Central Limit Theorem:
If X_{1} + X_{2} + ... + X_{n} is a random
sample from a distribution with mean
m and finite variance,
s^{2},
then for any constant k:
Note that this theorem holds for all distributions, continuous
and discrete.
A useful alternative way of writing the formula is: P{[(å_{i=1,n}
X_{i}) - nm]/
sn^{1/2}
£ k} »
F(k)
For finite sample size this is an approximation. If the distribution from
which the random sample is drawn is unimodal and symmetric, then the sample
size can be "small" and the approximation will be fairly good. If the
distribution is skewed, then the sample size must be "larger" for the
approximation to be fairly good. There are no hard and fast rules.
Example: Sampling from a Bernoulli distribution.
Suppose we flip a coin 900 times. What is the probability that the number
of heads will be greater than 495?
This is taking a random sample of size 900 from a Bernoulli distribution
with parameter p = 1/2. Recall that:
æ p x = 1
ç
f(x) = ç 1-p x = 0
ç
è 0 otherwise
Where E(X) = p = m and
VAR(X) = p(1 - p) =
s^{2}
In this case
m = p = 1/2 and
s^{2} = p(1 - p) = 1/4
Hence: P[(å_{i=1,900} X_{i}) > 495] =
P{[(å_{i=1,900} X_{i}) - 900*1/2]/(1/2)*30 > (495 - 450)/15} =
P(Z > 3) = 1 - F(3) = .00135
Problem 7.27 p.308
Let X_{i} = "Life of Lamp i". We are given
m = 50
hours, and
s = 4 hours. We are to calculate: P(System will work longer than 1300 Hours) =
P[(å_{i=1,25} X_{i}) > 1300] =
P{[(å_{i=1,25} X_{i}) - 25*50]/(4*5) >
(1300 - 1250)/20} =
P(Z > 2.5) = 1 - F(2.5) = .0062
So, 62 out of every 10,000 times you came back from a trip to Paris a
lamp would still be burning in the greenhouse!
Problem 7.32 p.308
Let X_{i} = "Time Person i Waits to go Through the Checkout
Once he/she has reached it". We are given
m = 2.5
minutes, and
s = 2 minutes.
We are to calculate: P(It will take more than 4 hours to serve a 100 people) =
P[(å_{i=1,100} X_{i}) > 240] =
P{[(å_{i=1,100}
X_{i}) - 100*2.5]/(2*10) > (240 - 250)/20} =
P(Z > -1/2) = 1 - F(-1/2) =
1 - .3085 = .6915
Graduation Problem
Suppose we have 600 graduating seniors. Based upon records we know that,
on average, 1/3 of seniors have 2 parents at their graduation, 1/3 of seniors
have 1 parent at their graduation, and 1/3 of seniors have no parents at
their graduation. What is the probability that less than 650 parents show
up at the graduation?
Let X_{i} = "Number of Parents for the ith Senior". To get
m and
s we have to get the distribution of
X_{i}.
Clearly f(x_{i}) is a discrete uniform distribution. That is: