CHAPTER 6 NORMAL DISTRIBUTION CONTINUOUS RANDOM VARIABLES 1

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. 2

Continuous Probability Distribution • • • 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 Table 1: Frequency and Relative Frequency Distribution of Student Heights Height of a Student (inches), x f RF 60 to less than 62 150. 0750 62 to less than 64 320. 1600 64 to less than 66 480. 2400 66 to less than 68 600. 3000 68 to less than 70 250. 1250 70 to less than 72 200. 1000 =1. 0000 3

Continuous Probability Distribution Table 1: Frequency and Relative Frequency Distribution of Student Heights Height of a Student (inches), x f RF 60 to less than 62 150. 0750 62 to less than 64 320. 1600 64 to less than 66 480. 2400 66 to less than 68 600. 3000 68 to less than 70 250. 1250 70 to less than 72 200. 1000 =1. 0000 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 4 sample data.

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. 5

Continuous Probability Distribution Let us re-create Table 1 and add a new column for area. Table 2: Frequency and Relative Frequency Distribution of Student Heights Height of a Student (inches), x 1. Column 4 shows that the sum of the areas of all the rectangles is greater than 1. 0. 2. 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. 3. The area of each rectangle is now calculated as relative frequency density times the corresponding class width. Area f RF 60 to less than 62 150. 0750. 150 62 to less than 64 320. 1600. 320 64 to less than 66 480. 2400. 480 66 to less than 68 600. 3000. 600 68 to less than 70 250. 1250. 250 70 to less than 72 200. 1000. 200 = 2. 00 6

Continuous Probability Distribution Table 3: Frequency, Relative Frequency, and Relative Frequency Density Distribution of Student Heights Height of a Student (inches), x f RF RF Density Area 60 to less than 62 150 . 0750. 0375 . 0750 62 to less than 64 320 . 1600. 0800 . 1600 64 to less than 66 480 . 2400. 1200 . 2400 66 to less than 68 600 . 3000. 1500 . 3000 68 to less than 70 250 . 1250. 0625 . 1250 70 to less than 72 200 . 1000. 0500 . 1000 =1. 0000 7

Continuous Probability Distribution A probability distribution of a continuous random variable must satisfy the following two conditions: 1. 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. 2. 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. 8

Continuous Probability Distribution General Statement: 1. 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 2. So, the probability that the height of a randomly selected student lies between 68 to 70 is given by the area under the curve from x = 68 to x = 70. 68 70 9

Continuous Probability Distribution General Statement: 3. 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) = 0 Area of a line is zero as shown in the figure to the right. 4. Since P(x=a) = 0 and P(x=b) = 0, then, 5. 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. 10

6. 1 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. 1. The total area under the curve is 1. 0. 2. 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. 11

The Normal distribution 3. 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. 4. From the three graphs, we need only both the mean, and standard deviation, , to find the area under a normal distribution curve. 5. Each set of result in a family of normal distribution curves as shown to the right. 12

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. 1. 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. 2. Table IV of Appendix C lists the area under the curve to the left of z values from -3. 49 to 3. 49. Note Although the values of z are negative to the left of the mean, the area under the curve is positive. 13

The Standard Normal Distribution To read the area under the curve from Table IV of Appendix C, we perform the following steps: 1. Divide the z value into two portions. 2. The first portion consists of: • • • 1 st digit to the left decimal point, The decimal point followed by The 1 st digit to the right of the decimal point 3. The second portion is made-up of: • Decimal point, followed by a zero & • The 2 nd digit after the decimal point. 4. For example, given z = 1. 85, then • The 1 st portion is 1. 8 and • The second portion is. 05. 5. To find the required area, locate the 1 st portion under the column for z, and the 2 nd portion in the row of z. Then the entry where the row for the 1 st portion intersect the column for the 2 nd portion is the required area to the left of the z value. 14

