edm_hw01

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Iowa State University *

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305

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Statistics

Date

Feb 20, 2024

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pdf

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2

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Meyers Stat 305 Spring 2024 Homework 01 Directions: Put work and answers on a separate sheet of paper. Show your work to receive partial credit. Clearly indicate your final answer for each question. Due: 11:59 PM on Friday 2024-02-02 1. A certain factory operates three different shifts. Accidents in this factory are categorized as related or unrelated to working conditions. Based on historical data, the follow table describes the probabilities of these types of accidents. Note: In real world situations, these probabilities would fluctuate over time based on different employees, managers, equipment, etc. For the purpose of this question, assume these probabilities are constant over time. Unsafe Conditions Unrelated to Conditions Day 0 . 10 0 . 35 Shift Swing 0 . 08 0 . 20 Night 0 . 05 0 . 22 (a) For a random accident, what is the probability that it was on the swing shift and was unrelated to conditions? P(SS1) + P(SS2) = P(SS) , .08 + .20 = .28 (b) For a random accident, what is the probability that it was unrelated to conditions? P(D2) + P(SS2) + P(N2) = P(UTC) , .35 + .20 + .22 = .77 (c) For a random accident, what is the probability that it was on the swing shift given it was unrelated to conditions? P(SS|UTC) = P(SS UTC) / P(UTC) , .20 / .77 = .26 (d) For a random accident, what is the probability it was on the night shift? P(N1) + P(N2) = P(N) , .05 + .22 = .27 (e) For a random accident, what it is the probability it was related to unsafe conditions given it occurred on the night shift? P(UC|N) = P(UC N) / P(N) , .05 / .27 = .19 2. The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.65, the analogous probability for the second signal is .45, and the probability that he must stop at at least one of the two signals is 0.8. (a) What is the probability he must stop at both signals. P(S1) x P(S2) = P(P1 P2) , .65 x .45 = .29 (b) What is the probability he must stop at the second signal but not the first signal? P(S1 ) x P(S2) = P(S1 S2) , .35 x .45 = .16 (c) What is the probability he must stop at exactly one signal? P(S1 S2) + P(S1 S2 ) , .16 + .36 = .52 3. A company uses three different assembly lines to manufacture a particular component. Of those manufactured on line 1, 5% need rework to remedy a defect, whereas 8% of line 2 components need rework and 10% of line 3 need rework. Suppose that 50% of all components are produced by line 1, 30% are produced by line 2, and 20% come from line 3. (a) What is the probability that a randomly selected component needs to be reworked? P(L1R L2R L3R) , (.5 x .05) + (.3 x .08) + (.2 x .1) = .025 + .024 + .02 = .069 If a randomly selected component needs rework, what is the probability that it came from line 1? From line 2? From line 3? .025 , .024 , and .02 respectively
4. Suppose Meyersville is partitioned into four regions - north, south, east and west. By par- tition, we mean that the four regions are disjoint and together (union) they make up all of Meyersville. Suppose that the percent of the population that live in these regions is 30% , 35% , 20% , and 15% respectively. Within these regions, the percent of the population that is Norwegian is 40% , 4% , 50% , and 1% respectively. (a) Suppose an individual is selected completely at random. What is the probability that they are from the east and they are Norwegian? P(E NR3) = P(E) x P(NR3) , .2 x .5 = .1 (b) Suppose an individual is selected completely at random. What is the probability they are Norwegian? P(N NR1) + P(S NR2) + P(E NR3) + P(W NR4) = P(NR) , (.3 x .4) + (.35 x .04) +(.2 x .5) + (.15 x .01) = .12 + .014 + .1 + .0015 = .24 (c) Suppose an individual is selected completely at random. What is the probability they are Norwegian given they are from the east? P(NR3|E) = P(NR3 E) / P(E) , .1 / .2 = .5 (d) Suppose an individual is selected completely at random. What is the probability they are from the east, given they are Norwegian? P(E|NR) = P(E NR) / P(NR) , (.2 x .24) / .24 = .2
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