# Mathematical Probability Ch 7 Exercises

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MAT 1600 β Chapter 7 Problem Set 1. Compute the following probabilities for a standard normal random variable [Using Normal table attached below] a. π(π β€ 0) = 0.5 b. π(π β€ 1.62) = 0.9474 c. π(π β€ β1.89) = 0.0294 d. π(π β₯ 1.05) = 1 β π(π β€ 1.05) = 1 β 0.8531 = 0.1469 e. π(π β₯ β1.22) = 1 β π(π β€ β1.22) = 1 β 0.1112 = 0.8888 f. π(0.65 β€ π β€ 1.76) = π(π β€ 1.76) β π(π β€ 0.65) = 0.9608 β 0.7422 = 0.2186 g. π(β1.55 β€ π β€ 1.88) = π(π β€ 1.88) β π(π β€ β1.55) = 0.9699 β 0.0606 = 0.9093 h. π(β1.71 β€ π β€ β1.33) = π(π β€ β1.33) β π(π β€ β1.71) = 0.0918 β 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 π(π > 30) = π(π β₯ 30) =? π β π 30 β 30 0 π= = = =0 π 5 5 π(π β₯ 0) = 1 β π(π β€ 0) = 1 β 0.5 = 0.5 b. More than 35 inches π(π β₯ 35) =? π β π 35 β 30 5 π= = = =1 π 5 5 π(π β₯ 1) = 1 β π(π β€ 1) = 1 β 0.8413 = 0.1587 c. Less than 25 inches π(π β€ 25) =? π β π 25 β 30 β5 π= = = = β1 π 5 5 π(π β€ β1) = 0.1587 d. Less than 40 inches π(π β€ 40) =? π β π 40 β 3 ...
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