Understanding Continuous Random Variables and Probability Distributions

1. Chapter 6 Continuous Random Variables

2. Continuous Random Variables 6.1 Continuous Probability Distributions 6.2 The Uniform Distribution (Optional) 6.3 The Normal Probability Distribution 6.4 Approximating the Binomial Distribution by Using the Normal Distribution (Optional) 6.5 The Exponential Distribution (Optional) 6.6 The Normal Probability Plot (Optional)

3. Two Types of Random Variables • Random variable: a variable that assumes numerical values that are determined by the outcome of an experiment • Discrete • Continuous • Discrete random variable: Possible values can be counted or listed; doesn’t take values on an interval of the real line. • The number of defective units in a batch of 20 • Toss a coin. Let x=1 if we have a head, x=0 if we have a tail

4. Random Variables Continued • Continuous random variable: May assume any numerical value in one or more intervals • The waiting time for a credit card authorization • The interest rate charged on a business loan

5. Continuous Probability Distributions • A continuous random variable may assume any numerical value in one or more intervals • To compute probabilities about the range of values that a continuous random variable x might attain, we use continuous probability distribution (or called probability density function), denoted by f(x).

6. Properties of Continuous Probability Distributions • Properties of probability density function f(x): f(x) is a continuous function such that • f(x)  0 for all x • The total area under the curve of f(x) is equal to 1 • The probability of X to be between a and b, P(a<X<b), is the area under the curve between a and b.

7. Properties of Density Function P(a ≤ x ≤ b) is the area colored in blue under the curve and between the values x = a and x = b Note: P(a ≤ x ≤ b) =P(a<x<b)

8. Area and Probability • The blue area under the curve f(x) from x = a to x = b is the probability that x could take any value in the range a to b • Symbolized as P(a  X  b) • Note for continuous variable x , P(X=a)=0; P(a  X  b)=P(a<X<b), Or as P(a < X < b), because each of the interval endpoints has a probability of 0 • P(X<b) is the total area under curve to the left of b; and P(X>b) is the total area under curve to the right of b.

9. The Normal Probability Distribution • A random variable with normal distribution is uniquely determined by its mean m and standard deviation s. Or equivalently a normal density function is uniquely determined by its mean m and standard deviation s. • The normal curve is bell-shaped symmetrical about its mean m • So m is also the median • The curve of f(x) is tallest over its m. • The area under the entire normal curve is 1 • What is the total area under the normal curve to the right of m? What does it mean?

10. The Normal Probability Distribution Continued

11. Gauss

12. Means and Standard Deviations • A normal distribution can have any mean and any positive standard deviation. • The mean gives the location of the line of symmetry. • The standard deviation describes the spread of the data. μ = 3.5 σ = 1.5 μ = 3.5 σ = 0.7 μ = 1.5 σ = 0.7

13. Example: Understanding Mean and Standard Deviation • Which curve has a greater mean? Solution: Curve A has the greater mean (The line of symmetry of curve A occurs at x = 15. The line of symmetry of curve B occurs at x = 12.)

14. How to decide the Position and Shape of the Normal Curve • The mean m positions the peak of the normal curve over the real axis • The variance s2 measures the width or spread of the normal curve

15. Properties of Normal Distributions • The mean, median, and mode are equal, µ • The normal curve is bell-shaped and symmetric about the mean µ. • The total area under the curve is equal to one. • The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean. Total area = 1 x μ

16. Properties of the Normal Distribution • There are an infinite number of normal curves • The shape of any individual normal curve depends on its specific mean m and standard deviation s • The highest point is over the mean • Any normal distribution can always be converted to a standard normal distribution. All problems concerning a normal random variable can always be solved by a standard normal random variable.

17. The Standard Normal Distribution z 0 1 2 3 3 2 1 Standard normal distribution • A normal distribution with a mean of 0 and a standard deviation of 1. Area = 1

18. The Standard Normal Table • The standard normal table is a table that lists the area under the standard normal curve to the left of the z (z score) value of interest, that is P(Z<z) • See Table 6.1, Table A.3 in Appendix A, and the table on the back of the front cover • This table is so important that it is repeated 3 times in the textbook! • Always look at the accompanying figure for guidance on how to use the table

19. Example: Using The Standard Normal Table x μ =0 1.15 Find the area to the left of 1.15. Solution: Find 1.1 in the left hand column. P(Z<1.15)=? The area to the left of z = 1.15 is 0.8749. This means the probability that for a standard normal random variable Z the probability P(Z 1.15)=P(Z<1.15)=0.8749 Question: What is P(Z>1.15)?1-0.8749

20. The Standard Normal Table Continued • In this course to find out P(Z<z), always round z to the nearest one hundredth. • Then find the one hundredth digit in the first row of the table, and then find the remaining digits in the first column. • At the intersection of the column and row you find in the last step, you find out P(Z<z) or P(Z≤z), which is equivalent to the areas under the normal curve from the leftmost part up to any value of z are given in the body of the table.

