uestion: 1 pt ve of US adults complete parte through cement Assuming the sample

Question

uestion: 1 pt ve of US adults complete parte through cement Assuming the sample is representati nd the probatiity that both aduts dre out more hananne per week probatiny that bon adult dine more than once per week is ree decima places as needed. the probabilty hat felher adut dines out more than onou per week. probability tha neither adult dnes out more han orce per week is und three deomalplaces as needed to one of the two aduts dines out more than once per week End the probablity thatatlea e probabity that anleait one the two aduts eines out move than once per weekin of 10 select your answeris

Explanation / Answer

Solution:-

In a sample of 1000 adults, 219 dine out at a resturant.

So, the probability of dinning out is 219 / 1000 = 0.219

(a) Probability that both dine out = 0.219 * 0.219 = 0.047961 or 0.480

(b) Probability that neither adult dines out = (1 - 0.219) * (1 - 0.219)

= 0.781 * 0.781

= 0.609961 or 0.610

(c) Probability that at least one of the two dines out = 0.219 * (1 - 0.219) + (1 - 0.219) * 0.219

= 0.219 * 0.781 + 0.781 * 0.219

= 0.171039 + 0.171039

= 0.342078 or 0.342

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