NORMAL APPROXIMATION. In the last chapter we discovered that, when sampling from almost any distribution, e r2 2 rdrdϕ = 2π e u du =2π.

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1 NOMAL APPOXIMATION Standardized Normal Distribution Standardized implies that its mean is eual to and the standard deviation is eual to. We will always use Z as a name of this V, N (, ) will be our symbolic notation for the corresponding distribution. X µ σ/ n In the last chapter we discovered that, when sampling from almost any distribution, t has a sampling distribution whose MGF is e. We will show that this corresponds to: f(z) =c e z,where <z<. cis a constant whose value is a reciprocal of I e z dz.it is actually easier (an understatement) to compute: I = e x dx e y dy = and c = e x +y x y plane π : (a symmetric bell-shaped curve): dx dy = π e r rdrdϕ = π e u du =π. Thus I = π f(z) = π e z <z< To verify that this is the correct answer: M(t) = π e z e zt dz = π e z +zt dz =

2 π e t + t + ³ t e (z t) dz = π e t e u du = e t (check). From the expansion M(t) = +... we can immediately establish that µ =(all odd moments eual to ), σ =, and the kurtosis of Z is eual to 3. There is no analytic expression for F (x), but Maple has no difficulty evaluating any probability we need numerically, e.g. Pr(.3 <Z<.5) = π.5.3 z )dz =.5947 General Normal distribution We define Z as the following linear transformation of Z X = σz + µ where σ > and µ are two constants. From what we know about linear transformations, E(X) =µ, Var(X) =σ,andm x (t) =e µt M z (σt) =e σ t +µt. The shape of the PDF remains the same, only the scale changes. EXAMPLE: If M(t) =e t+t, what is the distribution? Answer: N (, ). Note that any further linear transformation of X N (µ, σ), such as Y = ax + b, keeps the result Normal. Also: when X and X are independent Normal Vs with any (mismatched) parameters, i.e. X N (µ,σ ) and X N (µ,σ ), then their sum X + X is also Normal, which follows from M X +X (t) =e (µ +µ )t+ σ +σ t. To find f x (x), we do this: F x (x) =Pr(X<x)=Pr(σZ + µ<x)=pr(z< x µ )= σ F x µ z σ. Differentiating with respect to x yields: fx (x) = f x µ σ z σ = ³ σ exp (x µ) π σ <x< EXAMPLE: Knowing that f(x) = 3 x +4x +4), identify the distribution. Answer: N (, π 8 3).

3 Computing probabilities is easy. EXAMPLE: If X N (7, 3), Pr( <X<) = ³ 3 exp (x 7) dx =.835 π 3 For a Normally distributed V, the µ±σ interval contains 8.% of the total probability, µ ± σ contains 95.44%, andµ ± 3σ raises this to a near certain 99.74% (for any practical purpose, the range is finite ). Applications of Central Limit Theorem: Finally, we can apply our knowledge that the distribution of X µ σ/ n is, approximately, N (, ), thebiggern, the better the approximation. This can be re-stated as: X N (µ, σ n ) or, euivalently: X + X X n N (nµ, nσ). EXAMPLES: oll a die times, what is the probability of getting more than sixes? B X has the binomial distribution with n =and p =. Since n is large, its distribution will be uite close to N ( 5 3, 5 9 ). Thus Pr( <X Binomial) Pr(.5 <X Normal )= 5/9 π.5 (x 5/3) )dx =5.8% (the exact answer is 5.9% -inthiscase, 5/9 one would expect to be within.5% of the exact answer). Note the continuity correction, 3

4 clear from: If X has the Poisson distribution with λ = 35.4, approximate Pr(X 3). B Since X N (35.4, 35.4), Pr(X Normal < 3.5) =.9% (the exact answer is.%) π (x 35.4) 35.4 )dx = Consider rolling a die repeatedly until obtaining sixes. What is the probability that this will happen in fewer than 7 rolls? B The exact distribution of X is Negative Binomial, with p = and k =. Pr(X Normal < 99.5) = 9.54% (the exact answer is 9.%) (x ) 3 π 3 )dx = If 5 cards are deal from a standard deck of 5, repeatedly and independently times, what is the probability of dealing at least 5 aces in total? B We need Pr(X + X X 5), where the X i s are independent, hypergeometric, with N = 5, K = 4 and n = 5 (µ = 5 and σ = 5 47 each). For their sum, µ sum = 5 5 and σ 3 sum = 47 = 94. The answer is, approximately, (s 5/3) )ds =.8% (the exact answer is 3.%). 94/873 π /873 4

5 Consider a random independent sample of size form the uniform distribution ³ U(, ). Find Pr(.49 X.5). B We know that X N.5,, which implies that the above probability is, approximately,.5 π/4.49 ( x.5) /4 )d x = 37.58%.(the exact answer is 37.5% - the approximation is now a lot more accurate for two reasons: the uniform distribution is continuous and symmetric). Consider a random independent sample of size from X = Prob: 3 What is the probability that the sample total will be negative (losing money, if this represents a game)? B First we compute the distribution s µ = and Var(X) = 3+4 = 4 P, then we introduce S = X 3 3 i and find µ s = 4 and σ s =. 3 Answer: Pr(S <).5 )ds =93.5%. (the exact value 4/3 π is 93.%). i= (s+/) 4/3 Pay $ to play the following game: 5 cards are dealt from a standard deck, and you receive $ for each ace and $5 for each king, ueen and jack. First, find the expected value and standard deviation of your net win. B W =X +5Y (where X is the number of aces dealt, Y correspondingly counts the total of kings, ueens and jacks). Thisimpliesthatµ w = = 5 3 dollars, Var(W )= 5( ) 47 5 = and σ w = = $.773. Secondly, compute the (approximate) probability of losing more that $5 after 7 rounds of this game. B Defining S 7 P W i, we get µ s = and σ s =.773 i= 7 = Thus, Pr(S < 5) 5.5 Pr(SNormal < 5.5) = (t+5.385) )dt = π % (the exact answer is 5.% - not bad considering that the individual proba- 5

6 bilities are still well over %). Note the unusual continuity correction!