Chapter 5 Continuous Random Variables and Probability Distributions Statistics for Business and Economics 8th Edition Chapter 5 Continuous Random Variables and Probability Distributions

Continuous Probability Distributions A continuous random variable is a variable that can assume any value in an interval thickness of an item time required to complete a task temperature of a solution height, in inches These can potentially take on any value, depending only on the ability to measure accurately. Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Continuous Probability Distributions Two characteristics The probability that x assumes a value in any interval lies in the range 0 to 1 The total probability of all the (mutually exclusive) intervals within which x can assume a value of 1.0

Figure 6.3 Area under a curve between two points.

Figure 6.4 Total area under a probability distribution curve.

Figure 6.5 Area under the curve as probability.

Figure 6.6 Probability that x lies in the interval 65 to 68 inches.

Expectations for Continuous Random Variables The mean of X, denoted μX , is defined as the expected value of X The variance of X, denoted σX2 , is defined as the expectation of the squared deviation, (X - μX)2, of a random variable from its mean Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Linear Functions of Variables Let W = a + bX , where X has mean μX and variance σX2 , and a and b are constants Then the mean of W is the variance is the standard deviation of W is Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Example 5.2: (b.p-185) home heating costs A homeowner estimates that within the range of likely temperatures her January heating bill, Y, in dollars, will be Y = 290 – 5 T Where T is the average temperature for the month, in degrees Fahrenheit. If the average January temperature can be represented by a random variable with mean 24 and S.D. 4, find the mean and S.D. of this home owner’s January heating bill. Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Exercise 5.10 The profit for a production process is equal to $1000 minus 2 times the number of units produced. The mean and variance for the number of units produced are 50 and 90, respectively. Find the mean and variance of the profit. Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

THE NORMAL DISTRIBUTION Normal Probability Distribution A normal probability distribution , when plotted, gives a bell-shaped curve such that: The total area under the curve is 1.0. The curve is symmetric about the mean. The two tails of the curve extend indefinitely.

Figure 6.11 Normal distribution with mean μ and standard deviation σ.

The Normal Distribution (continued) ‘Bell Shaped’ Symmetrical Mean, Median and Mode are Equal Location is determined by the mean, μ Spread is determined by the standard deviation, σ The random variable has an infinite theoretical range: +  to   f(x) σ x μ Mean = Median = Mode Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Properties of Normal Distribution Suppose that the Random variable X follows a normal distribution with parameters μ and σ2. Then the following properties hold: The mean of random variable is μ: E(X) = μ The variance of random variable is σ2: Var(X) = E [(X-μ)2] = σ2 The shape of the probability density function is a symmetric bell-shaped curve centered on the mean (see Fig. 6.8) We define the normal distribution using the notation X~N(μ,σ2). Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

The Normal Distribution (continued) The normal distribution closely approximates the probability distributions of a wide range of random variables Distributions of sample means approach a normal distribution given a “large” sample size Computations of probabilities are direct and elegant The normal probability distribution has led to good business decisions for a number of applications Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

The Normal Distribution Shape By varying the parameters μ and σ, we obtain different normal distributions Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

The Normal Distribution Shape f(x) Changing μ shifts the distribution left or right. Changing σ increases or decreases the spread. σ μ x Given the mean μ and variance σ we define the normal distribution using the notation Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Cumulative Normal Distribution For a normal random variable X with mean μ and variance σ2 , i.e., X~N(μ, σ2), the cumulative distribution function is f(x) x0 x Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Finding Normal Probabilities The probability for a range of values is measured by the area under the curve x a μ b Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Finding Normal Probabilities (continued) x a μ b x a μ b x a μ b Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

THE STANDARD NORMAL DISTRIBTUION z Values or z Scores Definition The units marked on the horizontal axis of the standard normal curve are denoted by z and are called the z values or z scores. A specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation.

The Standardized Normal Any normal distribution (with any mean and variance combination) can be transformed into the standardized normal distribution (Z), with mean 0 and variance 1 Need to transform X units into Z units by subtracting the mean of X and dividing by its standard deviation f(Z) 1 Z Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Example If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100. Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Comparing X and Z units 100 200 X 2.0 Z (μ = 100, σ = 50) 2.0 Z (μ = 0, σ = 1) Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z) Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Example 5.3 (b.p-191): investment portfolio value A client has an investment portfolio whose mean value is equal to $500,000 with a S.D. of $15,000. She has asked you to determine the probability that the value of her portfolio is between $485,000 and $530,000. Also draw the normal distribution diagram for this problem. Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Example 5.4 (b.p-192): normal probabilities If X~N(15,16), find the probability that X is larger than 18. Show relevant diagram What will be the probability that X is less than 18? Use Table 1 given in the Appendix (will be given in the exam) Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Probability as Area Under the Curve The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below f(X) 0.5 0.5 μ X Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Appendix Table 1 The Standardized Normal table in the textbook (Appendix Table 1) shows values of the cumulative normal distribution function For a given Z-value a , the table shows F(a) (the area under the curve from negative infinity to a ) a Z Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

The Standardized Normal Table Appendix Table 1 gives the probability F(a) for any value a .9772 Example: P(Z < 2.00) = .9772 Z 2.00 Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

The Standardized Normal Table (continued) For negative Z-values, use the fact that the distribution is symmetric to find the needed probability: .9772 .0228 Example: P(Z < -2.00) = 1 – 0.9772 = 0.0228 Z 2.00 .9772 .0228 Z -2.00 Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

General Procedure for Finding Probabilities To find P(a < X < b) when X is distributed normally: Draw the normal curve for the problem in terms of X Translate X-values to Z-values Use the Cumulative Normal Table Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.

Finding an x Value for a Normal Distribution For a normal curve, with known values of μ and σ and for a given area under the curve to the left of x, the x value is calculated as x = μ + zσ

Example 6-19 (similar to Ex-5.7 in book pg-194) Almost all high school students who intend to go to college take the SAT test. In a recent test, the mean SAT score (in verbal and mathematics) of all students was 1020. Debbie is planning to take this test soon. Suppose the SAT scores of all students who take this test with Debbie will have a normal distribution with a mean of 1020 and a standard deviation of 153. What should her score be on this test so that only 10% of all examinees score higher than she does?

Example 6-19: Solution Area to the left of the x value = 1.0 - .10 = .9000 Look for .9000 in the body of the normal distribution table. The value closest to .9000 in Table IV is .8997, and the z value is 1.28. x = μ + zσ = 1020 + 1.28(153) = 1020 + 195.84 = 1215.84 ≈ 1216 Thus, if Debbie scores 1216 on the SAT, only about 10% of the examinees are expected to score higher than she does.

Finding an x value.