Transcript Welcome Back

• HW solutions are on the web.
• See website for how to calculate
probabilities with minitab, excel, and TI
calculators.
Determining normal probabilities:
• Suppose X has a normal distribution with
mean 5 and std dev 2.
• Notation X~N(5,4)
[notation uses N(mean,variance)]
• What’s the probability that X is less than
4?
0.05
0.10
Pr(X<4) = area under
curve to left of x=4
0.0
density
0.15
0.20
Normal Density
0
5
x
10
7
• What’s Pr(X < 4)?
• Draw (previous page)
• Center and scale:
– Pr(X<4)
= Pr( (X-5)/2 < (4-5)/2 )
= Pr( Z < -1/2 )
• Look up (appendix 1)
= Pr(Z<-1/2)
= 0.3085
“Centering and Scaling?”
• Suppose X~N(mu,sigma^2).
• Why does (X-mu)/sigma have a N(0,1)
distribution? (X-mu) part is “centering” and
/sigma part is “scaling”.
• Idea: All normal distributions have the
same shape. Centering and scaling just
relabels the x and y axes. The area under
the curve (and the probabilities) remains
the same.
0.05 0.10 0.15 0.20
0.0
density
Pr(X<4) = area under
curve to left of x=4
4
0.4
0
10
0.1
0.2
0.3
Pr(Z<-0.5) = area under
curve to left of -0.5
Same area as above
0.0
Density
5
x
-4
-2
-0.5
0
Centered and scaled X
2
4
Want area in
between these
bars
0.08
0.06
0.0
0.02
0.04
density
X ~ N(2,9)
What’s Pr(1<X<4)?
0.10
0.12
Example 2
-5
0
5
X
First let’s do this with the tables.
10
Purpose of all
this is to get to
an expression
that only uses
Pr(Z<a) where
Z~N(0,1). All
because tables
have Pr(Z<a).
Pr(1<X<4)
=Pr[(1-2)/3<Z<(4-2)/3] (where Z~N(0,1))
=Pr[Z<(4-2)/3] –Pr[Z<(1-2)/3]
=Pr(Z < 2/3) – Pr(Z < -1/3)
= 0.7475 - 0.3694
= 0.3781
EVEN IF YOU’LL ALWAYS USE CALCULATORS,
MINITAB, EXCEL, OR MATLAB TO DO THESE
PROBLEMS, YOU’LL NEED TO KNOW HOW TO DO
CALCULATIONS LIKE THE ONES ABOVE…
Using excel or minitab, the only step that is necessary is to
get the probability in terms of CDFs (i.e. Pr[X <= k]).
Pr(1<X<4)
= Pr(X<4) – Pr(X<1), where X ~N(2,32)
= 0.748 – 0.369 = 0.378
(Do demo in class)
Three probabilities to memorize:
Pr(Z < 2.33) = 99%
Pr(Z < 1.96) = 97.5%
Pr(Z < 1.28) = 90%
Remember: Z~N(0,1) “the standard normal”
1.0
0.8
0.6
0.4
0.2
Let X~N(10,16). Find an
a such that
Pr(X < a) = 0.80.
0.0
are on this
axis
CDF
Later in the course, we
will need to be able to
do things like the
Probabilities
following:
Plot of x versus Pr(X<x)
when X~N(10,42)
0
5
10
x
a is this number here
15
20
Let X~N(10,16). Find an a such that
Pr(X < a) = 0.80.
Pr[(X-10)/4 < (a-10)/4] =
=Pr(Z < (a-10)/4] =0.80
Using the table “backwards” we find that
Pr(Z < 0.84) = 0.80
As a result, (a-10)/4 = 0.84
So, a = 13.36
This is called an inverse probability problem.
The Normal Distribution is Pervasive
• Examples of things that are normally distributed:
– Heights, weights, abilities, many, many other
measurements
– In general, when a quantity is the result of a
combination of many factors and influences, samples
of that quantity are very likely to be approximately
normally distributed.
– Why?
• A store keeps track of the average amount
spent by people each day.
• Let Xi = average amount spent on day i
• It turns out that there is a good reason to
believe that Xi has a normal distribution!!!
• This reason is the
CENTRAL LIMIT THEOREM
Central Limit Theorem
• Let X1,…,Xn be n independent random
variables each with constant mean m and
constant variance s2.
• Then, as n gets large,
(X1+…+Xn)/n ~ N(m, s2/n)
and
(X1+…+Xn) ~ N(nm, ns2)
What does “large n” mean
What “large” is depends on the distribution
of Xi.
•
•
•
If Xi’s are already normal, then the result
is true for any n
If Xi’s have a symmetric distribution, then
n at least 3 is probably large enough
If Xi’s have a skewed distribution, the n of
20 or 30 is probably large enough
Example
• Suppose the amount of potassium in a banana
is normally distributed with mean 630mg and
standard deviation 40mg.
• You eat 3 bananas a day. Let T = amount of
potassium you eat. What is the probability that T
< 1800mg?
• By central limit theorem, T~N[3*630,3*(402)].
Want Pr(T < 1800)
= Pr[ (T-1890)/(sqrt(3)*40) < (1800 –
1890)/(sqrt(3)*40)]
= Pr[ Z < -1.33] = 0.0981 (see p672-3 in book for
table or use Calculator or Excel or Minitab)
0.005
0.004
0.003
0.002
0.001
0.0
density
Area under curve
to left of line is
Pr(T < 1800)
1600
1700
1800
1900
T
2000
2100