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X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

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Page 1: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 2: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

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X = 2*Bin(300,1/2) – 300E[X] = 0

Page 3: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

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Y = 2*Bin(30,1/2) – 30E[Y] = 0

Page 4: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

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Z = 4*Bin(10,1/4) – 10E[Z] = 0

Page 5: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

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W = 0E[W] = 0

Page 6: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

• Is there a good parameter that allow to distinguish between these distributions?

• Is there a way to measure the spread?

Page 7: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

• The variance of X, denoted by Var(X) is the mean squared deviation of X from its expected value = E(X):

Var(X) = E[(X-)2]. The standard deviation of X, denoted

by SD(X) is the square root of the variance of X.

SD(X) = Var(X).

Page 8: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

2 2Var(X) = E(X ) E(X) .Proof:

E[ (X-)2] = E[X2 – 2 X + 2]

E[ (X-)2] = E[X2] – 2 E[X] + 2

E[ (X-)2] = E[X2] – 22+ 2

E[ (X-)2] = E[X2] – E[X]2

Claim: Var(X) = E[(X-)2].

Page 9: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

1. Claim: Var(X) >= 0.

Pf: Var(X) = (x-)2 P(X=x) >= 0

2. Claim: Var(X) = 0 iff P[X=] = 1.

Page 10: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 11: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 12: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

Theorem: X is random variable on sample space S, and P(X=r) it’sprobability distribution. Then for any positive real number r:

(proof in book)

2

( )(| ( ) | )

V XP X E X r

r

In words: the probability of finding a value of X farther away from the meanthan r is smaller than the variance divided by r^2.

r

Page 13: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

What is the probability that with 100 Bernoulli trials we find more than89 or less than 11 successes when the prob. of success is ½.

X counts number of successes.EX=100 x ½ =50

V(X) = 100 x ½ x ½ = 25.

P(|X-50|>=40)<=25/40^2 = 1/64

Page 14: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 15: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

“k standard deviations”

Page 16: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 17: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 18: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 19: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 20: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

Let be random variables that for

Page 21: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0
Page 22: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

, ,

Page 23: X = 2*Bin(300,1/2) – 300 E[X] = 0 Y = 2*Bin(30,1/2) – 30 E[Y] = 0

What is the probability of getting 25 or fewer heads in 100 coin tosses?