# NIPRL Chapter 5. Normal Distribution 5.1 Probability Calculation Using the Normal Distribution 5.2 Linear Combinations of Normal Random Variables 5.3 Approximating.

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NIPRL Chapter 5. Normal Distribution 5.1 Probability Calculation Using the Normal Distribution 5.2 Linear Combinations of Normal Random Variables 5.3 Approximating Distributions with the Normal Distribution 5.4 Distributions Related to the Normal Distribution

NIPRL 5.1 Probability Calculation Using the Normal Distribution 5.1.1 Definition of the Normal Distribution 5105 N(5,4) N(5,0.25) N(10,4) N(5,4)

NIPRL 5.1.2 The Standard Normal Distribution 0 0.5 1 N(1,0) x Ф(x) x

NIPRL 5.1.3 Probability Calculation for General Normal Distributions

NIPRL 5.1.4 Examples of the Normal Distributions (1/3) Example 18: Tomato Plant Heights heights of tomato plants : mean=29.4cm, standard deviation=2.1cm (1) Using Chebychev’s inequality at least 75% has a height within the interval [25.2, 33.6]. (2) Under the normal assumption, the interval of X with 1-α probability: Therefore, the interval of X with 90% coverage is [25.95, 32.85] using z α =z 0.05 =1.645 (3) Probability that a height is between 29cm and 30cm is

NIPRL 5.1.4 Examples of the Normal Distributions (2/3) Example 37: Concrete Block Weights weight of concrete block~N(11.0, 0.3 2 ) (1) Interval of X with 99% coverage: [μ-σz 0.005, μ+σz 0.005 ]=[10.23, 11.77]. (2) The probability that a concrete block weighs less than 10.5kg is

NIPRL 5.1.4 Examples of the Normal Distributions (3/3) Stock Prices annual return~N(8.0, 1.5 2 ) ‘satisfactory’ : return > 5% ‘excellent’ : return > 10% (1) probability of unsatisfactory (2) probability of excellent

NIPRL 5.2 Linear Combinations of Normal Random Variables 5.2.1 The Distribution of Linear Combinations of Normal Random Variables (1/2) Linear Functions of a Normal Random Variable

NIPRL 5.2.1 The Distribution of Linear Combinations of Normal Random Variables (2/2) Properties of independent Normal Random Variables

NIPRL 5.2.2 Examples of Linear Combinations of Normal Random Variables (1/6) Example 23: Piston Head Construction X 1 ~N(30.00, 0.05 2 ),: radius of piston X 2 ~N(30.25, 0.06 2 ) : radius of cylinder (1) distribution of Y=X 2 -X 1 Y~N(30.25-30.00, 0.05 2 + 0.06 2 )=N(0.25, 0.0061) (2) probability that a piston head will not fit within a cylinder (3) probability that Y is between 0.10mm and 0.35mm

NIPRL 5.2.2 Examples of Linear Combinations of Normal Random Variables (2/6) Example 18: Tomato Plant Heights (1) distribution of average height (2) interval of X with 95% coverage (z 0.025 =1.96)

NIPRL 5.2.2 Examples of Linear Combinations of Normal Random Variables (3/6) Example 37: Concrete Block Weights

NIPRL 5.2.2 Examples of Linear Combinations of Normal Random Variables (4/6) Example 35: Stock Prices X A ~N(8.0, 1.5 2 )=N(8.0, 2.25) : annual return of company A X B ~N(9.5, 4.00) : annual return of company B (1) probability that B’s stock proves to be unsatisfactory (2) probability that B’s stock proves to be excellent

NIPRL 5.2.2 Examples of Linear Combinations of Normal Random Variables (5/6) What is the probability that Company B’s stock performs better than Company A’s stock? If Y=X B -X A, Y~N(9.5-8.0, 4.00+2.25)=N(1.5, 6.25) (1) The required probability is (2) probability that B’s stock performs at least two percentages better than A’s stock is

