chap 6 - CHAPTER 6 5. The information below is the number...

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CHAPTER 6 5. The information below is the number of daily emergency service calls made by the volunteer ambulance service of Walterboro, South Carolina, for the last 50 days. To explain, there were 22 days on which there were 2 emergency calls, and 9 days on which there were 3 emergency calls. a.) Convert this information on the number of calls to a probability distribution. x Frequency P(x) x * (P)x (x-u)^2 * P(x) 0 8 0.16 0 0.4624 1 10 0.2 0.2 0.098 2 22 0.44 0.88 0.0396 3 9 0.18 0.54 0.3042 4 1 0.02 0.08 0.1058 50 1.7 1.01 b.) Is this an example of a discrete or continuous probability distribution ? Discrete distribution c.) What is the mean number of emergency calls per day? 1.7 x P(x) x * (P)x 0 0.16 0 1 0.2 0.2 2 0.44 0.88 3 0.18 0.54 4 0.02 0.08 1.7 d.) What is the standard deviation of the number of calls made daily? 1.005 Square root of 1.01 = 1.005 x P(x) x- u (x-u)^2 (x-u)^2 * P(x) 0 0.16 0-1.7 2.89 0.4624 1 0.2 1-1.7 0.49 0.098 2 0.44 2-1.7 0.09 0.0396 3 0.18 3-1.7 1.69 0.3042 4 0.02 4-1.7 5.29 0.1058 1.01 15. Industry standards suggest that 10 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 12 Nissans yesterday. n = 12 Probability = .1
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a.) What is the probability that none of these vehicles requires warranty service? . 2824 (12 C 0) (.1)^0 (1-.1)^(12-0) = (1) (1) (.2824) b.) What is the probability exactly one of these vehicles requires warranty service? . 3765 (12 C 1) (.1)^1 (1-.1)^(12-1) = (12) (.1) (.3138) = .3765 c.) Determine the probability that exactly two or these vehicles require warranty service. .2301 (12 C 2) (.1)^2 (1-.1)^(12-2) = (66) (.01) (.3487) = .2301 d.) Compute the mean and standard deviation of this probability distribution. Mean
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This note was uploaded on 10/15/2009 for the course ECON 2302 taught by Professor Parker during the Spring '09 term at University of Texas-Tyler.

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chap 6 - CHAPTER 6 5. The information below is the number...

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