At a particular airport, 81% of the flights arrived on time last year. If 15 flights are randomly selected, find the probability that 11 or fewer flights are on time.
The correct answer and explanation is :
To solve this problem, we can model the number of flights arriving on time as a binomial random variable, because the outcome for each flight is either “on time” or “not on time,” and each flight is independent of the others.
Step 1: Define the variables
We know that:
- The probability of a flight arriving on time is $p = 0.81$.
- The probability of a flight not arriving on time is $q = 1 – p = 0.19$.
- The number of flights being considered is $n = 15$.
- We want to find the probability that 11 or fewer flights are on time, so we need to calculate $P(X \leq 11)$, where $X$ is the number of flights that arrive on time.
Step 2: Binomial probability formula
The probability mass function (PMF) for a binomial distribution is given by:
$$
P(X = k) = \binom{n}{k} p^k q^{n-k}
$$
Where:
- $\binom{n}{k}$ is the binomial coefficient, which calculates how many ways $k$ successes can occur in $n$ trials.
- $p^k$ is the probability of $k$ successes (flights on time).
- $q^{n-k}$ is the probability of $n-k$ failures (flights not on time).
To calculate $P(X \leq 11)$, we need to compute the sum of the individual probabilities from $X = 0$ to $X = 11$:
$$
P(X \leq 11) = \sum_{k=0}^{11} P(X = k) = \sum_{k=0}^{11} \binom{15}{k} (0.81)^k (0.19)^{15-k}
$$
Step 3: Using a normal approximation
Since the values of $n$ and $p$ are large enough, we can use the normal approximation to the binomial distribution for easier computation. The normal approximation to a binomial distribution $X$ is given by:
$$
X \sim N(\mu, \sigma^2)
$$
Where:
- $\mu = np = 15 \times 0.81 = 12.15$
- $\sigma = \sqrt{npq} = \sqrt{15 \times 0.81 \times 0.19} \approx 1.77$
We can approximate $P(X \leq 11)$ using the normal distribution with continuity correction. The continuity correction accounts for the fact that the binomial distribution is discrete, while the normal distribution is continuous. So, we will calculate:
$$
P(X \leq 11) \approx P\left( Z \leq \frac{11.5 – 12.15}{1.77} \right)
$$
Where $Z$ is the standard normal variable.
Step 4: Z-score calculation
$$
Z = \frac{11.5 – 12.15}{1.77} = \frac{-0.65}{1.77} \approx -0.37
$$
Using the standard normal table or a calculator, we find the probability corresponding to a Z-score of $-0.37$ is approximately $0.355$.
Conclusion
Thus, the probability that 11 or fewer flights are on time is approximately 0.355 or 35.5%.
This solution uses the normal approximation, which is valid for large $n$ and when $p$ is not too close to 0 or 1. The result shows that there is a 35.5% chance that 11 or fewer of the 15 randomly selected flights will be on time.