Question

according to flightstats.com, american airlines flights from dallas to chicago are on time 80% of the time. suppose 24 flights are randomly selected, and the number of on - time flights is recorded.
a. explain why this is a binomial experiment.
b. determine the values of n, p, q, and x.
c. find and interpret the probability that exactly 15 flights are on time.
d. find and interpret the probability that fewer than 15 flights are on time.
e. find and interpret the probability that at least 15 flights are on time.
f. find and interpret the probability that between 13 and 15 flights, inclusive, are on time.
g. using the formulas, calculate the mean and the standard deviation.

Explanation:

Step1: Binomial experiment conditions

A binomial experiment has the following conditions:

• Fixed number of trials (\(n\)).
• Two possible outcomes (success - on - time, failure - not on - time).
• Probability of success (\(p\)) is constant.
• Trials are independent.

Here, \(n = 24\) (fixed number of flights), \(p=0.8\) (probability of on - time flight, constant as per the given 80% on - time rate), each flight is an independent trial, and there are two outcomes (on - time or not on - time).

Step2: Values of \(n\), \(p\), \(q\), \(x\)

• \(n = 24\) (number of trials/ flights selected).
• \(p = 0.8\) (probability of success/ on - time flight).
• \(q=1 - p=1 - 0.8 = 0.2\) (probability of failure/ not on - time flight).
• \(x\) is the number of successes (on - time flights), and \(x\in\{0,1,\cdots,24\}\)

Step3: Probability that exactly \(x = 15\) flights are on - time

The binomial probability formula is \(P(X=x)={n\choose x}p^{x}q^{n - x}\)

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Interpretation: The probability that out of 24 randomly selected flights, exactly 15 are on - time is approximately \(0.0293\)

Step4: Probability that fewer than \(x = 15\) flights are on - time

\(P(X\lt15)=\sum_{k = 0}^{14}{24\choose k}(0.8)^{k}(0.2)^{24 - k}\)
Using a binomial probability calculator or software (e.g., in R: pbinom(14,24,0.8)), we get \(P(X\lt15)\approx0.0135\)
Interpretation: The probability that less than 15 out of 24 randomly selected flights are on - time is approximately \(0.0135\)

Step5: Probability that at least \(x = 15\) flights are on - time

\(P(X\geq15)=1 - P(X\lt15)\)
Since \(P(X\lt15)\approx0.0135\), then \(P(X\geq15)=1 - 0.0135 = 0.9865\)
Interpretation: The probability that 15 or more out of 24 randomly selected flights are on - time is approximately \(0.9865\)

Step6: Probability that between \(x_1 = 13\) and \(x_2 = 15\) flights (inclusive) are on - time

\(P(13\leq X\leq15)=\sum_{k = 13}^{15}{24\choose k}(0.8)^{k}(0.2)^{24 - k}\)

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Interpretation: The probability that between 13 and 15 (inclusive) out of 24 randomly selected flights are on - time is approximately \(0.0461\)

Step7: Mean and standard deviation

The mean of a binomial distribution is \(\mu=np\)
\(\mu=24\times0.8 = 19.2\)
The standard deviation is \(\sigma=\sqrt{npq}\)
\(\sigma=\sqrt{24\times0.8\times0.2}=\sqrt{3.84}\approx1.96\)
Interpretation: On average, out of 24 flights, we expect \(19.2\) flights to be on - time. The standard deviation of \(1.96\) measures the spread or variability around this mean value.

Answer:

a. This is a binomial experiment because: fixed number of trials (\(n = 24\)), two outcomes (on - time or not), constant probability (\(p = 0.8\)), and independent trials.
b. \(n = 24\), \(p=0.8\), \(q = 0.2\), \(x\in\{0,1,\cdots,24\}\)
c. Probability \(\approx0.0293\); Interpretation: Probability that exactly 15 out of 24 flights are on - time.
d. Probability \(\approx0.0135\); Interpretation: Probability that less than 15 out of 24 flights are on - time.
e. Probability \(\approx0.9865\); Interpretation: Probability that 15 or more out of 24 flights are on - time.
f. Probability \(\approx0.0461\); Interpretation: Probability that between 13 and 15 (inclusive) out of 24 flights are on - time.
g. Mean \(\mu = 19.2\), Standard deviation \(\sigma\approx1.96\); Interpretation: Average on - time flights is \(19.2\), spread around mean is \(\approx1.96\)