Probability and Statistics The Binomial Probability Distribution and Related Topics Chapter 5 Section 1 Introduction to Random Variables and Probability Distributions Essential Question: How are the mean and the standard deviation determined from a discrete probability distribution? Student Objectives: The student will distinguish between random and discrete random variables. The student will graph discrete probability distributions. The student will compute the mean and standard deviation for a discrete probability distribution. The student will compute the mean and standard deviation for a linear function of a random variable x. The student will compute the mean and standard deviation for a linear combination of two independent random variables. Terms: Continuous random variable Discrete random variable Linear function of a random variable Linear function of two independent random variables Mean ( µ ) Probability Distribution Random variable Standard deviation ( )
Equations: Mean: Expected value for a discrete probability distribution: µ = $ Standard Deviation: = x " µ Linear function: L = a + bx ( ) 2 # P( x) Mean of a linear function: µ L = a + bµ Variance of a linear function: L 2 = b 2 2 Standard deviation for a linear function: L = b 2 2 = b "( x P( x) ) Linear combination: W = ax 1 + bx 2 Mean of a linear combination: µ W = aµ 1 + bµ 2 Variance of a linear combination: 2 W = a 2 2 1 + b 2 2 2 Standard deviation for a linear function: W = a 2 2 1 + b 2 2 2 Graphing Calculator Skills: Calculating the values of the mean and standard deviation for a discrete probability distribution: 1-Var Stats L 1,L 2 where L 1 contains the x-values and L 2 contains the probabilities. Sample Questions: Directions: In problems #1-7, identify each of the following as either a discrete or continuous random variable. 1. The number of people who are in a car. 2. The number of miles you drive in one week. 3. The weight of a box of cereal. 4. The number of boxes of cereal you buy in one year.
5. The length of time you spend eating your lunch. 6. The number of patients on a psychiatric ward in one day. 7. The volume of blood that is transfused during an operation. 8. In a personality inventory test for passive-aggressive traits, the possible scores are: 1 = extremely passive 2 = moderately passive 3 = neither 4 = moderately aggressive 5 = extremely aggressive The test was administered to a group of 110 people and the results were as follows: x (score) 1 2 3 4 5 f (frequency) 19 23 32 26 10 Construct a probability distribution table, calculate the expected value (the mean) and the standard deviation. Use a histogram to graph the probability distribution. x f P(x) x P( x) x µ ( x µ ) 2 ( x µ ) 2 " P( x) 1 19 2 23 3 32 4 26 5 10 Sum:
9. A local automotive repair shop has two work centers. The first center examines the car with a computer diagnostic machine and the second center repairs the car. Let x 1 and x 2 be random variables representing the time in minutes to diagnose and repair the car. Assume x 1 and x 2 are independent random variables. A recent study of the repair center produced the following data: Computer diagnose x 1 : µ 1 = 18.5 minutes and 1 = 5.6 minutes Repair x 2 : µ 2 = 137.75 minutes and 2 = 17.2 minutes a. Suppose it costs $1.25 per minute to diagnose the automobile and $0.95 per minute to repair the automobile (without parts). Compute the linear combination equation, mean, variance, and the standard deviation of W. b. The repair shop charges a flat rate of $0.75 per minute to diagnose the car and if no repairs are needed, there is an additional $25.00 service charge. Compute the linear equation, mean, variance, and standard deviation of L. Homework: Pages 190-195 Exercises: #1-19 odd Exercises: #2-18 even
ANSWERS Directions: In problems #1-7, identify each of the following as either a discrete or continuous random variable. 1. The number of people who are in a car. DISCRETE 2. The number of miles you drive in one week. CONTINUOUS 3. The weight of a box of cereal. CONTINUOUS 4. The number of boxes of cereal you buy in one year. DISCRETE 5. The length of time you spend eating your lunch. CONTINUOUS 6. The number of patients on a psychiatric ward in one day. DISCRETE 7. The volume of blood that is transfused during an operation. CONTINUOUS
8. In a personality inventory test for passive-aggressive traits, the possible scores are: 1 = extremely passive 2 = moderately passive 3 = neither 4 = moderately aggressive 5 = extremely aggressive The test was administered to a group of 110 people and the results were as follows: x (score) 1 2 3 4 5 f (frequency) 19 23 32 26 10 Construct a probability distribution table, calculate the expected value (the mean) and the standard deviation. Use a histogram to graph the probability distribution. x f P(x) x P( x) x µ ( x µ ) 2 ( x µ ) 2 " P( x) 1 19 0.1727 0.1727-1.8636 3.4731 0.5999 2 23 0.2091 0.4182-0.8636 0.7459 0.1560 3 32 0.2909 0.8727 0.1364 0.0186 0.0054 4 26 0.2364 0.9455 1.1364 1.2913 0.3052 5 10 0.0909 0.4545 2.1364 4.5640 0.4149 Sum: 110 1.0000 2.8636 1.4814 µ = 2.8636 = 1.4814 = 1.2171
9. A local automotive repair shop has two work centers. The first center examines the car with a computer diagnostic machine and the second center repairs the car. Let and be random variables representing the time in minutes to diagnose and repair the car. x 1 x 2 Assume and are independent random variables. A recent study of the repair center produced the following data: Computer diagnose x 1 : µ 1 = 18.5 minutes and 1 = 5.6 minutes Repair x 2 : µ 2 = 137.75 minutes and 2 = 17.2 minutes x 1 x 2 a. Suppose it costs $1.25 per minute to diagnose the automobile and $0.95 per minute to repair the automobile (without parts). Compute the linear combination equation, mean, variance, and the standard deviation of W. W = 1.25x 1 + 0.95x 2 µ W = 1.25( 18.5) + 0.95( 137.75) µ W = 23.125 +130.8625 µ W = 153.9875 µ W = $153.99 2 W = ( 1.25) 2 2 1 + ( 0.95) 2 2 2 W 2 = 1.5625 1 2 + 0.9025 2 2 2 W = 1.5625( 5.6) 2 + 0.9025( 17.2) 2 2 W = 1.5625( 31.36) + 0.9025( 295.84) W 2 = 49 + 266.9956 W 2 = 315.9956 W 2 = $316.00 W = 315.9956 W = 17.7763 W = $17.78 b. The repair shop charges a flat rate of $0.75 per minute to diagnose the car and if no repairs are needed, there is an additional $25.00 service charge. Compute the linear equation, mean, variance, and standard deviation of L. L = 0.75x 1 + 25 µ L = 25 + 0.75( 18.5) µ L = 25 +13.875 µ L = 38.875 µ L = $38.88 2 L = ( 0.75) 2 2 1 L 2 = 0.5625 1 2 ( ) 2 ( ) L 2 = 0.5625 5.6 L 2 = 0.5625 31.36 L 2 = 17.64 L 2 = $17.64 L = 17.64 L = 4.2 L = $4.20