ST 361 Normal Distribution - Department of Statistics
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
➢ Shape:
[pic]
➢ Symmetric, centering at [pic] also the median and mean.
➢ Can be fully specified via two parameters: [pic] and [pic]. The distribution is denoted by [pic], It can be shown [pic] is the variance of x ([pic]is the standard deviation of x)
Ex.
[pic]
2. Density function (for your reference):
[pic]
• 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?
[pic]
That is, we want to find [pic] such that [pic]
(2) What are the most extreme 5% of the scores?
[pic]
That is, we want to find [pic] such that [pic]
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 [pic] 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 [pic]1.25) =0.8944
2. P(Z [pic]-1.25) =0.1056 ( = 1 - 0.8944)
3. P(Z [pic]-1.25) = 1 - P(Z [pic]-1.25) =1 - 0.1056 = 0.8944
4. P(-.38 [pic] Z [pic].25) =P[Z[pic]0.25] – P[Z[pic]-0.38] = 0.5987 – 0.3520 = 0.2467
In general, P[a [pic]Z[pic]b] = P[Z[pic]b] – P[Z[pic]a].
5. P(Z [pic]-6) < P[Z [pic] -3.89] = 0.0000
6. P(Z [pic]2) = 1 – P[Z < 2] = 1 – P[Z [pic] 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(Z[pic]z*) = 0.1
[pic] (the exact value is -1.281552)
2. P(Z[pic]z*) = 0.5
[pic]
3. P(Z[pic]z* or Z[pic]-z*) = 0.1
By symmetry of N(0,1), P[Z[pic]- 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 [pic]z*] = 0.025, or P[Z [pic] 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* [pic] Z [pic] z*] = 0.68 [pic] P[Z [pic] - z*] = (1 - 0.68)/2 = 0.16 [pic] - z* = - 0.995,
z = 0.995.
III. General Normal Distribution[pic][pic]
• If X has a normal distribution with mean [pic]and SD [pic], then we can standardize X to Z by
[pic] has the standard normal distribution
• Therefore,
[pic] where [pic] and [pic]
• Calculating probability and percentiles
Ex. A variable X ~ N(100, 5). Calculating the following probabilities:
1. P(90 [pic] X [pic]125) =[pic], where
[pic]. So
P(90 [pic] X [pic]125) = P(-2 [pic] Z [pic] 5) = P(Z [pic] 5) – P(Z [pic] -2) = 1 – 0.0228 = 0.9772.
2. P( X[pic]98 ) = P[Z [pic] [pic]] = 1 – P[Z [pic] -0.4] = 1- 0.3446 = 0.6554
3. Find the x* such that P( X [pic]x* )=0.1
[pic]. But P[Z [pic] -1.28] = 0.1, so
[pic], 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
[pic]
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)= [pic]
2. P(575 [pic]X [pic] 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 x*] = 0.05 or equivalently P[X < x*] = 0.95. Since P[Z ................
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