Answer: P=0
Step-by-step explanation:
Let X be a random variable that denotes the number of right handed students in the class.
X follows binomial distribution with n=230 and
p=1−0.15=0.85
Using Normal approximation to approximate the required binomial probability:
Mean, μ=np=230×0.85=195.5
Standard deviation, σ=√np(1−p)=√230×0.85(1−0.85)=5.42
Probability that a right-handed student in one of these classes is forced to use a left arm seat
We are given that 30 seats are designed for lefties, the rest of the 220 seats are for righties.
Incase there are more than 220 righties, then one or more righties are forced to use the left handed tablets.
Hence the required probability is;
P(X>220)=P((x−μ)/σ < (220.5−195.5)/5.42)
=P(Z>4.62)=1−P(Z<4.62). (Using standard normal distribution table)
P=1- 1= 0
P= 0