In a survey of consumers aged 12 and older respondents me asked how many cell ph

Question

In a survey of consumers aged 12 and older respondents me asked how many cell phones me in use by the household (No two respondents were from the household) Among the respondents 204 answered 'none ' 291 sari 'one' 364 said 'two ', 140 sari 'three ' and 111 responded with four or more A survey respondent is selected at random Find the probability that his/her household has four or more cell phones in use Is it unlikely for a household lo have four or more Ml phones In use? Consider an event to be unlikely it its probability is less than or equal to 0 05 P(four or more cell phones) (Round to three decimal places as needed) Is it unlikely for a household to have low or more cell phones in use? No because the probability of a respondent with four or more cell phones in use is less than or equal to 0 05 Yes. because the probability of a respondent with low or more cell phones in use is less than or equal to 0 05 No because the probability of a respondent with four or more cell phones in use is greater than 0 05 Yes because the probability of a respondent with four or more cell phones in use is greater than 0 05

Explanation / Answer

Yes you are correct, Please see below the updated solution

Type1= Respondent Answered with No cell phones= 204
Type2= Respondent Answered with one cell phones=291
Type3=Respondent Answered with two cell phones=364
Type4= Respondent Answered with three cell phones=140
Type5= Respondent Answered with four or more cell phones=111

A Survey respondent is selected at random. Find the probablity that his/her household has four or more cell ohones.

i.e. we need to find P(Respondent Answered with four or more cell phones/ given one respondent)
i.e. P(Type5/given one respondent)

Applying Bayes theorem

P(Type5/given one respondent)= {P(Type5)* P(respondent/Type5)}/{P(Type5)* P(respondent/Type5)+P(Type4)* P(respondent/Type4)+P(Type3)* P(respondent/Type3)+P(Type2)* P(respondent/Type2)+P(Type1)* P(respondent/Type1)}

We have P(Type5)= 111/1110
P(Type4)=140/1110
P(Type3)=364/1110
P(Type2)=291/1110
P(Type1)=204/1110

And P(respondent/Type5)= Probablity of respondent given this is from type5= 1/111
P(respondent/Type4)= Probablity of respondent given this is from type4= 1/140
P(respondent/Type3)= Probablity of respondent given this is from type3= 1/364
P(respondent/Type2)= Probablity of respondent given this is from type2= 1/291
P(respondent/Type1)= Probablity of respondent given this is from type2= 1/204

Putting these value in P(Type5/given one respondent), we get

=(111/1110)*(1/111)/{(111/1110)*(1/111)+(140/1110)*(1/140)+(364/1110*1/364)+(291/1110*1/291)+(204/1110*1/204)}
=1/1110/{1/1110+1/1110+1/1110+1/1110+1/1110}
=1/5
=0.2

P(Type5/given one respondent)=0.2 which is greater than 0.05 , so that given event is not unlikley i.e. not unlikley for household to have four or more cell phone in use.

So Answer is C- because the probability of a respondent with four or more cell phone in use is greater than 0.05

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