CHAPTER 1. PROBABILITY
I. Solution Steps
+ Step 1: Define the events from the problem.
+ Step 2: Define the probabilities and conditional probabilities for the events defined in Step 1.
+ Step 3: Find the system of events which is both mutually exclusive and collectively exhaustive (compute the complement if needed).
+ Step 4: Apply the formula.
II. Probability rules
( )
1. Complement rule: P A = 1 − P ( A) .
2. Addition rule: P ( A B ) = P ( A) + P ( B ) − P ( A B ) .
In particular, if A, B are mutually exclusive, then P ( A B ) = P ( A) + P ( B ) .
3. Conditional probability: P ( A | B ) =
P ( A B)
.
P ( B)
4. Multiplication rule: P ( A B ) = P ( A) P ( B | A) = P ( B ) P ( A | B )
Events A, B are said to be statistically independent if and only if P ( A B ) = P ( A) P ( B ) .
5. Total probability:
Given the system of events A1 , A2 ,..., An that are both mutually exclusive and collectively exhaustive, then
n
P ( B ) = P ( Ai ) P ( B Ai )
i =1
= P ( A1 ) P ( B | A1 ) + P ( A2 ) P ( B | A2 ) + ... + P ( An ) P ( B | An )
or
( ) (
P ( B ) = P ( A) P ( B | A) + P A P B | A
)
CHAPTER 2. DISCRETE RANDOM VARIABLES
I. Distribution table
Given the distribution table:
X
x1
x2
….
xn
P
p1
p2
….
pn
Then
0 pi 1
1. n
.
pi = 1
i =1
2. P(a X b) = pi .
a xi b
n
3. = E ( X ) = pi xi .
i =1
n
4. 2 = Var ( X ) = xi2 pi − 2 .
i =1
In particular, if Y = aX + b then Y = a X + b; Y2 = a 2 X2 .
II. Binomial Distribution
Suppose that
• a random experiment can result in two possible outcomes, “success” and “failure,”
• and that p is the probability of a success in a single trial.
Let X be the the number of resulting successes in n independent trials.
The probability distribution of X is called binomial distribution.
P( X = k ) = Cnk p k (1 − p)n−k
E ( X ) = np
Var ( X ) = np(1 − p)
III. Poisson Distribution
Let X be the number of occurrences in a given continuous interval (such as time, surface area, or length). Then the probability distribution of
X is called the Poisson distribution.
P( X = k ) =
e− k
,
k!
E ( X ) = V (X ) =
CHAPTER 3. CONTINUOUS RANDOM VARIABLES
I. Density Function
Given the density function f ( x ) of a continuous random variable X . Then
i)
f ( x ) 0, x R
+
ii)
f ( x ) dx = 1
−
b
iii) P ( a X b ) = P ( a X b ) = P ( a X b ) = P ( a X b ) = f ( x ) dx
a
x
iv) F ( x ) = f ( t ) dt
−
+
v) E ( X ) = x f ( x)dx
−
+
vi) V ( X ) = ( x − E ( X ) ) f ( x)dx
−
2
II. Normal Distribution
If X follows the normal distribution with the mean and variance 2 , then
b−
a−
P (a X b) = F
−F
The values of F ( x ) can be found in Appendix Table 1 with the notices
i) F ( − x ) = 1 − F ( x ) .
ii) x 4 : F ( x ) = 1
iii) x −4 : F ( x ) = 0
(
Moreover, if X N X , X2
) and Y N ( , ) , then aX + bY also follows the normal distribution
Y
2
Y
aX + bY N ( a X + bY ; a 2 X2 + b2 Y2 )
CHAPTER 4. SAMPLING DISTRIBUTION
Let the random variables X 1 , X 2 ,..., X n denote a random sample from a population.
I. Sample mean
- Sample mean: X =
X 1 + X 2 + ... + X n
.
n
- If the parent population distribution is normal (or n 25 ) then
2
X −
N ( 0; 1) .
X N ;
or Z =
n
/
n
II. Sample proportion
X
- Sample proportion: p =
n
where n is the sample size and X is the number of objects having the characteristic of interest.
- When n is large ( nP (1 − P ) 5 ), we have
p N P; P (1 − P ) or Z =
n
p − P
P (1 − P )
n
N ( 0; 1)
III. Sample variance
( X − X ) + ( X − X ) + ... + ( X − X ) .
