# Ch 1 - 28E

## Problem 28E Chapter 1

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Problem 28E

28E

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Answer: Step1: Given the data is Deposited Frequency Relative frequency energy (mj) 1.0-2.0 5 5/804 = 0.0062 2.0-2.4 11 11/804 = 0.0136 2.4-2.6 13 13/804 = 0.0161 2.6-2.8 30 30/804 = 0.0373 2.8-3.0 46 0.0572 3.0-3.2 66 0.0820 3.2-3.4 133 0.1654 3.4-3.6 141 0.1753 3.6-3.8 126 0.1567 3.8-4.0 92 0.1144 4.0-4.2 73 0.0907 4.2-4.4 38 0.0472 4.4-4.6 19 0.0236 4.6-5.0 11 0.0136 N = 804 Total = 1 Step 2: b). To calculate the proportion of these ignition trials resulted in a deposited energy of less than 3 mj. Proportion of energy level less than 3 mj = 0.0062 + 0.0136 + 0.0161 + 0.0373 + 0.0572. = 0.1304 Therefore the Proportion of energy level less than 3 mj is 0.1304. Step 3: c). To calculate the proportion of these ignition trials resulted in a deposited energy of at least 4 mj. Proportion of energy level at least 4 mj = 0.0907 + 0.0472 + 0.0236 + 0.0136 = 0.1751 Therefore the Proportion of energy level at least 4 mj is 0.1751.

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