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statistics cheat sheet

Nasser M. Abbasi

Sumemr 2008 page compiled on July 2, 2015 at 6:07pm

problem: phone calls received at rate λ = 2  per hr. If person wants to take 10 min shower, what is probability a phone will ring during that time?

answer: first change to       10     10-
ω =  λ60 = 2 60 = .3333  , now we want P(X  ≥ 1 ) = 1 − P (X ≤ 1) = 1 − P (0)

but          k
P (k) = λk!e−λ  , but remember, we are using ω  , so           k
P (k) = ωk! e−ω  so             0
P (0) = .33303!-e−.3333 = 0.777

so P (X  ≥ 1) = 1 − .777 = 0.283  , so 28%  change the phone will ring.

How long can shower be if they wish probability of receiving no phone calls to be at most 0.5?

              ω0 −ω           −ω
P (0) = 0.5 =  0! e  →  0.5 = e  hence ln0.5 = − ω →  ω = 0.693  , so   x
λ 60-=  0.693 →  x = 20.7  minutes

To find quantile, say 1
4   , first find an expression for F (x)  as function of x  , then solve for x  in F (x ) = .25

For median, solve for x  in F (x) = .5

properties of CDF: 1. Show F (x) ≥ 0  for all x.  Do this by showing  ′
F (x) ≥ 0  , and show limit F (x ) → 1  as x →  ∞ and limit F (x) →  0  as x →  − ∞ . And P (k1 ≤ T < k2) = F (k2) − F (k1)

properties of pdf:

1.
piecewise continuous
2.
pdf(x ) ≥ 0
3.
∫ ∞
 − ∞ pdf (x) = 1

remember d-tan− 1x =  -1-2
dx           1+x

The geometric distribution is the only discrete memoryless random distribution. It is a discrete analog of the exponential distribution. continuous

Some relations  

∑n  k = 1-n(n + 1)
k=1     2

Geometric sum

∑n      1 − rn+1
   rk = ---------
k=0       1 − r

if − 1 < r < 1  , then

 ∞
∑    k   --1--
    r =  1 − r
 k=0

if the sum is from 1 then

∑n              n+1
   rk =  r(1-−-r---)-
k=1         1 − r

if − 1 < r < 1  , then

∑∞         r
    rk = -----
 k=1      1 − r

        ∫ ∞
Γ (x) =     ux− 1e−udu
         0

Γ (n) = (n − 1)!

d          1
---ln(x) = --
dx         x

∫
   ln(y)dy =  − y + yln(y)

∫  1
   -dy = ln (y)
   y

(            )
       n       =  ----n!-----
  n1  n2  n3      n1! n2! n3!

If given joint density f   (x,y)
 XY  and asked to find conditional P (X |Y ) = fXY(x,y)
            fY(y)   so need to find marginals. Marginal is found from         ∫
fY (y) = x fXY (x,y) dx  , and          ∫
fX (x ) = y fXY (x,y) dy

To convert from x,y  to polar, example: given              -------------
f (x,y) = c∘ 1 − (x2 + y2)  find c  , where x2 + y2 ≤ 1  , then write

 ∫  𝜃= π∫  r=1√ ------
c              1 − r2rdrd𝜃
   𝜃= −π  r=0

Use identity above.

law of total probablity: if we know Y |X  and X  and want to know distribution of Y  , then          ∫
f (Y ) =  ∞  fY|X (y|x )fX (x)dx
          −∞

PIC

Z =  ¯Xn-−√--μ-→  N (0,1)
     σ ∕  n

     X¯n-−--μ-
T =  Sn∕ √n--→  T(n )

where S
  n  is std  of the sample.

Note Var(sample) has chi square (n) distribution.

CI for T:

pict