Thread: Normal approximation

Could someone help me solve this please

Suppose that a particular aircraft flight has 282 passenger seats in the aircraft, and that passengers will turn up independently of each other to take the flight with probability p. The airline overbooks the flight with excess of 20 more passengers than there are seats in the aircraft. Using a normal approximation, what is the value of p such that the probability that the flight is overbooked is equal to 0.01?

Do you use X~Bin(282,p) or X~Bin(302,p) and then P(x>282) or P(282<X<302)?

Originally Posted by Jpmps

Could someone help me solve this please

Suppose that a particular aircraft flight has 282 passenger seats in the aircraft, and that passengers will turn up independently of each other to take the flight with probability p. The airline overbooks the flight with excess of 20 more passengers than there are seats in the aircraft. Using a normal approximation, what is the value of p such that the probability that the flight is overbooked is equal to 0.01?

Do you use X~Bin(282,p) or X~Bin(302,p) and then P(x>282) or P(282<X<302)?

The probability that the plane is overbooked is where The normal approximation is that:



and the required probability using this approximation is that

CB