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ENGR 371 Formula sheet...

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CribSheetsforENGR371 n! ( n − r )!

•

The number of permutations of n distinct objects taken r objects at a time: Prn =

•

The number of permutations of n objects of which n1 are of one kind, n2 are of second kind, …. nk are of kth kind:

n! , where n1 + n2 +iii+ nk = n . n1 ! n2 !iii n k ! ⎛n⎞

n!

•

The number of combinations of n distinct objects taken r objects at a time: ⎜ ⎟ = ⎝ r ⎠ r !( n − r ) !

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Multiplication rule: an operation has k steps, and the number of ways for completing step k is nk, the total number of way for completing the operation is n1 × n 2 ×iii×n k

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Probability of a union: P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) , where A and B are two events

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Conditional probability & Bayes theorem: P ( A B ) =

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Total probability: the sample space S constitutes a partitions of B1, B2 and B3, the probability of any event A of S, where A is overlapped with parts of B1, B2 and B3,

P (A ∩ B ) P (B )

=

P ( B A) P ( A) P (B )

, P (B) ≠ 0 .

P ( A) = P ( A ∩ B1 ) + P ( A ∩ B 2 ) + P (A ∩B 3 ) = P (A B 1 )P (B 1 ) +P (A B 2 )P (B 2 ) +P (A B 3 )P (B 3 )

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Bayes theorem: the sample space S constitutes a partitions of B1, B2 and B3, then for any event A of S, P ( Bk A ) =

P ( A Bk ) P ( Bk ) P (A)

, k=1, 2, or 3

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Independence: P ( A B ) = P ( A ) , P ( A ∩ B ) = P ( A ) P ( B )

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Cumulative probability distribution: F ( x ) = P ( X ≤ x ) , where X is a random variable.

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Mean of a random variable X: μ = E ( X ) , E (X ) =

∑ xf (x ) for a discrete random variable, all x

E( X ) =

∫ xf ( x) dx

for a continuous random variable, where f(x) is probability density function

all

•

Mean of a random function g(X): E ⎡⎣ g ( X )⎤⎦ =

∑ g (x ) f ( x ) for a discrete random variable X, all x

E ⎡⎣ g ( X )⎤⎦ =

∫ g ( x ) f ( x) dx

for a continuous random variable X.

all

•

Variance of a random variable X: σ 2 = E ⎡⎢ (X − μ )2 ⎤⎥ ⎣ ⎦

•

Binominal distribution: probability function: ⎜ ⎟ p x (1 − p)

⎛n ⎞ ⎝x ⎠

number of successes in n trials

1 

n −x

, μ = np and σ 2 = np (1 − p ) , x is the

•

⎛ x −1 ⎞ r x−r ⎟ p (1 − p ) , μ = r / p and r 1 − ⎝ ⎠

Negative binominal distribution: probability function: ⎜

σ 2 = r (1 − p ) p 2 , x is the number of trials for r successes

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⎛ K ⎞⎛ N − K ⎞ ⎜ ⎟⎜ ⎟ x ⎠⎝ n − x ⎠ ⎝ , where N objects contain K objects Hypergeometric distribution: probability function ⎛N⎞ ⎜ ⎟ ⎝n⎠

as success, a random sample of n objects selected from N objects, x is the number of successes ⎛ N −n ⎞ ⎟ ⎝ N −1 ⎠

μ = np and σ 2 = np (1 − p ) ⎜

e −λλ x , μ = λ and σ 2 = λ , x is the number of events x!

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Poisson distribution:

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Normal or Gauss distribution: ⎛ x2 exp ⎜⎜ − 2π ⎝ 2

1

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⎛ ( x − μ )2 exp ⎜ − ⎜ 2σ 2 σ 2π ⎝ 1

⎞ ⎟⎟ ⎠

X is a binominal random variable with parameter n and p, the probability of X can be approximated by standard Normal distribution with using Z =

•

• • • •

⎞ ⎟ , and standard Normal distribution: ⎟ ⎠

X − np np (1 − p )

X is a Poisson random variable with parameter λ , the probability of X can be approximated by X−λ standard Normal distribution using Z = λ ⎧1 ⎛ x ⎞ x >0 ⎪ exp ⎜− ⎟ , β > 0 , μ = β and σ 2 = β Exponential distribution: f ( x ) = ⎨β ⎝ β ⎠ ⎪ elesewhere 0 ⎩

Covariance between random variables X and Y: σ XY = E ⎡⎣( X − μ X )(Y − μ Y )⎤⎦ = E ( XY ) − μ X μ Y σ Correlation between random variables X and Y: ρ XY = XY σ XσY Marginal probability function of X and Y: fx ( X ) =