The Standard Normal Distribution Example #1 – Problem 6. 10 Example #1 – Solution For the standard normal From the given information, distribution, find the area within z = -1 and z = 1. one standard deviation of the mean Then the interval of the required area is, -that is, the area between -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) -1 1 1 z z =. 8413 -. 1587= 0. 6826 -1 z 15

The Standard Normal Distribution 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 0 2. 34 z -2. 58 0 z 16

The Standard Normal Distribution Example #2 – Solution . 84 1. 95 z -2. 49 -2. 15 z -. 57 1. 87 z 17

The Standard Normal Distribution 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 1. 36 z z -1. 97 18

The Standard Normal Distribution Example #3 – Solution z -2. 05 1. 76 19

The Standard Normal Distribution Example #4 – Problem 6. 22 Determine the following probabilities for the standard normal distribution Example #4 – Solution -2. 46 z 1. 88 0 1. 96 z 20

The Standard Normal Distribution Example #4 – Solution -2. 58 z 0 z. 73 21

6. 2 APPLICATIONS OF NOERMAL DISTRIBUTION 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, 22

Standardizing a Normal Distribution Note 1. The z value should be rounded up to two decimal places. 2. From the given formula for z: a. b. c. 3. 4. 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 Normal distribution 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. x=a x Standard distribution z=0 z z=b 23

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 -1. 33 z z=0 22 z=0 2 x z 24

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

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 -2. 12 z=0 z 26

Standardizing a Normal Distribution Example #6 – Solution 14. 48 0 1. 24 16. 54 2. 27 x z 10. 06 8. 22 -1. 89 -. 97 0 z 27

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 -. 92 z 0 28

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

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 30

Application of the Normal Distribution Example #9 – Problem 6. 42 The Bank of Connecticut issues Visa and Master. Card 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 31

6. 3 Central Limit Theorem Example: p According to the National Center for Biotechnology Information, the average of Americans when Alzheimer’s disease is first diagnosed is 74. 7 years. Assume that the population standard deviation is 8. 6 years. If a random sample of 35 patients who have been diagnosed with Alzheimer’s disease is selected, what is the probability that the mean age of the sample is greater than 71 years? p 32

What Question is Asked? We must pay particular attention to the exact question that is being asked. We are not asked to find a probability associated with a single value rather we are asked to find a probability associated with a sample size of 35. In order to answer this question, we must first understand the Central Limit Theorem.

Central Limit Theorem The Central Limit Theorem states that as the sample size increases without limit the shape of the distribution of the sample means will approach a normal distribution. It is important to remember two things when using the Central Limit Theorem. 1. If Population normally distributed then Sampling distribution normally distributed regardless of sample size. 2. If Population not normally distributed then Sampling distribution is approximately normally distributed if sample size is at least 30.

Find the z Score p (1)

Find the z Score p (2)

Find the z Score p (3)

Probability Under the Curve It is always helpful to draw an appropriately shaded curve when finding probabilities associated with a normal distribution. Our z score of -2. 55 will be between z = -3 and z = -2. We are to find the probability that the z score for the sample mean would exceed -2. 55. We will shade the region to the right of z = -2. 55.

Use Table or Calculator to Find Probability Use a tool such as a normal probability table or a graphing calculator to find probability. We would find that the proportion of the distribution occupied by that shaded area is 0. 9946. The probability that the z score would be greater than -2. 55 is 0. 9946. Another way to state this result is by saying 99. 46% of all samples of size 35 among people diagnosed with Alzheimer’s disease would have an average of diagnosis that is greater than 71 years.

Summary In this Power. Point, we learned how to calculate a probability associated with a sample using the central limit theorem.

6. 4 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 as • We 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. 41

THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION Steps for Using Normal Distribution for Binomial Distribution 1. Determine 2. Compute 3. 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 4. Note: Correction for Continuity For, 42

THE NORMAL APPROXIMATION OF THE BINOMIAL DISTRIBUTION Example #12 – Problem 6. 66 For a binomial probability distribution, n=20 and p=. 60. a. Find the probability P(x = 14) by using the table of binomial probabilities. b. 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 43

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 44