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Introduction to Continuous Random Variables and the Normal Distribution

Understand probability distributions for continuous random variables and the characteristics of the normal probability distribution.

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Introduction to Continuous Random Variables and the Normal Distribution

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  1. CHAPTER 6 CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION

  2. CONTINUOUS PROBABILITY DISTRIBUTION Introduction • As stated in Chapter 5, a random variable can be either a discrete or a continuous random variable. We also learned how to handle probability distribution for discrete random variable. • In this chapter, we are going to learn about probability distribution for continuous random variable. Definition: • Once again, continuous random variables are defined as variables that can assume any values within one or more intervals. • These values are not countable. • These variables can assume infinite number of values because there are infinite number of values in any given interval.

  3. Continuous Probability Distribution Table 1: Frequency and Relative Frequency Distribution of Student Heights • Now suppose we randomly selected 2,000 Cal Poly students and collected their heights. • Table 1 is the frequency and relative frequency distribution of student heights. • The height of each student is a continuous random variable. So let x = continuous random variable for the height of students

  4. Continuous Probability Distribution Table 1: Frequency and Relative Frequency Distribution of Student Heights Note: • In Chapter 4, we learned that if an experiment is performed n times and an Event A occurred f times, then using the relative frequency concept of probability, the probability that Event A will occur the next time is • Therefore, we can use the relative frequencies in Table 1 as the probability of each class. • The probabilities are not exact because we are using relative frequencies which are based on sample data.

  5. Continuous Probability Distribution • The histogram and polygon of the distribution in Table 1 is shown to the right. • As stated in Chapter 2, as the class width gets smaller and the number of classes increases, the polygon becomes a smooth curve. • The smooth curve is an approximation of the probability distribution curve of a continuous random variable.

  6. Continuous Probability Distribution • Column 4 shows that the sum of the areas of all the rectangles is greater than 1.0. • To make the sum of the areas to be 1.000, we have to define a new term, relative frequency density. It is obtained by dividing each relative frequency by the corresponding class width. • The area of each rectangle is now calculated as relative frequency density times the corresponding class width. Let us re-create Table 1 and add a new column for area. Table 2: Frequency and Relative Frequency Distribution of Student Heights

  7. Continuous Probability Distribution Table 3: Frequency, Relative Frequency, and Relative Frequency Density Distribution of Student Heights

  8. Continuous Probability Distribution A probability distribution of a continuous random variable must satisfy the following two conditions: • The total probability of all intervals within which x can assume a value is 1.0. That is, the total area under the probability distribution curve of a continuous random variable is always 1.0. • The probability that x assumes a value in any of the intervals (class) ranges from 0 to 1. In other words, the area under a probability distribution curve of a continuous random variable between any two points is between 0 and 1.

  9. Continuous Probability Distribution General Statement: • The probability that x assumes a value within a certain interval is equal to the area under the curve between the endpoints of the interval. For example, let • So, the probability that the height of a randomly selected student lies between 68 to 70 (5th class) is given by the area under the curve from x = 68 to x = 70. 70 68

  10. Continuous Probability Distribution General Statement: • The probability of a continuous random variable is always calculated for an interval. Therefore, the probability that x assumes a single value is zero. P(x=a) = 0 P(x=b) = 0Area of a line is zero as shown in the figure to the right. • Since P(x=a) = 0 and P(x=b) = 0, then, • In other words, the probability that x assumes a value in an interval from a to b is the same whether or not a and b are included.

  11. THE NORMAL DISTRIBUTION Introduction A continuous random variable can exhibit a normal probability distribution. A continuous random variable that possess a normal probability distribution is called a normal random variable Definition: A normal probability distribution has a mean, , and a standard deviation, , and is characterized by its symmetric bell-shaped curve that satisfy the following conditions. • The total area under the curve is 1.0. • The curve is symmetric about the mean. That is, the mean line divides the shaded region into two equal parts; 50% to the left and 50% to the right of the mean.

  12. The Normal distribution • The two tails of the curve extend indefinitely in both directions. The curve never touches but get very close to the horizontal line such that the area under the region beyond on the left and on the right is considered to be zero. • From the three graphs, we need only both the mean, and standard deviation, , to find the area under a normal distribution curve. • Each set of result in a family of normal distribution curves as shown to the right.

