formula sheet final exam fall2011

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Formula Sheet for ENGR 371 The number of permutations of n distinct objects taken r at a time is n P r = n! (nr)! . The number of distinct permutations of n things of which n 1 are of one kind, n 2 are of a second kind, . . . , n k are of a k th kind is n! n1!n2!n k ! : The number of combinations of n distinct objects taken r at a time is n r = n! r!(nr)! : P (A [ B [ C)= P (A)+ P (B)+ P (C) P (A \ B) P (A \ C) P (B \ C)+ P (A \ B \ C); where A; B; and C are any three events. P (AjB)= P (A\B) P (B) if P (A) 6=0: If the events B 1 ;B 2 ;:::;B k constitute a partition of a sample space S such that P (B i ) 6=0 for i =1; 2;:::;k; then for any event A of S; P (A)= k P i=1 P (B i \ A)= k P i=1 P (AjB i )P (B i ): Bayess Rule: If the events B 1 ;B 2 ;:::;B k constitute a partition of a sample space S such that P (B i ) 6=0 for i =1; 2;:::;k; then for any event A of S such that P (A) 6=0 P (B r jA)= P (B r \ A) P k i=1 P (B i \ A) = P (AjB r )P (B r ) P k i=1 P (AjB i )P (B i ) ; for r =1; 2;:::;k F (x)= P (X x) where X is a random variable and F (x) is its cumulative distribution function. f (yjx)= f (x;y) g(x) where f (x; y) is the joint probability distribution function of X and Y; and g(x) is the marginal probability distribution function of X; and g(x) > 0: E[X]= P all x xf (x) if X is a discrete random variable. E[X]= R 1 1 xf (x)dx if X is a continuous random variable. E[g(X)] = P all x g(x)f (x) if X is a discrete random variable. E[g(X)] = R 1 1 g(x)f (x)dx if X is a continuous random variable. 2 = E[(X X ) 2 ] is the variance of X where X is a random variable. Binomial distribution: f (x)= n x p x q nx ;x =0; 1; 2;:::;n with = np and 2 = npq: Geometric distribution: f (x)= pq x1 ;x =1; 2; 3;::: with =1=p and 2 = (1 p)=p 2 : Negative banomial: f (x)= x1 r1 p r q xr ;x = r; r +1;r +2;::: with = r=p and 2 = r(1 p)=p 2 : Hypergeometric: f (x)= ( K x )( NK nx ) ( N n ) ;x = maxf0;n+K N g to minfK; n) with = np and 2 = npq Nn N1 : Poisson: f (x)= e x x! ;x =0; 1; 2;::: with with = and 2 = Normal distribution: f (x)= 1 p 2 e (x) 2 =2 2 The distribution of an exponential random variable X is given by f (x)= e x ; x> 0 0; elsewhere where 1= > 0: The mean of an exponential random variable X is 1= and its variance is 1= 2 : 1

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Page 1: Formula Sheet Final Exam Fall2011

Formula Sheet for ENGR 371

• The number of permutations of n distinct objects taken r at a time is nPr = n!(n−r)! .

• The number of distinct permutations of n things of which n1 are of one kind, n2 are of a second kind, . . . ,nk are of a kth kind is n!

n1!n2!···nk! .

• The number of combinations of n distinct objects taken r at a time is(nr

)= n!

r!(n−r)! .

• P (A ∪B ∪ C) = P (A) + P (B) + P (C)− P (A ∩B)− P (A ∩ C)− P (B ∩ C) + P (A ∩B ∩ C), where A, B,and C are any three events.

• P (A|B) = P (A∩B)P (B) if P (A) 6= 0.

• If the events B1, B2, . . . , Bk constitute a partition of a sample space S such that P (Bi) 6= 0 for i = 1, 2, . . . , k,

then for any event A of S, P (A) =k∑i=1

P (Bi ∩A) =k∑i=1

P (A|Bi)P (Bi).

• Bayes’s Rule: If the events B1, B2, . . . , Bk constitute a partition of a sample space S such that P (Bi) 6= 0for i = 1, 2, . . . , k, then for any event A of S such that P (A) 6= 0

P (Br|A) =P (Br ∩A)∑ki=1 P (Bi ∩A)

=P (A|Br)P (Br)∑ki=1 P (A|Bi)P (Bi)

, for r = 1, 2, . . . , k

• F (x) = P (X ≤ x) where X is a random variable and F (x) is its cumulative distribution function.