NIPRL 5.2.2 Examples of Linear Combinations of Normal Random Variables (6/6) Example 38: Chemical Concentration Level C is measured in two methods. X A ~N(C, 2.97) : Method A X B ~N(C, 1.62) : Method B 99.7% coverage intervals: [C-5.17, C+5.17] : Method A [C-3.82, C+3.82] : Method B Combine the two method to Y so that it can minimize the variability. After that, interval of Y with 99.7% coverage is [C-3.07, C+3.07]

NIPRL 5.3 Approximating Distributions with the Normal Distribution 5.3.1 The Normal Approximation to the Binomial Distribution 118 X~B(16,0.5) Y~N(8,4) 11.57.5

NIPRL 5.3.2 The Central Limit Theorem

NIPRL 5.3.3 Examples of Employing Normal Approximations (1/6) Example 17: Milk Container Contents X~B(20, 0.261) : the number of underweight container Y~N(20x0.261, 20x0.261x(1-0.261))=N(5.22, 3.86) : approximation (1) probability that a box contains no more than 3 underweight containers Now suppose X~B(500, 0.261). Then Y~N(500x0.261, 500x0.261x(1-0.261))=N(130.5, 96.44) (2) probability at least 150 out of 500 are underweight

NIPRL 5.3.3 Examples of Employing Normal Approximations (2/6) Example 39: Cattle Inoculations probability of provoking a serious adverse reaction : 0.0005 X : the number of animals that will suffer an adverse reaction X~B(500000, 0.0005)

NIPRL 5.3.3 Examples of Employing Normal Approximations (3/6) Example 27: Glass Sheet Flaws X : the number of flaws in a glass sheet X~Poisson(λ=0.5) (1) distribution of the total number of X in 100 sheets of glass (2) probability that there are less than 40 flaws in 100 sheets

NIPRL 5.3.3 Examples of Employing Normal Approximations (4/6) (3) distribution of average of X and probability that this average is between 0.45 and 0.55

NIPRL 5.3.3 Examples of Employing Normal Approximations (5/6) Example 30: Pearl Oyster Farming The probability that an oyster produces a pearl with a diameter of at least 4mm is 0.6 How many oysters does an oyster farmer need to farm in order to be 99% confident of having at least 1000 pearls? X : the number of pearls X~B(n, 0.6) ⇒ Y~N(0.6n, 0.24n)

NIPRL 5.3.3 Examples of Employing Normal Approximations (6/6) In conclusion, the farmer should farm about 1750 oysters in order to be 99% confident of having at least 1000 peals. The expected number of pearls and its standard deviation are Recall the diameter of pearl has mean 5.0 and variance 8.33. If 1750 pearls are obtained, the average diameter has mean 5.0 and variance 0.00476(=8.33/1750). Interval of the average diameter with 99.7% coverage is

NIPRL 5.4 Distributions Related to the Normal Distribution 5.4.1 The Lognormal Distribution μ=1,σ=0.5 μ=σ=1 μ=2,σ=1

NIPRL 5.4.2 The Chi-Square Distribution ν=5ν=5 χν2χν2 ν=10 ν=15 α χ2α,νχ2α,ν

NIPRL Chi-Square Distribution: Let Probability density function: Mean and variance:

NIPRL Example: Suppose that the coordinate errors are independent normal random variables with mean 0 and standard deviation 2. Find the probability that the distance between the point chosen and the target exceeds 3. (Sol) Let where Then, Therefore,

NIPRL 5.4.3 The t-Distribution 0 t v distribution N(0,1) 0 α tα,νtα,ν

NIPRL The t-density is symmetric about zero. If becomes larger, it becomes more and more like a standard normal density since The mean and variance: A constant such that is defined. Then, from the symmetry about zero, That is,

NIPRL 5.4.4 The F-Distribution ν 1 =v 2 =25 ν 1 =5, v 2 =25 ν 1 =v 2 =5 F α,ν1,ν2 α F ν1,ν2 distribution

NIPRL Let A constant such that Is defined. Then, That is, By definition,

NIPRL 5.4.5 The Multivariate Normal Distribution Bivariate Normal Distribution for special case: How does the shape of function change as varies ? Marginal distributions of X and Y ? General Multivariate Normal Distribution

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