  13. THE STANDARD NORMAL DISTRIBTUION Definition A standard normal probability is a normal distribution of a continuous random variable having a mean of zero and a standard deviation of 1. These types of random variables are denoted by z and called z values or z scores. Note Although the values of z are negative to the left of the mean, the area under the curve is positive. • A value of z indicates the distance between the mean and the point represented by z in terms of standard deviation. For example, a point with a z score of 2 is to the right of the mean. A point with a z score of -3 is to the left of the mean. • Table IV of Appendix C lists the area under the curve to the left of z values from -3.49 to 3.49.

  14. The Standard Normal Distribution To read the area under the curve from Table IV of Appendix C, we perform the following steps: • Divide the z value into two portions. • The first portion consists of: • 1st digit to the left decimal point, • The decimal point followed by • The 1st digit to the right of the decimal point • The second portion is made-up of: • Decimal point, followed by a zero & • The 2nd digit after the decimal point. • For example, given z = 1.85, then • The 1st portion is 1.8 and • The second portion is .05. • To find the required area, locate the 1st portion under the column for z, and the 2nd portion in the row of z. Then the entry where the row for the 1st portion intersect the column for the 2nd portion is the required area to the left of the z value.

  15. z -1 z z 1 -1 1 The Standard Normal Distribution Example #1 – Problem 6.10 Example #1 – Solution For the standard normal distribution, find the area within one standard deviation of the mean-that is, the area between From the given information, z = -1 and z = 1. Then the interval of the required area is, -1 < z < 1 From Table IV, The area to the left of z=-1 is 0.1587 The area to the left of z = 1 is 0.8413 Area between is P(-1<z<1) = P(z<1) – P(z<-1) =.8413 - .1587= 0.6826

  16. The Standard Normal Distribution z 0 2.34 Example #2 – Problem 6.16 Find the area under the standard normal curve a) from z=0 to z=2.34 b) between z=0 and z=-2.58 c) from z=.84 to z=1.95 d) between z=-.57 and z=-2.49 e) between z=-2.15 and z=1.87 Example #2 – Solution z 0 -2.58

  17. The Standard Normal Distribution z 1.95 .84 z -.57 -2.49 z 1.87 -2.15 Example #2 – Solution

  18. The Standard Normal Distribution z 1.36 z -1.97 Example #3 – Problem 6.17 Find the area under the standard normal curve a) to the right of z=1.36 b) to the left of z=-1.97 c) to the right of z=-2.05 d) to the left of z=1.76 Example #3 – Solution

  19. The Standard Normal Distribution z -2.05 Example #3 – Solution 1.76

  20. The Standard Normal Distribution z -2.46 1.88 Example #4 – Problem 6.22 Determine the following probabilities for the standard normal distribution Example #4 – Solution z 1.96 0

  21. The Standard Normal Distribution Example #4 – Solution z 0 -2.58 z .73

  22. STANDARDIZING A NORMAL DISTRIBUTION Introduction • Not all normal random variables possess standard normal distribution. That is, there are normal random variables with mean and standard deviation other than 0 and 1, respectively. • However, we can still use Table IV of Appendix C to find the area under a normal distribution curve, provided we convert the normal distribution to standard normal distribution. • In other words, we have to convert the normal random variable, x, to its equivalent z value or z score. Formula • The formula for converting a normal random variable to a z score is,

  23. Normal distribution x x=a Standard distribution z z=b z=0 Standardizing a Normal Distribution Note • The z value should be rounded up to two decimal places. • From the given formula for z: • z value = 0 if x = • z value is negative if x is to the left of • z value is positive if x is to the right of • The z value for the mean of a normal distribution is always 0. • The z value for an x-value indicates the distance between the mean and the x value in terms of the standard deviation.

