Final answer:
The probability of 10 or fewer passengers from a sample of 125 being on United Airlines, given the airline controls 15% of the market, is calculated to be approximately 0.0076 using normal approximation. This is not one of the provided options, suggesting there might be a mistake in the choices given or a miscalculation.
Step-by-step explanation:
The probability that from a random sample of 125 passengers, 10 or fewer were on United Airlines flights when the airline controls 15% of the domestic market can be calculated using the normal approximation to the binomial distribution. First, we calculate the mean (μ) and the standard deviation (σ) of the binomial distribution:
- μ = n * p = 125 * 0.15 = 18.75
- σ = √(n * p * (1 - p)) = √(125 * 0.15 * 0.85) ≈ 3.6056
We then standardize the value for 10 passengers using the z-formula: z = (X - μ) / σ. Substituting the values:
z = (10 - 18.75) / 3.6056 ≈ -2.43
Using standard normal distribution tables or a calculator, we find the probability of z being less than or equal to -2.43, which is approximately 0.0076. This probability indicates that it's quite unlikely for there to be 10 or fewer passengers from United Airlines in a random sample of 125 passengers.
However, this value is not among the provided choices. It seems there might have been a mistake as none of the provided options closely match the calculated probability. Always ensure to double-check calculations and consider using software tools or calculators designed for statistical analysis for confirmation.