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 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
• 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
• 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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