an airline company knows that the chabxd of a passenger not showing up for a flight is 4.37%. a flight on an Airbus 431 is fully booked with 185 passengers. find the mean and standard deviation of the number of passengers who do not show up for their flight

1 answer

To find the mean and standard deviation of the number of passengers who do not show up for their flight, we can model the situation using a binomial distribution.

Let:

  • \( n = 185 \) (the total number of passengers),
  • \( p = 0.0437 \) (the probability that a passenger does not show up).

Mean

The mean \( \mu \) of a binomial distribution is given by the formula: \[ \mu = n \cdot p \] Substituting the values, we have: \[ \mu = 185 \cdot 0.0437 = 8.0945 \] So the mean number of passengers who do not show up is approximately 8.09.

Standard Deviation

The standard deviation \( \sigma \) of a binomial distribution is given by the formula: \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \] Calculating \( 1 - p \): \[ 1 - p = 1 - 0.0437 = 0.9563 \]

Now substituting the values into the standard deviation formula: \[ \sigma = \sqrt{185 \cdot 0.0437 \cdot 0.9563} \] Calculating this step-by-step:

  1. Calculate \( n \cdot p \cdot (1 - p) \): \[ 185 \cdot 0.0437 \cdot 0.9563 \approx 185 \cdot 0.0418 \approx 7.733 \]

  2. Now, take the square root: \[ \sigma \approx \sqrt{7.733} \approx 2.78 \]

Summary

  • Mean number of passengers who do not show up: approximately 8.09
  • Standard deviation of the number of passengers who do not show up: approximately 2.78.