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Mathematical Probability Ch 7 Exercises

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MAT 1600 – Chapter 7 Problem Set 1. Compute the following probabilities for a standard normal random variable [Using Normal table attached below] a. 𝑃(𝑍 ≀ 0) = 0.5 b. 𝑃(𝑍 ≀ 1.62) = 0.9474 c. 𝑃(𝑍 ≀ βˆ’1.89) = 0.0294 d. 𝑃(𝑍 β‰₯ 1.05) = 1 βˆ’ 𝑃(𝑍 ≀ 1.05) = 1 βˆ’ 0.8531 = 0.1469 e. 𝑃(𝑍 β‰₯ βˆ’1.22) = 1 βˆ’ 𝑃(𝑍 ≀ βˆ’1.22) = 1 βˆ’ 0.1112 = 0.8888 f. 𝑃(0.65 ≀ 𝑍 ≀ 1.76) = 𝑃(𝑍 ≀ 1.76) βˆ’ 𝑃(𝑍 ≀ 0.65) = 0.9608 βˆ’ 0.7422 = 0.2186 g. 𝑃(βˆ’1.55 ≀ 𝑍 ≀ 1.88) = 𝑃(𝑍 ≀ 1.88) βˆ’ 𝑃(𝑍 ≀ βˆ’1.55) = 0.9699 βˆ’ 0.0606 = 0.9093 h. 𝑃(βˆ’1.71 ≀ 𝑍 ≀ βˆ’1.33) = 𝑃(𝑍 ≀ βˆ’1.33) βˆ’ 𝑃(𝑍 ≀ βˆ’1.71) = 0.0918 βˆ’ 0.0436 = 0.0482 2. Assume that the annual rainfall in New York is normally distributed with a mean of 30 inches per year, with a standard deviation of 5 inches per year. Compute the following probabilities for the rainfall in New York over the coming year: a. More than 30 inches 𝑃(𝑋 > 30) = 𝑃(𝑋 β‰₯ 30) =? 𝑋 βˆ’ πœ‡ 30 βˆ’ 30 0 𝑍= = = =0 𝜎 5 5 𝑃(𝑍 β‰₯ 0) = 1 βˆ’ 𝑃(𝑍 ≀ 0) = 1 βˆ’ 0.5 = 0.5 b. More than 35 inches 𝑃(𝑋 β‰₯ 35) =? 𝑋 βˆ’ πœ‡ 35 βˆ’ 30 5 𝑍= = = =1 𝜎 5 5 𝑃(𝑍 β‰₯ 1) = 1 βˆ’ 𝑃(𝑍 ≀ 1) = 1 βˆ’ 0.8413 = 0.1587 c. Less than 25 inches 𝑃(𝑋 ≀ 25) =? 𝑋 βˆ’ πœ‡ 25 βˆ’ 30 βˆ’5 𝑍= = = = βˆ’1 𝜎 5 5 𝑃(𝑍 ≀ βˆ’1) = 0.1587 d. Less than 40 inches 𝑃(𝑋 ≀ 40) =? 𝑋 βˆ’ πœ‡ 40 βˆ’ 3 ...
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