An airline finds that 5% of the persons who make reservations on a certain flight do not show up for the flight. if the airline
sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?
Probability of success (showing up) = 1-0.05=0.95 is constant and known. Trials are Bernoulli (show or no show). Trials are independent and random (assumed from context) Number of trials is known, n=160. All being satisfied, we can then model with binomial distribution, where P(x)=C(n,x)p^x*(1-p)^(n-x) where C(n,x)=n!/(x!(n-x)!)
Here we look for P(X<=155)=P(X=0)+P(X=1)+P(X=2)+...+P(X=155) =0.9061461 (using technology, or add up 156 values calculated, or read from binomial distribution table).
Alternatively, the normal approximation can be used, when n is large. mean=np=160*0.95=152 standard deviation=sqrt(np(1-p))=2.75681 Apply continuity correction, x=155.5 Z=(155.5-152)/2.75681=1.26958 P(z<=Z)=0.89788 (read from normal distribution tables) Error=(0.89788-0.9061461)*100%=-0.83% The approximation is considered good considering p=0.95 is quite skewed, but compensated by n>>50.
An object that is cooler, or colder, than a second object means that the first object has a lower, or lesser, temperature than the second object. So, we write the inequality that states that -5°C is less than 4°C.