Question

A lecture hall has 180 seats with folding arm​ tablets, 30 of which are designed for​ left-handers. The average size of classes that meet there is 165​, and we can assume that about 14​% of students are​ left-handed. What's the probability that a​ right-handed student in one of these classes is forced to use a lefty arm​ tablet?

Answer 1

Probability that the students are right-handed = 1 - 0.14 = 0.86

Let X be the number of right-handed student in the class. Then X ~ Binomial(n = 165, p = 0.86)

Number of seats designed for​ right-handers = 180 - 30 = 150

Probability that a​ right-handed student in one of these classes is forced to use a lefty arm​ tablet

= P(X > 150)

Using Normal approximation to Binomial distrinution, X will follow Normal distribution with mean = np = 165 * 0.86 = 141.9 and standard deviation = = = 4.457

P(X > 150) = P[X > 150.5] (Using Continuity correction)

= P[Z > (150.5 - 141.9)/4.457]

= P[Z > 1.93]

= 0.0268