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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) 1 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. 4