JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 1

Continuous Probability Distributions

• Many continuous probability distributions, including: UniformNormalGammaExponentialChi-SquaredLognormalWeibull

JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 2

Normal Distribution

• The “bell-shaped curve”• Also called the Gaussian distribution• The most widely used distribution in statistical

analysis forms the basis for most of the parametric tests we’ll

perform later in this course.describes or approximates most phenomena in

nature, industry, or research

• Random variables (X) following this distribution are called normal random variables. the parameters of the normal distribution are μ and σ

(sometimes μ and σ2.)

JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 3

Normal DistributionThe density function of the normal random variable X, with mean μ and variance

σ2, has the following properties: peak is both the mean and the mode and occurs at x = μ curve is symmetrical about a vertical axis through the mean total area under the curve and above the horizontal axis = 1. points of inflection are at x = μ + σ

Normal Distribution

00.05

0.10.150.2

0.25

0.30.35

0 2 4 6 8 10 12

x

P(x

)

(μ = 5, σ = 1.5)

JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 4

Standard Normal Random Variable

• Note: the probability of X taking on any value between x1 and x2 is given by:

• To ease calculations, we define a normal random variable

where Z is normally distributed with μ = 0 and σ2 = 1

dxedxxnxXxPx

x

xx

x

2

1

2

22

1

2

)(

212

1),;()(

X

Z

JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 5

Standard Normal Distribution

Standard Normal Distribution

-5 -4 -3 -2 -1 0 1 2 3 4 5

Z

• Table A.3: “Areas Under the Normal Curve”

JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 6

Applications of the Normal Distribution• A certain machine makes electrical resistors having a mean

resistance of 40 ohms and a standard deviation of 2 ohms. What percentage of the resistors will have a resistance less than 44 ohms?

• Solution: X is normally distributed with μ = 40 and σ = 2 and x = 44

P(X<44) = P(Z< +2.0) = 0.9772

Therefore, we conclude that 97.72% will have a resistance less than 44 ohms. What percentage will have a resistance greater than 44 ohms?

-5 0 5

2

4044

Z

XZ

JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 7

The Normal Distribution “In Reverse”• Example:

Given a normal distribution with = 40 and σ = 6, find the value of X for which 45% of the area under the normal curve is to the left of X.

• Solution: If P(Z < k) = 0.45, what is the value of k?

45% of area under curve less than kk = -0.125 from table in back of book

Z = (x- μ) / σ

Z = -0.125 = (x-40) / 6

X = 39.25

-5 0 5

-5 0 5

39.25 40

X values

Z values

JMB Ch6 Lecture2 Review EGR 252 Spring 2011 Slide 8

Your Turn• P(Z ≤ 1) =

Area to the left of the blue line

• P(Z ≥ -1) =Area to the right of the blue line

• P(-2.5 ≤ Z ≤ 2.8) = Area between the two lines

• For a normally distributed variable for which the mean is 30 and standard deviation is 10, P(X < 40) =

Area to the left of the blue line

• P(Z ≥ -1) =

-5 0 5

-5 0 5

-5 0 5

-5 0 5

30 40 x values for µ = 30