- Sample variance: S =
2
2
1
2
2
2
n
n −1
- If the parent population distribution is normal then
( n − 1) S 2 2
2
n −1
CHAPTER 5. CONFIDENCE INTERVAL ESTIMATION
I. Formulas
Confidence Interval Estimation with the sample n and the confidence level (1 − )
Population mean
Estimation Given
+ X N , 2 (or n is
for
large)
+ 2 is known
(
)
Confidence
interval
x z /2
Margin of
error
ME = z /2
Width
w = 2 z /2
Upper
confidence
limit
Lower
confidence
limit
n
n
(
large)
+ 2 is unknown
w = 2 tn−1, /2
n
Given
+ X N , 2
)
ME = tn−1, /2
n
LCL = x − z /2
(
s
n
p z
/2
s
n
ME = z /2
s
n
UCL = x + tn−1, /2
LCL = x − tn−1, /2
w = 2 z /2
s
n
s
n
Population variance 2
Population proportion P
Given
Given
+ X N , 2 (or n is + n is large
x tn−1, /2
n
UCL = x + z /2
Population mean
(
p 1 − p
)
( n − 1) s 2 ( n − 1) s 2
; 2
2
n
−
1,
/2
n
−
1,1
−
/2
n
(
p 1 − p
(
n
p 1 − p
UCL = p + z /2
LCL = p − z /2
)
(
n
p 1 − p
n
ME =
)
n
p 1 − p
(
)
1
1
1
− 2
( n − 1) s 2 2
2
n
−
1,1
−
/2
n
−
1,
/2
1
1
w = ( n − 1) s 2 2
− 2
n −1,1− /2 n −1, /2
)
n − 1) s 2
(
UCL =
)
n − 1) s 2
(
LCL =
n2−1,1− /2
n2−1, /2
II. Critical values
1. Critical values for the standard normal distribution:
z is a value such that P ( Z z ) =
Some critical values: z0,05 = 1,645; z0,025 = 1,96; z0,01 = 2,33; z0,005 = 2,58
Given the critical value z , we can determine the significant level (or confidence level 1 − ) by the following formula
= 1 − F ( z )
where F ( z ) can be found in Appendix Table 1.
2. Critical values for the Student’s t distribution
tn−1, is a value such that P(Tn−1 tn−1, ) =
The value tn−1, can be found in Appendix Table 8.
3. Critical values for the Chi-squared distribution
n2−1, is a value such that P( n2−1 n2−1, ) =
The value n2−1, can be found in Appendix Table 7a and 7b.
CHAPTER 6. HYPOTHESIS TESTS
I. Solution steps
+ Step 1: State the null and alternative hypotheses.
+ Step 2: Summarize all information.
+ Step 3: Decision rule: reject H 0 if …… (Formula)
+ Step 4: Random sample => conclusion
II. Decistion rule
For population mean
X N ( , ) (or n is large)
2
2 is known
Reject H 0 if x 0 − z /2
or x 0 + z /2
X N ( , ) (or n is large)
2
2 is unknown
H 0 : = 0
H1 : 0
H 0 : = 0
H1 : 0
n
Reject H 0 if x 0 + z
s
n
Reject H 0 if x 0 + tn−1,
n
Reject H 0 if x 0 − z
n
n
Reject H 0 if x 0 − tn−1, /2
or x 0 + tn−1, /2
H 0 : = 0
H1 : 0
s
n
s
n
Reject H 0 if x 0 − tn−1,
s
n
H 0 : P = P0
H1 : P P0
For population
proportion
nP0 (1 − P0 ) 5
or p P0 + z /2
P0 (1 − P0 )
n
2
2
H0 : = 0
2
2
H1 : 0
For population
variance
X N ( , 2 )
P0 (1 − P0 )
n
Reject H 0 if p P0 − z /2
n − 1) s 2
(
2
Reject H if
0
or
( n − 1) s 2
2
0
02
n −1, /2
n2−1,1− /2
H 0 : P = P0
H1 : P P0
H 0 : P = P0
H1 : P P0
Reject H 0 if
Reject H 0 if
p P0 + z
P0 (1 − P0 )
n
p P0 − z
H 0 : 2 = 02
2
2
H1 : 0
2
2
H 0 : = 0
2
2
H1 : 0
n − 1) s 2
(
2
Reject H if
0
02
P0 (1 − P0 )
n
n −1,
n − 1) s 2
(
2
Reject H if
0
02
n −1,1−
0