  24. Standardizing a Normal Distribution Example #5 – Problem 6.28 Determine the z value for each of the following x values for a normal distribution with Example #5 – Solution x 12 z -1.33 z=0 x 22 z z=0 2

  25. Standardizing a Normal Distribution Example #5 – Solution x 19 z z=0 1 x 13 z -1 z=0

  26. Standardizing a Normal Distribution Example #6 – Problem 6.30 Find the following areas under a normal distribution curve with Example #6 – Solution x 7.76 z z=0 -2.12

  27. Standardizing a Normal Distribution Example #6 – Solution x 10.06 8.22 16.54 14.48 z -1.89 -.97 z 0 2.27 1.24 0

  28. Standardizing a Normal Distribution Example #7 – Problem 6.35 Let x be a continuous random variable that is normally distributed with a mean of 80 and a standard deviation of 12. Find the probability that x assumes a value Example #7 – Solution x 69 z 0 -.92

  29. Standardizing a Normal Distribution Example #7 – Solution x 73 z 0 -.58 29

  30. APPLICATIONS OF THE NORMAL DISTRIBUTION This section deals with how to use what we have learned to solving real problems. Example #8 – Problem 6.40 Tommy W., a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. In fact, his times between pitches are normally distributed with a mean of 36 seconds and a standard deviation of 2.5 seconds. What percentage of his times between pitches are Example #8 – Solution

  31. Application of the Normal Distribution Example #9 – Problem 6.42 The Bank of Connecticut issues Visa and MasterCard credit cards. It is estimated that the balance on all Visa credit cards issued by the bank have a mean of $845 and a standard deviation of $270. Assume that the balances on all these Visa cards follows a normal distribution Example #9 – Solution

  32. DETERMINING THE z AND x VALUES WHEN AN AREA UNDER THE NORMAL DISTRIBUTION CURVE IS KNOWN So far, we have learned how to determine an area under a normal distribution curve given the z value. What about the reverse? This section deals with locating the z value or z score given the area under the normal distribution curve. How to find z and x values given area To find the value of z given the area under the standard normal curve: • Locate the position of the given area in Table IV of Appendix C • Add the corresponding values under the z column and z row. • This is the value of the required z value. • To find the value of x from the z value, we use the formula

  33. Determining the z and x Values Given the Area Example #10 – Problem 6.54 Find the value of z so that the area under the standard normal curve Example #10 – Solution

  34. Determining the z and x Values Given the Area Example #10 – Solution

  35. Determining the z and x Values Given the Area Example #10 – Solution

  36. Determining the z and x Values Given the Area Example #11 – Problem 6.63 A study has shown that 20% of all college textbooks have a price of $90 or higher. It is known that the standard deviation of the prices of all college textbooks is $9.50. Suppose the prices of all college textbooks have a normal distribution. What is the mean price of all college textbooks? Example #11 – Solution

  37. THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION Introduction • In Chapter 5, we discussed the binomial distribution for discrete random variables. We presented and discussed the formula for calculating the binomial probability of x successes in n trials asWe also stated that this formula could be tedious and complicated to use for very large values of n. We can use normal distribution to approximate the probability for binomial distributions. • The probability of binomial distribution obtained by using normal distribution is very close to exact probability when n is large and probability of success, p, is very close 0.50. General Statement • Usually, the normal distribution is used to approximate the binomial distribution when • We will establish these conditions as the basis for using normal distribution to estimate binomial distribution.

  38. THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION Steps for Using Normal Distribution for Binomial Distribution • Determine • Compute • Convert discrete random variable to continuous random variable. This is done by adding 0.5 to and subtracting 0.5 from the value of x. This process is called correction for continuity. • Compute the required probability that x assumes a value in the interval obtained in (3) by using normal distribution Note: Correction for Continuity For,

  39. THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION Example #12 – Problem 6.66 For a binomial probability distribution, n=20 and p=.60. • Find the probability P(x = 14) by using the table of binomial probabilities. • Find the probability P(x = 14) by using the normal distribution as an approximation to the binomial distribution. What is the difference between this approximation and the exact probability calculated in part a? Example #12 – Solution

  40. THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION Example #13 – Problem 6.72 According to a May 27, 2009 Minneapolis Star Tribune article, 78% of U.S. households have at least one credit card. Find the probability that in a random sample of 500 U.S. households, 375 to 385 households have at least one credit card. Example #13 – Solution

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