PROBLEM SET-3 Discrete Probability - Solutions

# PROBLEM SET-3 Discrete Probability - Solutions - STATISTICS...

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Unformatted text preview: STATISTICS FOR SOCIAL SCIENCES BUS 152 ___________________________________________________________________________ PROBLEM SET 3 - SOLUTIONS SOME BASIC DISCRETE PROBABILITY DISTRIBUTIONS Problem 1 (5.1) Given the following probability distribution: Distribution A Distribution B X P(X) X P(X) 0.50 0.05 1 0.20 1 0.10 2 0.15 2 0.15 3 0.10 3 0.20 4 0.05 4 0.50 a. Compute the expected value for each distribution. b. Compute the standard deviation for each distribution. c. Compare and contrast the results of distributions A and B . SOLUTION: (a) Distribution A Distribution B X P(X) X*P(X) X P(X) X*P(X) 0.50 0.00 0.05 0.00 1 0.20 0.20 1 0.10 0.10 2 0.15 0.30 2 0.15 0.30 3 0.10 0.30 3 0.20 0.60 4 0.05 0.20 4 0.50 2.00 1.00 1.00 1.00 3.00 μ = 1.00 μ = 3.00 (b) Distribution A X (X– μ ) 2 P(X) (X– μ ) 2 *P(X) (–1) 2 0.50 0.50 1 (0) 2 0.20 0.00 2 (1) 2 0.15 0.15 3 (2) 2 0.10 0.40 4 (3) 2 0.05 0.45 1.50 σ = ∑ ( X – μ ) 2 ⋅ P ( X ) = 1.22 Distribution B X (X– μ ) 2 P(X) (X– μ ) 2 *P(X) (–3) 2 0.05 0.45 1 (–2) 2 0.10 0.40 2 (–1) 2 0.15 0.15 3 (0) 2 0.20 0.00 4 (1) 2 0.50 0.50 1.50 σ = ∑ ( X – μ ) 2 ⋅ P ( X ) = 1.22 (c) Distribution A: Becaus e the mean of 1 is greater than the median of 0, the distribution is right-skewed. Distribution B: Becaus e the mean of 3 is less than the median of 4, the distribution is left- skewed. The means are different but the variance s are the sam e. _________________________________________________________________________________ Problem 2 (5.3) Using the company records for the past 500 working days, the manager of Konig Motors, a suburban automobile dealership, has summarized the number of cars sold per day into the following table: Number of Cars Sold per Day Frequency of Occurrence 40 1 100 2 142 3 66 4 36 5 30 6 26 7 20 8 16 9 14 10 8 11 2 Total 500 a. Form the probability distribution for the number of cars sold per day? b. Compute the mean or expected number of cars sold per day? c. Compute the standard deviation? SOLUTION: a) X 1 2 3 4 5 6 7 8 9 10 11 P(X) 0.080 0.200 0.284 0.132 0.072 0.060 0.052 0.040 0.032 0.028 0.016 0.004 1 2 3 4 5 6 7 8 9 10 11 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Probability Distribution (a)-(c) X n P(X) X*P(X) (X- μ ) 2 (X- μ ) 2 * P(X) 40 40/500=0.080 0.000 9.339136 0.74713088 1 100 0.200 0.200 4.227136 0.84542720 2 142 0.284 0.568 1.115136 0.31669862 3 66 0.132 0.396 0.003136 0.00041395 4 36 0.072 0.288 0.891136 0.06416179 5 30 0.060 0.300 3.779136 0.22674816 6 26 0.052 0.312 8.667136 0.45069107 7 20 0.040 0.280 15.555136 0.62220544 8 16 0.032 0.256 24.443136 0.78218035 9 14 0.028 0.252 35.331136 0.98927181 10 8 0.016 0.160 48.219136 0.77150618 11 2 0.004 0.044 63.10713663....
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## This note was uploaded on 12/22/2011 for the course BUSINESS bus 152 taught by Professor Çeşmecibaşı during the Spring '11 term at Middle East Technical University.

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PROBLEM SET-3 Discrete Probability - Solutions - STATISTICS...

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