0.75
0.80 0.75
)
0.0433
0.0433
= P (Z 1.1547) = 0.1241
P ( 0.80) = P (
p
(c) If test contains 250 questions, what is probability student scores 80% or higher?
p = p = 0.75, p =
.75(.25)
= 0.0274
250
0.80 0.75
)
0.0274
= P (Z 1.8248) = 0.0340
P ( 0.80) = P (Z
p
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Example: 40% of cars turn right at a fork in a
road. Of 200 cars arriving at the intersection
let X = # turning right. Find P (75 X 85).
X is binomial with n = 200, p = 0.40.
X = np = 200(.40) = 80,
X = np(1 p) = 200(0.40)(0.60) = 6.93.
X 80
85 80
75 80
P (75 X 85) = P
6.93
6.93
6.93
P (0.72 Z 0.72)
= 0.7642 0.2358 = 0.5284
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You nd 100 of the next 200 cars turned
right? Is this sucient reason to believe that the assumed 40% is wrong?
After all, this result might just be due
to chance.
To answer this question,
nd the probability that X is 100 or
larger if p = 0.4 is true.
X 80
100 80
P (X 100) = P
6.93
6.93
P (Z 2.89) = 1 0.9981
= 0.0019
Since this probability is very small, then
probably p is actually greater than 0.4.
This is an example of a statistical inference called an hypothesis test.
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Continuity correction. Example 4.1-4 p 233.
X is binomial n = 16, p = 0.5. Use normal approximation to nd P (X = 7).
X = np = 16(.5) = 8,
X =
np(1 p) =
16(0.5)(0.5) = 2. Naive approximation gives
answer 0 because normal is continuous and
area above the point x = 7 under the normal
density curve is 0. Instead write
P (X = 7) = P (6.5 < X 7.5)
7 .5 8
6.5 8
P(
<Z
)
2
2
= (.25) (.75)
= 0.4013 0.2266 = 0.1747
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Exact value.
16 7
.5 (1 .6)9
P (X = 7) =
7
= 0.1740
Read entire example in text. Note that
there is an error in text but the nal
answer 0.1747 is correct.
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