Question

12. In a survey of 1364 people, 1001 people said they voted in a recent presidential election. Voting records show that 71​% of eligible voters actually did vote. Given that 71​% of eligible voters actually did​ vote, (a) find the probability that among 1364 randomly selected​ voters, at least 1001 actually did vote (b). What do the results from part​ (a) suggest?

P(X≥1001)= ____

(Round to four decimal places as​ needed.)

13. In a study of 218,247 cell phone​ users, it was found that 51 developed cancer of the brain or nervous system. Assuming that cell phones have no​ effect, there is a 0.000286 probability of a person developing cancer of the brain or nervous system. We therefore expect about 63 cases of such cancer in a group of 218,247 people. Estimate the probability of 51 or fewer cases of such cancer in a group of 218,247 people. What do these results suggest about media reports that cell phones cause cancer of the brain or nervous​ system? ​

​P(X≤51)=____

​(Round to four decimal places as​ needed.)

14. Based on a recent​ survey, 20​% of adults in a specific country smoke. In a survey of 100 students, it is found that 17 of them smoke.

Find the probability that should be used for determining whether the 20​% rate is correct for students.

? (Round to four decimal places as​ needed.)

What can be​concluded?

(Round to four decimal places as​ needed.)

Answer 1

Q 12 : Answer is 0.0260 . Because , mean = 1364*0.71 = 968.44 and standard deviation is 16.75 (according to binomial distribution (npq)^1/2 = 16.75 )

we use normal distrbution , atleast 1001 probability is 0.0260 .

Q 13 : Answer is 0.0741 . Let , n= 218247 . p = 0.000286

mean =np = 218247*0.000286 = 62.4

standard deviation = 7.89

we use normal distrbution , atmost 51 probability is 0.0741 .

Q 14 : Answer is 0.2711 . Let n= 100 , p = 0.20 . x = 17 . probabilty = 0.2711

using binomial distribution . we get probability = 0.2711 .