Asked by Jess
                Suppose the number of cell phones in a household has a binomial distribution with parameters n=24 and p=25%.
Find the probability of a household having:
(a) 11 or 16 cell phones
(b) 14 or fewer cell phones
(c) 18 or more cell phones
(d) fewer than 16 cell phones
(e) more than 14 cell phones
            
        Find the probability of a household having:
(a) 11 or 16 cell phones
(b) 14 or fewer cell phones
(c) 18 or more cell phones
(d) fewer than 16 cell phones
(e) more than 14 cell phones
Answers
                    Answered by
            mathhelper
            
    same type of question as the previous one, except even more 
tedious button pushing.
e.g.
c) 18 or more
= C(24,18) (.25^18) (.75^6) + C(24,19)(.25^19)(.75^5) + ....
.... + C(24,24)(.25^24)(.75^0)
= ....
good grief!, hope you have a good calculator
    
tedious button pushing.
e.g.
c) 18 or more
= C(24,18) (.25^18) (.75^6) + C(24,19)(.25^19)(.75^5) + ....
.... + C(24,24)(.25^24)(.75^0)
= ....
good grief!, hope you have a good calculator
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