solution

a. The probability that a randomly selected coffee drinker would take no coffee breaks during the day is given by 

$p(0)=0.28$

b. The probability that a randomly selected coffee drinker would take more than 2 coffee breaks during the day is given by $p(3) + p(4) + p(5)= 0.18$

Answer

a. 0.28  b. 0.18 c. $\mu =1.32, ~ \sigma \approx 1.2$  d. 0.94