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

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**.