Lecture 26 Section 6.3.1 – 6.3.2 Tue, Mar 2, 2004 Normal Percentiles Lecture 26 Section 6.3.1 – 6.3.2 Tue, Mar 2, 2004

Standard Normal Percentiles Given a value of Z, we know how to find the area to the left of that value of Z. The problem of finding a percentile is exactly the reverse: Given the area to the left of a value of Z, find that value of Z? That is, given the percentage, find the percentile.

Standard Normal Percentiles What is the 90th percentile of Z? That is, find the value of Z such that the area to the left is 0.9000. Look up 0.9000 as an entry in the standard normal table. Read the corresponding value of Z. Z = 1.28.

Practice Find the 99th percentile of Z. Find the 1st percentile of Z. Find Q1 and Q3 of Z. What value of Z cuts off the top 20%? What values of Z determine the middle 30%?

Standard Normal Percentiles on the TI-83 To find a standard normal percentile on the TI-83, Press 2nd DISTR. Select invNorm. Enter the percentile as a decimal (area). Press ENTER.

Standard Normal Percentiles on the TI-83 invNorm(0.99) = 2.236. invNorm(0.01) = -2.236. invNorm(0.50) = 0. Q1 = invNorm(0.25) = -0.674. Q3 = invNorm(0.75) = 0.674. invNorm(0.80) = 0.8416. invNorm(0.35) = -0.3853. invNorm(0.65) = 0.3853.

Normal Percentiles To find a percentile of a variable X that is N(, ), Find the percentile for Z. Use the equation X =  + Z to find X.

Example Let X be N(30, 5). Find the 95th percentile of X. The 95th percentile of Z is 1.64. Therefore, X = 30 + (1.64)(5) = 38.2. 38.2 is 1.64 standard deviation above average.

Normal Percentiles on the TI-83 Find the standard normal percentile and use the equation X =  + Z. Or, use invNorm and specify  and . invNorm(0.95, 30, 5) = 38.2.

Uniform Distributions A uniform distribution is one in which all values within a specified interval are equally likely. The distribution must be over a bounded interval [a, b]. That is, it cannot be infinite in either direction. This distribution is denoted U(a, b).

Uniform Distributions The graph of U(a, b) is a horizontal straight line. a b

Uniform Distributions What is the height of the graph? ? a b

Uniform Distributions The area must be 1, so the height times the width equals 1. ? a b b - a

Uniform Distributions The height is 1/(b – a). 1/(b – a) a b b - a

Uniform Distributions Check that the area is 1. 1/(b – a) Area = 1 a b b - a

Uniform Distributions The mean of U(a, b) is  = (a + b)/2, the midpoint of the interval [a, b]. The standard deviation of U(a, b) is  = (b – a)/12, but we won’t need to know that.

Example: Waiting Times Suppose that a traffic light stays red for exactly 30 seconds before turning green. You arrive at a “random” moment. What is the distribution of waiting times? It ought to be uniform on [0, 30].

Example: Waiting Times Let X be the waiting time. Then X is U(0, 30). 1/30 30

Example: Waiting Times What proportion of the time must you wait at least 25 seconds? 1/30 30

Example: Waiting Times What proportion of the time must you wait at least 25 seconds? 1/30 25 30

Example: Waiting Times That is, you must wait from 25 to 30 seconds. 1/30 25 30

Example: Waiting Times Proportion = area = (30 – 25) x (1/30) = 5/30 = 1/6. 1/30 25 30

Assignment Page 341: Exercises 6, 8, 15, 16, 19, 21.