2 In a large university 75 of students live in dormitories A
2. In a large university, 75% of students live in dormitories. A random sample of 5 students is selected. a. What is the probability that the sample contains exactly three students who live in the dormitories? b. What is the probability that the sample contains no students who lives in the dormitories? c. What is the probability that the sample contains more than three students who live in the dormitories? d. What is the expected number of students (in the sample) who live in the dormitories? e. What is the variance of students (in the sample) who live in the dormitories?
Solution
a)
Note that the probability of x successes out of n trials is          
           
 P(n, x) = nCx p^x (1 - p)^(n - x)          
           
 where          
           
 n = number of trials =    5      
 p = the probability of a success =    0.75      
 x = the number of successes =    3      
           
 Thus, the probability is          
           
 P (    3   ) =    0.263671875 [ANSWER]
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b)
Note that the probability of x successes out of n trials is          
           
 P(n, x) = nCx p^x (1 - p)^(n - x)          
           
 where          
           
 n = number of trials =    5      
 p = the probability of a success =    0.75      
 x = the number of successes =    0      
           
 Thus, the probability is          
           
 P (    0   ) =    0.000976563 [ANSWER]
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c)
Note that P(more than x) = 1 - P(at most x).          
           
 Using a cumulative binomial distribution table or technology, matching          
           
 n = number of trials =    5      
 p = the probability of a success =    0.75      
 x = our critical value of successes =    3      
           
 Then the cumulative probability of P(at most x) from a table/technology is          
           
 P(at most   3   ) =    0.3671875
           
 Thus, the probability of at least   4   successes is  
           
 P(more than   3   ) =    0.6328125 [ANSWER]
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d)
E(x) = n p = 5*0.75 = 3.75 [ANSWER]
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 e)
variance = np(1-p) = 5*0.75*(1-0.75) = 0.9375 [ANSWER]