21. Example: Using The Standard Normal Table Find the cumulative area that corresponds to a z-score of -0.24. P(Z<-0.24)=0.4052 P(Z>-0.24)=1-0.4052?

22. Z • P( a<Z <b) =P(Z<b)-P(Z<a) • P(-1<Z<2)=P(Z<2)-P(Z<-1) =0.9772-0.1577 • P(Z<2)=0.9772 • P(Z<-1)=0.1577

23. The Non-Standard Normal Distribution • If X is normally distributed with mean  and standard deviation , then the random variable zis normally distributed with mean 0 and standard deviation 1 • Then P(X  x)=P(Z (x-µ)/σ)

24. Probability and Normal Distributions μ = 500 σ = 100 P(x < 600) = Area x μ =500 600 • Find out P(X<600) for a normal random variable with =500 and standard deviation =100. P(X < x)=P(Z (x-µ)/σ) ; P(X < 600) = P(Z < 1)

25. Probability and Normal Distributions X has Normal Distribution μ = 500 σ = 100 P(x < 600) x μ =500 600 P(X < x)=P(Z (x-µ)/σ) ; P(X < 600) = P(Z < 1)=0.8413

26. Solution: Finding Probabilities for Normal Distributions P(24 < x < 54) x 24 45 Normal Distributionμ = 45 σ = 12 54 P(x1  X  x2)=P(x2-µ)/σ Z (x1-µ)/σ) ; P(24 < X < 54) = P(-1.75 < Z < 0.75) = 0.7734 – 0.0401 = 0.7333

27. The Properties of z-score • Z-score, (x-µ)/σ,measures the number of standard deviations that x is from the mean m • The algebraic sign on z indicates on which side of m is x • z is positive if x > m (x is to the right of m on the number line) • z is negative if x < m (x is to the left of m on the number line)

28. Finding a z-Score Given an area or probability (the inverse problem or percentile problem) 0.025 z z 0 Find the z score which gives you the given probability; if you can not find the exact value, find the closest; if you have a tie, take the average of the two z scores. Find the z-score that such that P(Z<z)=0.025. z=? The areas closest to 0.025 in the table is 0.025 with z = -1.96 . Then z-score is -1.96.

29. Example: Finding a z-Score Given an Area Find the z-score that corresponds to a cumulative area of 0.3632. 0.3632 z z 0

30. Solution: Finding a z-Score Given an Area • Locate 0.3632 in the body of the Standard Normal Table. The z-score is -0.35. • The values at the beginning of the corresponding row and at the top of the column give the z-score. • If you cannot find the exact given area, find the closest value in the table; in case of tie, take the average of the two z-scores.

31. Solution: Finding a z-Score Given an Area • Find z such that P(Z<z)=0.8923 • The closest number in the table to 0.8923 is 0.8925. The z-score is 1.24.

32. z 0 • Find the z-score that such that P(Z<z)=0.01. z=? 0.01 z z 0

33. Finding a z-Score Given an area or probability (the inverse problem or percentile problem) 0.05 z z 0 Find the z-score that corresponds to P(Z<z)=0.05. z=? The z-score corresponds to an area of 0.05. The areas closest to 0.05 in the table are 0.0495 (z = -1.65) and 0.0505 (z = -1.64). Because 0.05 is halfway between the two areas in the table, 0.0495 and 0.0505, and we have a tie, so the z score is the average of -1.64 and -1.65.

34. Finding Percentiles for Non-Standard Normal Random Variable with mean m sd s • Formulate the problem in terms of x and write is as P(X<x)=p • Calculate the corresponding z values such that P(Z<z)=p • x = μ + zσ Note: It is always useful to draw a picture showing the required areas before using the normal table

35. Example: Finding a Specific Data Value 5% x 75 Scores for a civil service exam are normally distributed, with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment? P(X<x)=0.95 An exam score in the top 5% is any score above the 95th percentile. Find the z-score that corresponds to a cumulative area of 0.95. 1 – 0.05 = 0.95 ?

36. Solution: Finding a Specific Data Value 5% From the Standard Normal Table, the areas closest to 0.95 are 0.9495 (z = 1.64) and 0.9505 (z = 1.65). Because 0.95 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65. That is, z = 1.645. • P(X<x)=0.95, X normal with mean 75 standard deviation 6.5 • Find z such that P(Z<z)=0.95, z=1.645 • Using the equation x = μ + zσ, x = 75 + 1.645(6.5) ≈ 85.69 z 1.645 0

37. Solution: Finding a Specific Data Value 5% x 85.69 75 Using the equation x = μ + zσ x = 75 + 1.645(6.5) ≈ 85.69 z The lowest score you can earn and still be eligible for employment is 86.