ST 361 Normal Distribution s1

Distribution of Continuous R.V.: Normal Distribution (Ch 1.4)

Topics:

§1.4 What is Normal Distribution, and its density function, mean, variance

Standard Normal Distribution: (a) Calculating Probability

(b) Calculating Percentile

General Normal Distribution : (a) Calculating Probability

(b) Calculating Percentiles

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I. Normal random variable/Normal Distribution

· A distribution for describing continuous random variables

· Two common ways to describe a Normal distribution

1. Density plot

Ø Shape:

Ø Symmetric, centering at also the median and mean.

Ø Can be fully specified via two parameters: and . The distribution is denoted by , It can be shown is the variance of x (is the standard deviation of x)

Ex.

2. Density function (for your reference):

· What problems are we interested in solving regarding normal distribution?

1. Know how to calculate probabilities from a given normal distribution

Ex. Test score X ~ N (75, 5). P(90< X < 100)? P(X < 60) ?

2. Be able to identify the percentiles of the population

Ex. Test score X ~ N (75, 5). (1) What are largest 10% of the scores?

That is, we want to find such that

(2) What are the most extreme 5% of the scores?

That is, we want to find such that

II. Standard Normal Distribution

· Normal distribution with mean=0 and SD=1. Denoted by N(0, 1).

· Usually use Z to denote a standard normal r.v.

· Why learn the standard normal distribution?

o Area under the normal curve can only be calculated numerically.

So statisticians have established a table that shows the left tail area under the standard normal curve of any given number (see the very first page of the textbook).

o Later we can use such table to solve for all normal distribution.

How? One can STANDARDIZE any given to N(0, 1), and then use the area table of standard normal to solve the problem (Your HW2, Question #2, 2.61)

· Use the area table of standard normal curve

(1) Calculate probability

Ex. A variable Z ~ N(0, 1). Calculating the following probabilities:

1. P(Z 1.25) =0.8944

2. P(Z -1.25) =0.1056 ( = 1 - 0.8944)

3. P(Z -1.25) = 1 - P(Z -1.25) =1 - 0.1056 = 0.8944

4. P(-.38 Z .25) =P[Z0.25] – P[Z-0.38] = 0.5987 – 0.3520 = 0.2467

In general, P[a Zb] = P[Zb] – P[Za].

5. P(Z -6) < P[Z -3.89] = 0.0000

6. P(Z 2) = 1 – P[Z < 2] = 1 – P[Z 2] = 1 – 0.9772 = 0.0228

(2) Obtain extreme values

Ex1. A variable Z ~ N(0, 1). Find the following z* that fulfills the probability:

1. P(Zz*) = 0.1

(the exact value is -1.281552)

2. P(Zz*) = 0.5

3. P(Zz* or Z-z*) = 0.1

By symmetry of N(0,1), P[Z- z*] = 0.05, -z* = -1.645, z* = 1.645.

Ex2. Consider a standard Normal r.v. Z~N(0,1). At what value of z*, the area to the right is 2.5%?

Want to find z* such that P[Z z*] = 0.025, or P[Z z*] = 0.975. The value of z* = 1.96

Ex3. Consider a standard Normal r.v . Z~N(0,1). At what value of z*, the area between –z* and z* is 68%?

P[-z* Z z*] = 0.68 P[Z - z*] = (1 - 0.68)/2 = 0.16 - z* = - 0.995,

z = 0.995.

III. General Normal Distribution

· If X has a normal distribution with mean and SD , then we can standardize X to Z by

has the standard normal distribution

· Therefore,

where and

· Calculating probability and percentiles

Ex. A variable X ~ N(100, 5). Calculating the following probabilities:

1. P(90 X 125) =, where

. So

P(90 X 125) = P(-2 Z 5) = P(Z 5) – P(Z -2) = 1 – 0.0228 = 0.9772.

2. P( X98 ) = P[Z ] = 1 – P[Z -0.4] = 1- 0.3446 = 0.6554

3. Find the x* such that P( X x* )=0.1

. But P[Z -1.28] = 0.1, so

, which gives x* = 100 + 5*(-1.28) = 93.6

4. Find the range that contains the MIDDLE 90% of the observations: Want to find a such that x is in [100 – a, 100 + a] with 90% probability

So, - a/5 = - 1.645, a = 5*1.645 = 8.225. The range is [100 – 8.225, 100 + 8.225] = [91.775, 108.225]

Ex. X is the diameter (in mm) of tires, normally distributed with mean 575 and SD 5.

1. P(575 < X < 579)=

2. P(575 X 579)=0.2881

3. Find the diameter x* such that there are only 1% tires longer than this diameter

That is, P[X > x*] = 0.01 or equivalently P[X < x*] = 0.99. Since P[Z<2.33] = 0.99, so x* = 575 + 5 * 2.33 = 586.65.

4. Find the (diameters of) tires that have most extreme 5% diameters.

That is, P[X > x*] = 0.05 or equivalently P[X < x*] = 0.95. Since P[Z<1.645] = 0.95, so x* = 575 + 5 * 1.645 = 583.225.

Putting everything together…. An overall example:

The diameter of a tire follows normally distribution with mean 575 and SD 5. We have 4 tires, and the diameters of these tires are independent of each other.

(a) What is the probability that a tire has its diameter between 570 and 580?

Let Xi be the diameter for tire i. Then P[570 < Xi < 580] = P[-1 < Z < 1] = 0.8413 – 0.1587 = 0.6826.

(b) What is the probability that all 4 tires have diameters between 570 and 580?

Let Ai = [570 < Xi < 580]. Then A1, A2, A3, A4 are independent. So

This probability = 1 – P[all tires are between 570 and 580]

= 1 - = 0.7829