Question

Assume that 23.6​% of people have sleepwalked. Assume that in a random sample of 1505 ​adults, 400 have sleepwalked.

1. a. Assuming that the rate of 23.6% is​ correct, the probability that 400 or more of the 1505 adults have sleepwalked is [ ]

b. Is that result of 611 or more significantly​ high?

c.  What does the result suggest about the rate of 23.6%?

2. In a study of 437 comma 485 cell phone​ users, it was found that 54 developed cancer of the brain or nervous system. Assuming that cell phones have no​ effect, there is a 0.000168 probability of a person developing cancer of the brain or nervous system. We therefore expect about 74 cases of such cancer in a group of 437 comma 485 people. Estimate the probability of 54 or fewer cases of such cancer in a group of 437 comma 485 people. What do these results suggest about media reports that cell phones cause cancer of the brain or nervous​ system?

(a) P(X≤54)=

3. In a survey of 1169 ​people, 807 people said they voted in a recent presidential election. Voting records show that 66​% of eligible voters actually did vote. Given that 66​% of eligible voters actually did​ vote, (a) find the probability that among 1169 randomly selected​ voters, at least 807 actually did vote.​ (b) What do the results from part​ (a) suggest?

(a) P(X≥807)

Answer 1

1a)

Sample size , n =    1505          
Probability of an event of interest, p =   0.236          
right tailed              
X >   400          
              
Mean = np =    355.18          
std dev ,σ=√np(1-p)=   16.4729          
              
P(X >   400   )      
              
Z=(X - µ ) / σ =        (400-355.18)/16.4729)=       2.721
              
=P(Z >   2.721   ) =    0.0033  

1b)

Sample size , n =    1505          
Probability of an event of interest, p =   0.236          
right tailed              
X >   611          
              
Mean = np =    355.18          
std dev ,σ=√np(1-p)=   16.4729          
              
P(X >   611   )      
              
Z=(X - µ ) / σ =        (611-355.18)/16.4729)=       15.530
              
=P(Z >   15.530   ) =    0.0000  
yes, result of 611 or more significantly​ high

c)

rate = 400/1505 = 0.2658

so, rate of 23.6 is significantly low