question archive Innocent until proven guilty? In Japanese criminal trials, about 95% of the defendants are found guilty

Innocent until proven guilty? In Japanese criminal trials, about 95% of the defendants are found guilty

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Innocent until proven guilty? In Japanese criminal trials, about 95% of the defendants are found guilty. In the United States, about 60% of the defendants are found guilty in criminal trials. (Source: The Book of Risks, by Larry Laudan, John Wiley and Sons) Suppose you are a news reporter following four criminal trials. (For each answer, enter a number.)

(a) If the trials were in Japan, what is the probability that all the defendants would be found guilty? (Round your answer to three decimal places.)

What is this probability if the trials were in the United States? (Round your answer to three decimal places.)

(b) Of the four trials, what is the expected number of guilty verdicts in Japan? (Round your answer to two decimal places.)

 verdicts

What is the expected number in the United Sates? (Round your answer to two decimal places.)

 verdicts

c) What is the standard deviation in Japan? (Round your answer to two decimal places.)

 verdicts

What is the standard deviation in the United States? (Round your answer to two decimal places.)

 verdicts

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Answer:

a)

P(all the dependents found guilty in japan)

= 0.815

P(all the dependents found guilty in US)

= 0.130

b) Expected number of guilty verdicts in Japan

= 3.80

Expected number of guilty verdicts in US

= 2.40

C)

the standard deviation in Japan

= 0.44

the standard deviation in US

= 0.98

Step-by-step explanation

Here given context is related to binomial distribution with parameters,

n = 4

a) in Japan, p = 0.95

P(all the dependents found guilty)

= (0.95)4

= 0.815

 

In US, p = 0.6

P(all the dependents found guilty)

= (0.60)4

= 0.130

 

b)

Expected number of guilty verdicts in Japan

= np

= 4*0.95

= 3.80

 

Expected number of guilty verdicts in US

= 4*0.60

= 2.40

 

c)

the standard deviation in Japan

= sqrt(np*1-p)

= sqrt(4*0.95*0.05)

= 0.44

 

the standard deviation in US

= sqrt(np*1-p)

= sqrt(4*0.60*0.4)

= 0.98