Problem 1SE

An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0.90. If the airline sells 105 tickets, what is the probability that everyone who appears for the flight will get a seat?

Solution 1SE

Step1 of 3:

We have An airplane and it has 100 seats of passengers.

Assume that the P( a person holding a ticket appears for the flight) = 0.90.

That is x = 100 p = 0.90.

We need to the probability that everyone who appears for the flight will get a seat when the airline sells 105 tickets.

Step2 of 3:

Here X be the random variable it represents the number of people out of 105 who appear for the flight. That is n = 105.

Let X follows binomial distribution with parameters “n and p ”.

Then the probability mass function of binomial distribution is given by:

, x = 0,1,2,...,n.

Where,

n = sample size

x = random variable

p = probability of success

q = 1 - p (probability of failure)

= 1 - 0.90

= 0.10

Now,

1).Mean of the binomial distribution is

=

= 94.5

Hence, = 94.5.

2).Standard deviation of binomial distribution is

=

=

= 3.0740

Hence, =3.0740.

Step3 of 3:

Consider the Z statistics

We need to find

Now,

=

=

=

value is obtained from standard normal table(area under normal curve) then

= 0.9744 (In area under normal curve we have to see row 1.9 under column 0.05)

Therefore, 0.9744.

Conclusion:

Therefore, the probability that everyone who appears for the flight will get a seat when the airline sells 105 tickets is 0.9744.