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• Probability distribution
1. Use probability distribution table to describe probability distribution of discrete
random number
– probability f (x) must be non-negative.
X
– sum of probability
f (x) = 1
2. Sample mean (Expected number), sample variance, and sample standard deviation
X
– Sample mean (Expected number): µ = E(X) =
xf (x)
X
– sample variance: σ 2 = VAR(X) =
(x − µ)2 f (x)
p
– sample std deviation: σ = s.d.(X) = VAR(X)
3. Properties of E(·) and VAR(·)
Given a, b constant
– E(aX + b) = aE(X) + b = aµ + b
– VAR(aX + b) = a2 VAR(X) = a2 σ 2
– s.d.(aX + b) = |a|s.d.(X) = |a|σ
• Bernoulli Trials and Binomial Distribution.
1. Bernoulli trials are independent and each trail has only two outcomes.
2. The probability distribution used to describe n bernoulli trials is binomial.
3. suppose P (S) = p, and X denote the number of successes in n bernoulli trials
then
X ∼ Binomial(n, p)
n
P (X = x) =
px (1 − p)n−x
x
4. Binomial table. If X ∼ Binomial(N, p) then we can use the table to find
P (X ≤ c)
To use the table we need to rewrite the probability in terms of P (X ≤ c)
(a)
(b)
(c)
(d)
P (X = 4) = P (X ≤ 4) − P (X ≤ 3)
P (X ≥ 4) = 1 − P (X < 4) = 1 − P (X ≤ 3)
P (X > 4) = 1 − P (X ≤ 4)
P (3 < X ≤ 5) = P (X ≤ 5) − P (X ≤ 3)
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5. Application of Binomial distribution. Suppose there are 23 boys and 21 girls in
a class. Randomly select 5 students, let X denote the number of boy selected
among these 5 students, then X ∼ Binomial(5, 23/44).
6. Normal Approximation.
If X ∼ Binomial(n, p), and n > 30, and 0.4 < p < 0.6,
p
then X ∼ N (np, np(1 − p)) approximately.
(a) To calculatepP (X = a), we calculate P (a − 0.5 < X < a + 0.5) under
X ∼ N (np, np(1 − p)).
p
(b) To calculate P (X ≤ a), we calculate P (X ≤ a+0.5 under X ∼ N (np, np(1 − p)).
p
(c) To calculate P (X < a), we calculate P (X ≤ a−0.5) under X ∼ N (np, np(1 − p)).
(d) P (a ≤ X ≤ b) = P (X ≤ b) − P (Xp< a), we calculate P (X ≤ b + 0.5) −
P (X ≤ a − 0.5) under X ∼ N (np, np(1 − p)).
• Normal distribution
1. If X ∼ N (µ, σ) then E(X) = µ, and s.d.(X) = σ
2. Standard normal distribution Z ∼ N (0, 1)
If X ∼ N (µ, σ), then
X −µ
∼ N (0, 1)
σ
3. Table Standard Normal distribution. If Z ∼ N (0, 1), then the table gives value
of
P (Z ≤ z0 )
To use the table, one need rewrite any probability in terms of P (Z ≤ z0 )
(a) P (Z = z0 ) = 0
(b) P (Z ≥ z0 ) = 1 − P (Z ≤ z0 )
(c) P (a ≤ Z ≤ b) = P (Z ≤ b) − P (z ≤ a)
On the other hand, given the probability P (Z ≤ z0 ) = a, we can use the table
to find out z0 .
4. For non-standard normal distribution X ∼ N (µ, σ), we should first standardize
X using
X −µ
∼ N (0, 1)
Z=
σ
X −µ
a−µ
a−µ
(a) P (X ≤ a) = P (
≤
) = P (Z ≤
), Z ∼ N (0, 1)
σ
σ
σ
a−µ
(b) P (X ≥ a) = 1 − P (X ≤ a) = 1 − P (Z ≤
)
σ
b−µ
a−µ
(c) P (a ≤ X ≤ b) = P (X ≤ b) − P (x ≤ a) = P (Z ≤
) − P (Z ≤
)
σ
σ
2
• Distribution of Sample mean X̄ =
X1 + X2 + ... + Xn
n
σ2
1. If E(X) = µ, s.d.(X) = σ, then E(X̄) = µ and VAR(X̄) = , and s.d.(X̄) =
n
σ
√
n
σ
2. If X ∼ N (µ, σ), then X̄ ∼ N (µ, √ )
n
3. Central Limit Theorem If X has mean µ and standard deviation σ, not
normal distributed, but n is large enough (> 30), then
σ
X̄ ∼ N (µ, √ )
n
• Statistical Inference Suppose given sample mean X̄ and sample standard deviation
s.
1. Using X̄ to estimate population mean µ, then the 1 − α percent error margin is
σ
error margin = zα/2 √
n
Which means there is 1 − α chance that the error term |X̄ − µ| is less than
s
zα/2 √ , i.e.,
n
σ
P (|X̄ − µ| < zα/2 √ ) = 1 − α
n
2. To make the error margin less than d, one should have a sample of size
z σ 2
α/2
n≥
d
3. The 1 − α percent confidence interval
σ
σ
(X̄ − zα/2 √ , X̄ + zα/2 √ )
n
n
There is 1 − α chance that the true value of µ falls into the confidence interval
4. If σ is not given, it should be replaced by the sample standard deviation
rP
(xi − x̄)2
s=
n−1
5. Test Hypothesis at α significance level. Let
Z=
3
X̄ − µ0
√
σ/ n
(a) H0 : µ = µ0 vs. H1 : µ 6= µ0 .
RR : {|Z| ≥ zα/2 }
p-value = 2P (Z > |Zobs |)
(b) H0 : µ = µ0 vs. H1 : µ > µ0 .
RR : {Z ≥ zα }
p-value = P (Z > Zobs )
(c) H0 : µ = µ0 vs. H1 : µ < µ0 .
RR : {Z ≤ −zα }
p-value = P (Z < Zobs )
6. p-value: the smaller the p-value is, the more significant the H1 is.
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