To solve this problem, we need to first determine the probability that some number of the booked passengers (out of the 185) do not show up for the flight. Given that the probability of a passenger not showing up is \( p = 0.0437 \) (or 4.37%), the probability of a passenger showing up is \( q = 1 - p = 0.9563 \).
Let \( X \) be the random variable representing the number of passengers who show up. Since we are dealing with a large number of passengers and a small probability of not showing up, we can use a normal approximation to the binomial distribution.
Step 1: Determine the mean and standard deviation of \( X \)
The number of passengers that show up (X) follows a binomial distribution with parameters \( n = 185 \) and \( p = 0.9563 \).
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The mean \( \mu \) of a binomial distribution is calculated as: \[ \mu = n \cdot q = 185 \cdot 0.9563 \approx 176.80 \]
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The variance \( \sigma^2 \) is calculated as: \[ \sigma^2 = n \cdot p \cdot q = 185 \cdot 0.0437 \cdot 0.9563 \approx 3.4272 \]
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Therefore, the standard deviation \( \sigma \) is: \[ \sigma \approx \sqrt{3.4272} \approx 1.85 \]
Step 2: Calculate the probability that 181 or more passengers show up
To find the probability that there are 4 or fewer empty seats, we need to find the probability that at least \( 181 \) passengers show up (which means \( 185 - 181 = 4 \) passengers could be accommodated in standby).
So, we need to find \( P(X \geq 181) \), which is equivalent to \( P(X \leq 180) \) using the complement:
Step 3: Standardize and use the normal distribution
We can transform our binomial random variable \( X \) to a standard normal variable \( Z \): \[ Z = \frac{X - \mu}{\sigma} \]
Now, we need to calculate \( P(X \leq 180) \): \[ P(X \leq 180) = P\left(Z \leq \frac{180 - 176.80}{1.85}\right) = P\left(Z \leq \frac{3.2}{1.85}\right) \approx P(Z \leq 1.73) \]
Step 4: Find the probability from the Z-table
Using the Z-table, we find: \[ P(Z \leq 1.73) \approx 0.9582 \]
Step 5: Calculate the probability for standby passengers
Since we are interested in the probability that at least 4 standby passengers are allowed on the plane, we want \( P(X \leq 180) \).
Therefore, the probability that 4 or fewer empty seats are available for standby passengers (which is equivalently that \( 181 \) or more passengers show up) is: \[ P(X \geq 181) = 1 - P(X \leq 180) = 1 - 0.9582 = 0.0418 \]
Final Answer
The probability that at least 4 standby passengers will be allowed on the plane is approximately 0.0418, or 4.18%.