According to American Airlines, Flight 215 from Orlando to Los Angeles is on time 90% of the time. Suppose 150 flights are randomly selected. Find the probability that few than 125 flights are on time.
To find the probability that fewer than 125 flights are on time, we need to use the binomial probability formula.
The binomial probability formula is given by:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- P(X=k) is the probability of getting exactly k successes
- n is the total number of trials or flights in this case (150)
- k is the number of successes (flights on time) we are interested in (fewer than 125 in this case)
- p is the probability of success (flight on time), which is 0.90 according to American Airlines
- (n choose k) is the binomial coefficient and can be calculated as n! / (k! * (n - k)!)
Now, let's calculate the probability using this formula:
P(X < 125) = P(X=0) + P(X=1) + P(X=2) + ... + P(X=124)
We can use a statistical calculator or software to simplify this calculation. However, I will provide the steps to calculate the probabilities manually.
First, calculate (n choose k) and p^k * (1-p)^(n-k) for each value of k from 0 to 124.
For k = 0:
(n choose k) = 150! / (0! * (150 - 0)!) = 1
p^k * (1-p)^(n-k) = 0.90^0 * 0.10^(150-0) = 0.10^150 = very close to 0
For k = 1:
(n choose k) = 150! / (1! * (150 - 1)!) = 150
p^k * (1-p)^(n-k) = 0.90^1 * 0.10^(150-1) = 0.90^1 * 0.10^149
Repeat this calculation for k = 2 to 124.
Next, add up all the individual probabilities (P(X=0), P(X=1), P(X=2), ..., P(X=124)) to get the final probability of fewer than 125 flights being on time.
P(X < 125) = P(X=0) + P(X=1) + P(X=2) + ... + P(X=124)
Again, this calculation can be time-consuming manually, so it's recommended to use a statistical calculator or software to simplify the process.
To find the probability that fewer than 125 flights are on time, we can use the binomial distribution formula.
The binomial distribution formula is given by P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where:
- P(X=k) is the probability of getting exactly k successes
- C(n, k) represents the combination of n items taken k at a time
- p is the probability of success on a single trial
- (1-p) is the probability of failure on a single trial
- n is the total number of trials
In this case, the probability of a flight being on time (success) is p = 0.9, since Flight 215 from Orlando to Los Angeles is on time 90% of the time. So, the probability of a flight being delayed (failure) is (1 - p) = 1 - 0.9 = 0.1.
We want to find the probability that fewer than 125 flights are on time, so we need to calculate P(X < 125), where X represents the number of flights on time.
Let's substitute these values into the formula:
P(X < 125) = sum of P(X=k) for k = 0 to 124
Now, let's calculate it step-by-step:
P(X < 125) = sum of P(X=k) for k = 0 to 124
P(X < 125) = Σ C(150, k) * (0.9)^k * (0.1)^(150-k) for k = 0 to 124
Since calculating this sum manually can be challenging, it's more convenient to use a statistical software or calculator to find the cumulative probability.