Statistics-Formula Sheet
Final Exam
Uniformin the interval [a, b]:
 1
a<x<b
f (x) = b − a
,
0
otherwise


0
x≤a

x − a
a<x<b
F (x) =

b−a


1
x≥b
Basics
Counting Permutations:
n!
.
n Pk =
(n − k)!
E[X] =
Counting Combinations:
n!
n
.
n Ck = k =
k!(n − k)!
a+b
2 ,
V (X) =
(a−b)2
12
Exponential pdf:
f (x; λ) = λe−λx , E[X] =
Gamma pdf:
f (x; α, β) =
V (X) = αβ 2 .
Conditional Probability Formula:
P [A ∩ B]
P [A|B] =
.
P [B]
Bayes Theorem (with partition B1 , · · · , Bk ):
P [Bi ] × P [A|Bi ]
P [Bi |A] = k
.
X
P [Bj ] × P [A|Bj ]
1
1
, V (X) = 2 .
λ
λ
1
α−1 e−x/β ,
β α Γ(α) x
E[X] = αβ,
Normal pdf:
1 x−µ 2
f (x) = σ√12π e− 2 ( σ ) , −∞ < x < ∞, E[X] =
µ, V (X) = σ 2 .
j=1
Chi-square pdf:
1
f (x, ν) = 2ν/2 Γ(ν/2)
xν/2−1 e−x/2 , E[X] = ν,
V (X) = 2ν.
Discrete Uniform:
Two Variables
Z ∞
p(x; N ) = 1/N for x = 1, · · · , N . E[X] =
2
Marginal Density of X: fX (x) =
f (x, y)dy
(N + 1)/2, V (X) = (N − 1)/12.
Random Variables
−∞
Binomial: b(x; n, p) =
np, V (X) = npq.
Poisson:
p(x; λ) =
λx −λ
,
x! e
n
x
px (1−p)n−x , E[X] =
Conditional Density of X given Y = y:
f (x, y)
fX|Y (x|y) =
fY (y)
Covariance:
Cov(X, Y ) = E[XY ] − E[X]E[Y ]
E[X] = λ, V (X) = λ.
Hypergeometric:
h(x; n, M, N ) =
V (X) = n ·
M
x
N −M
n−x
N
n
, E[X] = n ·
Correlation:
M
,
N
Corr(X, Y ) = ρ(X, Y ) =
M
M N −n
· (1 −
)·
.
N
N
N −1
Cov(X, Y )
σX σY
Confidence Intervals and Testing:
Population Mean:√
√
Neg. binom pmf:
C.I: x̄ ± (zα/2 )σ/ n Or x̄ ± (tα/2 )s/ n
r(1
−
p)
r
x
N B(x; r, p) = x+r−1
,
X̄ − µ0
r−1 p (1−p) , E[X] =
p
√
Test Statistic: Z =
σ/ n
r(1 − p)
V (X) =
.
p2
X̄ − µ0
√ , with n − 1 d.f.
T =
Geometric pmf:
s/
n
1
x−1
g(x; p) = p(1 − p) , E[X] = , V (X) =
p
Sample Size Calculation:
(1 − p)
When α, σ and the width w of the C.I. are all
.
p2
known:
1
n=
2z α2 σ
2
w
.
Population Proportion:
r
p̂q̂
C.I.: p̂ ± zα/2
,
n
Test Statistic:
p̂ − p0
Z=r
.
p 0 q0
n
Sample Size Calculation:
For C.I.: When α and the width w of the C.I.
are all known:
!2
2z α2 p(1 − p)
.
n=
w
Sample Size Calculation in Hypothesis Testing:
(When both α and β(µ0 ) are given)
σ(zα + zβ ) 2
n=
µ0 − µ0
OR σ(zα/2 + zβ ) 2
n=
.
µ 0 − µ0
2