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Statistics Assignment - Bayes Theorem

Added on - 23 Mar 2020

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SUBJECT: STATISTICSORDER ID: 62758Q1.Note that Q1 and Q2 does not apply Bayes Theorem but this concept is applicable in SQ3(a)Probability of getting a box from Thailand = 6000/10000=3/5Probability of getting a box from Malaysia=4000/10000=2/5Probability of getting a box from Thailand that is damaged=200/6000=1/30Probability of getting a box from Malaysia that is damaged=365/4000=73/800Therefore the probability of selecting a box at random and get a damaged fruit = (3/5*1/30) +(2/5*73/800)=113/2000(b)Probability of selecting a box at random that comes from Malaysia=2/5Probability of selecting a box at random with an overripe fruit that comes fromMalaysia=295/4000=59/800Thus the probability=2/5*59/800=59/2000Q2.25% emits excessive pollutants,75% does not emit excessive pollutants0.95 is the probability that a car emitting excessive pollutants fail0.1 is the probability that a car not emitting excessive pollutants failProbability of a car emitting excessive pollutants and fails the test is=0.25*0.95=0.2375Q3.Probability of selecting any machine at random=1/3
Machine 1 defective springs=0.02, produces 0.35 of the total springsMachine 2 defective springs=0.01, produces 0.25 of the total springsMachine 3 defective springs=0.03produces 0.4 of the total the springs(a)Probability of selecting a defective spring=(1/3*0.02)+(1/3*0.01)+(1/3*0.03)=1/50(b)Probability of(3/defective)=P(3)*P(D/3) divided by P(3)*P(D/3)+P(1)*P(1/D)+P(2)*P(2/)Where d= defective(0.03*0.04)/P(0.03*0.04)+(0.02*0.35)+P(0.01*0.25)=1.2*10^-3/0.0107=0.11215This is where Bayes theorem applies but question 1 and 2 are not using Bayes ConceptQ4.q^(n-x)*p^xp=0.15,q=1-p,1-0.3Q=0.7(a)P(x=5)0.7^(15-5)*0.3^515C5*0.7^10*0.3^5(15C5 all I mean is 15 combination )=0.2061(b)P(x<_5)
(15C0multiplied by 0.7^15multiplied by 0.3^0)+ (15C1multiplied by 0.7^14multiplied by 0.3^1) + (15C2othe product of 0.7^13o the product of 0.3^2) + (15C3multiplied by 0.7^12multiplied by 0.3^3) +(15C4multiplied by 0.7^11multiplied by 0.3^4) +0.2061=0.00475+0.0305+0.0916+0.1700+0.2186+0.2061=0.72155(c)E(X)=np=E(X^2) =np^2E(X) =0.3*15=4.5E(X^2) =0.3^2*15=1.35Q5.P(A=a) = (0.2)*(0.8^(x-1))Where a=1, 2, 3a=0.2*0.8^ (1-1) +0.2*0.8^ (2-1) +0.2*0.8^ (3-1)=0.2+0.16+0.128=0.488Y=X-1X=0.488x=0.2*0.8^1.1=0.2=0.488-0.2=0.288Q6.(a)1-p=q
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