• f(y|x) = f(x,y)g(x) where f(x, y) is the joint probability distribution function of X and Y, and g(x) is the

marginal probability distribution function of X, and g(x) > 0.

• E[X] =∑all x

xf(x) if X is a discrete random variable.

• E[X] =∫∞−∞xf(x)dx if X is a continuous random variable.

• E[g(X)] =∑all x

g(x)f(x) if X is a discrete random variable.

• E[g(X)] =∫∞−∞g(x)f(x)dx if X is a continuous random variable.

• σ2 = E[(X − µX)2] is the variance of X where X is a random variable.

• Binomial distribution: f(x) =(nx

)pxqn−x, x = 0, 1, 2, . . . , n with µ = np and σ2 = npq.

• Geometric distribution: f(x) = pqx−1, x = 1, 2, 3, . . . with µ = 1/p and σ2 = (1− p)/p2.

• Negative banomial: f(x) =(x−1r−1

)prqx−r, x = r, r + 1, r + 2, . . . with µ = r/p and σ2 = r(1− p)/p2.

• Hypergeometric: f(x) =(Kx)(N−K

n−x )(Nn)

, x = max{0, n+K−N} tomin{K,n) with µ = np and σ2 = npq(N−nN−1

).

• Poisson: f(x) = e−λλx

x! , x = 0, 1, 2, . . . with with µ = λ and σ2 = λ

• Normal distribution: f(x) = 1√2πσ

e−(x−µ)2/2σ2

• The distribution of an exponential random variable X is given by

f(x) =

{λe−λx , x > 00, elsewhere

where 1/λ > 0.

• The mean of an exponential random variable X is 1/λ and its variance is 1/λ2.1

Page 2: Formula Sheet Final Exam Fall2011

• The covariance between random variables X and Y is

σXY = E [(X − µX) (Y − µY )] = E [XY ]− µXµY

• The correlation between random variables X and Y is

ρXY =σXYσXσY

• If X1, X2, . . . , Xn represent a random sample of size n, then the sample variance is defined by the statistic

S2 =

∑ni=1(Xi − X̄)2

n− 1

where X̄ is the sample mean.

• Let X1, X2, . . . , Xn be independent random variables that are all normal with mean µ and standarddeviation σ. Then the random variable T = X̄−µ

S/√nhas t−distribution with v = n− 1 degrees of freedom.

• If x̄ is the mean of a random sample of size n from a population with known variance σ2, a (1 − α)100%confidence interval for µ is given by

x̄− zα/2σ√n< µ < x̄+ zα/2

σ√n

where zα/2 is the z-value leaving an area of α/2 to the right.

• One sided confidence bounds:

—A 100(1− α)% upper-confidence bound for µ is

µ ≤ x̄+ zασ√n

—A 100(1− α)% lower-confidence bound for µ is

x̄− zασ√n≤ µ

• In the case σ2 is unknown, the above expression becomes

x̄− tα/2,n−1s√n< µ < x̄+ tα/2,n−1

s√n

where s is the sample standard deviation.

• Test Statistic for mean, variance known

Z0 =X̄ − µ0

σ/√n

• Calculation of beta for a two sided test

β = Φ(zα/2 −δ√n

σ)− Φ(−zα/2 −

δ√n

σ)

• Calculation of sample size

n ≈(zα/2 + zβ)2σ2

δ2

• Test Statistic for mean, variance unknown

T0 =X̄ − µ0

S/√n

2

Page 3: Formula Sheet Final Exam Fall2011

• The random variable

X2 =(n− 1)S2

σ2

has χ2 distribution with n− 1 degrees of freedom.

• If s2 is the sample variance from a random sample of size n from a normal population with unknown varianceσ2, the a (1− α)100% confidence interval for σ2 is given by

(n− 1) s2

χ2α/2,n−1

≤ σ2 ≤ (n− 1) s2

χ2−α/2,n−1

.

• The 100(1− α)% lower and upper confidence bounds on σ2 are

(n− 1) s2

χ2α,n−1

≤ σ2 and σ2 ≤ (n− 1) s2

χ2−α,n−1

,

respectively.

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