∫

f ( x , y )dy and fy ( Y ) =

all y

•

•

2 

2

∫

f ( x, y )dx for

all x

continuous random variables X and Y; Marginal probability function of X and Y: f ( x , y ) and fy (Y ) = f ( x , y ) for discrete random variables X and Y fx(X) =

∑

∑

all y

all x

Conditional probability: f (Y X ) =

f (x , y ) fx ( x )

with

∫ f (Y X ) dy =1 or ∑ f (Y X ) =1 all Y

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Conditional mean: E (Y x ) = Y f (Y X ) dY for continuous random variables

•

Independence: f ( x , y ) = f ( x) f ( y ) , or f (Y X ) = f (Y ), f ( x ) and f ( y ) are marginal probability

∫

functions. •

Mean of a function h ( x , y ) , x and y are two random variables, E ⎡⎣h ( x , y ) ⎤⎦ = E ⎡⎣ h ( x, y )⎤⎦ =

y

Multinomial distribution: n trials in total, the success probability of class 1, 2 … k is p1, p2, p3…. pk, n! p1x1 p2x2 iii pkxk , where x1, x2 … and xk are the number of trials x1 ! x2 !iii xk !

the probability function:

•

corresponding to class 1, 2 ….. and k. Linear combination: Z=aX+bY+C, the mean of Z: E ( Z ) = aE ( X ) + bE (Y ) + C , the variance of Z: σ Z2 = a 2σ X2 + b 2σ Y2 + 2 abσ

XY

. n

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∑X Sample mean for sample size of n: X = n

∑( X

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Sample variance: S2 =

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Test statistic: Z =

−X)

n

2

i

n −1

X−μ

σ

i

i

i= 1

, for n random samples

has Normal distribution, for n random samples, where μ and σ are the

n

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population mean and standard deviation, respectively. X− μ has t-distribution with n-1 degrees of freedom, Test statistic: T = S

n

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Two independent populations with means μ1 and μ2 and variances σ12 and σ 22 , Test statistic Z =

X 1 − X 2 − ( μ1 − μ 2 )

σ12 n1

+

σ22

has standard Normal distribution, where X 1 are X 2 are two

n2

independent sample means from sample size of n1 and n2 •

ˆ )= θ . An unbiased point estimator Θˆ for a parameter θ must satisfy E ( Θ

•

If X is the sample mean with sample size n from a population with known variance σ 2 , a σ σ (1− α )100% confidence interval of the population mean: X − Z α 2 < μ < X + Z α 2 n

Note: (1 − α )100% confidence upper bound: μ < X + Zα 3 

or

∑∑ h ( x, y ) f ( x, y ) x

•

∫∫h ( x , y ) f ( x , y ) dxdy

σ n

n

(1 − α )100% confidence lower bound: •

μ > X − Zα

σ n

If X is the sample mean with sample size n from a population with unknown variance, a

(1− α )100% confidence interval of the population mean:

X − Tα

Note: (1 − α )100% confidence upper bound: μ < X + T α ,n −1

(1 − α )100% confidence lower bound: •

χα2 2

≤ σ2 ≤

,n−1

χ 2α 1−

2

•

S n S n

(n − 1)S 2 χα2,n − 1

≤σ 2

( n− 1) S2 χ12− α ,n−1

A ( 1− α ) 100% prediction interval on a single future observation from a Normal distribution: X − Tα

•

2,n −1S

1+

1 < X n +1 < X +Tα n

1 n

2,n −1S

1+

⎡

⎡ δ n⎤ δ n⎤ ⎥ − Φ ⎢ −Z α − ⎥ for two sided hypothesis, δ = μ − μ 0 . σ ⎥⎦ σ ⎥⎦ ⎢⎣ 2

Probability of type II error: β = Φ ⎢ Z α − ⎢⎣

2

⎡

Probability of type II error: β = Φ ⎢ Z α − ⎢⎣

2

δ n⎤ ⎥ for upper sided hypothesis σ ⎥⎦

⎡

Probability of type II error: β = 1 − Φ ⎢ − Zα − ⎢⎣

2

δ n⎤ ⎥ for lower sided hypothesis σ ⎥⎦

If the population variance is unknown, calculate β with replacement Z by T with T distribution for the mean hypothesis 2

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⎛ ⎜ Zα + Zβ ⎜ Sample size for a two-sided test: n ≈ ⎝ 2 2

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Hypothesis test for a population variance: test statistic: χ 2 =

4 

S n

,n−1

upper bound: σ 2 ≤

•

2 ,n−1

( n− 1 ) S2

Note lower bound:

•

S < μ < X +Tα n

If S 2 is the sample variance, a (1 −α )100% confidence interval for the population σ 2:

( n− 1 ) S2

•

μ > X − Tα ,n −1

2 ,n−1

⎞ 2 ⎟ σ ⎟ ⎠

δ

( n −1) S 2 σ2...