1. Statistical Inference
統計推論
parameter
size = N
Population
e.g. mean, s.d….
e.g. sample mean, sample s.d.
Sample
size = n
sample statistic
Random sample: P(a sample) =

2. Sample Mean Distribution
 and  are fixed,
but
and s are random variables,
So what’s the distribution of the sample mean ?

3. i.e. The distribution of The mean (i.e. expectation)
and the variance of
and 
can be found easily.
But we are not satisfied with only TWO single
information, we want to know ALL the situation of
i.e. The distribution of

4. If the distribution of X is normal, the distribution of
is also normal, since
is the linear combination of Xi.
BUT, how about if X is NOT normally distributed?

5. provided that n is sufficiently large.
Central Limit Theorem
Although
X ~ ?
we still have
approximately,
provided that n is sufficiently large.
In practice, n > 30.
Great!

6. ~N(1800,400) Thus, = 0.1056 Light bulbs,  = 1800 hrs,  = 200 hrs.
E.g. 5
Light bulbs,  = 1800 hrs,  = 200 hrs.
Find the prob. that a random sample of
100 bulbs will have average life > 1825 hrs
But, by CLT, we know the sample mean’s distribution!!
We don’t know what’s the distribution of bulbs’ life!!
~N(1800,400)
Thus, =

7. =30n Var(X) = nVar(Xi) =n82 By CLT, X~N(30n,64n) By question,
E.g. 7
A driver is allowed to work max. 10 hrs
per day. Journey time per delivery:
 = 30 min.,  = 8 min. How many
deliveries so as to less than 1/1000 chance
of exceeding 10 hrs.
=30n
Var(X) = nVar(Xi)
=n82
By CLT,
X~N(30n,64n)
By question,
By table,
P(X > 600)  0.001
P(z>3.09) = 0.001

8. To ensure the chance < 0.001, we can take:
 or 4.9
(rejected)
n  16
Hence we should schedule 16 or less deliveries per day.

9. Let X = total points = 3.5 By CLT, = 0.236
E.g. 10
A die is cast sixty times. Estimate the
probability that the total number of points is less than or equal to 200.
Let X = total points
= 3.5
E(X) = 603.5 = 210
By CLT,
Var(X) = 6035/12 = 175
X ~ N(210,175)
P(X  200) =
= 0.236