192 CHAPTER 4 Discrete Probability Distributions Frequency Distribution Score, x Frequency, f 1 24 2 33 3 42 4 30 5 21 Frequency Distribution Sales per day, x Number of days, f 0 16 1 19 2 15 3 21 4 9 5 10 6 8 7 2 Probability Score 0.05 0.10 0.15 0.20 0.25 0.30 1 2 3 4 5 x P(x) Passive-Aggressive Traits Constructing a Discrete Probability Distribution Let x be a discrete random variable with possible outcomes x1, x2, . . ., xn. 1. Make a frequency distribution for the possible outcomes. 2. Find the sum of the frequencies. 3. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies. 4. Check that each probability is between 0 and 1, inclusive, and that the sum of all the probabilities is 1. GUIDELINES Constructing and Graphing a Discrete Probability Distribution An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Each individual was given a whole number score from 1 to 5, where 1 is extremely passive and 5 is extremely aggressive. A score of 3 indicated neither trait. The results are shown at the left. Construct a discrete probability distribution for the random variable x. Then graph the distribution using a histogram. SOLUTION Divide the frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable. P112 = 24 150 = 0.16 P122 = 33 150 = 0.22 P132 = 42 150 = 0.28 P142 = 30 150 = 0.20 P152 = 21 150 = 0.14 The discrete probability distribution is shown in the table below. x 1 2 3 4 5 P1x2 0.16 0.22 0.28 0.20 0.14 Note that the probability of each value of x is between 0 and 1, and the sum of the probabilities is 1. So, the distribution is a probability distribution. The graph of the distribution is shown in the histogram at the left. Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome. Also, the probability of an event corresponds to the sum of the areas of the outcomes included in the event. For instance, the probability of the event “having a score of 2 or 3” is equal to the sum of the areas of the second and third bars. 11210.222 + 11210.282 = 0.22 + 0.28 = 0.50 Interpretation You can see that the distribution is approximately symmetric. TRY IT YOURSELF 2 A company tracks the number of sales new employees make each day during a 100-day probationary period. The results for one new employee are shown at the left. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram. Answer: Page A38 EXAMPLE 2
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