灵鹊做题网英文的统计学高数,超急
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英文的统计学高数,超急
QUESTION 4
(a) A student answers a multiple-choiceexamination question which offers four possible answers.Suppose that theprobability that the student knows the answer to the question is 0.8 and theprobability that he/she will guess is 0.2.Assume that,if the student guesses,the probability of guessing correctly is 0.25.If the student correctly answersthe question,what is the probability that he/she did so by guessing?（Hint：usingBayes’ Theorem）
(6 marks)
(b) In anexperiment,if a mouse is administered dosage level Aof a certain(harmless) hormone then thereisa0.2probability that the mouse will show signsofaggression within one minute.For dosage levels B and C,the probabilities are0.5and0.8,respectively.Ten mice are given exactly the same dosagelevelof the hormone and,of these,exactly6showssignsof aggression within one minuteof receiving thedose.
i) Calculate the probability ofthis happening for the dosage level A.(This is essentially a Binomial randomvariable problem)
(6 marks)
ii) Assuming that each of the threedosage levels was equally likely to have been administeredin thefirst place (withaprobabilityof1/3),use Bayes’ Theorem to evaluate thelikelihood of the dosage level A given that6 outofthe10micewereobservedtoreactinthisway.
(8 marks)
Total(20 marks)
QUESTION 4
(a) A student answers a multiple-choiceexamination question which offers four possible answers.Suppose that theprobability that the student knows the answer to the question is 0.8 and theprobability that he/she will guess is 0.2.Assume that,if the student guesses,the probability of guessing correctly is 0.25.If the student correctly answersthe question,what is the probability that he/she did so by guessing?（Hint：usingBayes’ Theorem）
(6 marks)
(b) In anexperiment,if a mouse is administered dosage level Aof a certain(harmless) hormone then thereisa0.2probability that the mouse will show signsofaggression within one minute.For dosage levels B and C,the probabilities are0.5and0.8,respectively.Ten mice are given exactly the same dosagelevelof the hormone and,of these,exactly6showssignsof aggression within one minuteof receiving thedose.
i) Calculate the probability ofthis happening for the dosage level A.(This is essentially a Binomial randomvariable problem)
(6 marks)
ii) Assuming that each of the threedosage levels was equally likely to have been administeredin thefirst place (withaprobabilityof1/3),use Bayes’ Theorem to evaluate thelikelihood of the dosage level A given that6 outofthe10micewereobservedtoreactinthisway.
(8 marks)
Total(20 marks)
先说清贝叶斯法则 P(A交B)=P(B|A)*P(A)=P(A|B)*P(B)
so P(A|B)=P(B|A)*P(A)/P(B)
Q4(a)问的是P(guess I correct)
P(correct I guess) = 0.25
P(guess) = 0.2
P(not guess)=0.8
P(correct I not guess)=1
0.2*0.25/(0.2*0.25+0.8*1) = 0.0588
b) 1) 二项式分布binomial distribution：C(10,6)*0.2^6*0.8^4 = 0.005505024
2) 问P(dosage A I 6 positive)
已知
P(dosage A) =1/3
P(dosage B) =1/3
P(dosage C) =1/3
P(6 positive I dosage A) = 0.005505024
同样可计算
P(6 positive I dosage B)= C(10,6)*0.5^6*0.5^4 = 0.205078125
P(6 positive I dosage C)= C(10,6)*0.8^6*0.2^4 = 0.088080384
答案 = (0.005505024*1/3)/(0.005505024*1/3+0.205078125 *1/3 + 0.088080384*1/3)
= 0.018432
so P(A|B)=P(B|A)*P(A)/P(B)
Q4(a)问的是P(guess I correct)
P(correct I guess) = 0.25
P(guess) = 0.2
P(not guess)=0.8
P(correct I not guess)=1
0.2*0.25/(0.2*0.25+0.8*1) = 0.0588
b) 1) 二项式分布binomial distribution：C(10,6)*0.2^6*0.8^4 = 0.005505024
2) 问P(dosage A I 6 positive)
已知
P(dosage A) =1/3
P(dosage B) =1/3
P(dosage C) =1/3
P(6 positive I dosage A) = 0.005505024
同样可计算
P(6 positive I dosage B)= C(10,6)*0.5^6*0.5^4 = 0.205078125
P(6 positive I dosage C)= C(10,6)*0.8^6*0.2^4 = 0.088080384
答案 = (0.005505024*1/3)/(0.005505024*1/3+0.205078125 *1/3 + 0.088080384*1/3)
= 0.018432