Questions: Assume that an airline operates a 158-seat McDonnel Douglas MD-90 on a particular route. Historically, the probability of a passenger showing up for a flight is 91%. 1. Assume that 158 tickets were sold. Let X be the number of passengers who showed up for the flight. a. Describe the distribution of X : X ~ ? (n=158, p=0.91) b. Find the probability that the flight is not full, in other words, find the probability that not all passengers will show up: P(X ≤ 157)=0.9 (Round the answer to 4 decimal places) c. Find the expected number of passengers who show up for the flight: E[X]=142 (Round the answer to the whole number) d. Find the expected number of empty seats by subtracting the E[X] from the plane capacity: 14 (Round the answer to the whole number) 2. Assume that the airline sells 7 more ticket(s). Let Y be the number of passengers who showed up for the flight. a. Describe the distribution of Y : Y ~ ? (n=, p=) b. Find the probability that more passengers will show up than the plane can carry: P(Y> )= (Round the answer to 4 decimal places)

Assume that an airline operates a 158-seat McDonnel Douglas MD-90 on a particular route. Historically, the probability of a passenger showing up for a flight is 91%. 1. Assume that 158 tickets were sold. Let X be the number of passengers who showed up for the flight. a. Describe the distribution of X : X ~ ? (n=158, p=0.91) b. Find the probability that the flight is not full, in other words, find the probability that not all passengers will show up: P(X ≤ 157)=0.9 (Round the answer to 4 decimal places) c. Find the expected number of passengers who show up for the flight: E[X]=142 (Round the answer to the whole number) d. Find the expected number of empty seats by subtracting the E[X] from the plane capacity: 14 (Round the answer to the whole number) 2. Assume that the airline sells 7 more ticket(s). Let Y be the number of passengers who showed up for the flight. a. Describe the distribution of Y : Y ~ ? (n=, p=) b. Find the probability that more passengers will show up than the plane can carry: P(Y> )= (Round the answer to 4 decimal places)
Transcript text: Assume that an airline operates a 158-seat McDonnel Douglas MD-90 on a particular route. Historically, the probability of a passenger showing up for a flight is $91\%$. 1. Assume that 158 tickets were sold. Let $X$ be the number of passengers who showed up for the flight. a. Describe the distribution of X : \[ X \sim ? \vee(n=158, p=0.91) \] b. Find the probability that the flight is not full, in other words, find the probability that not all passengers will show up: \[ P(X \leq 157)=0.9 \text { (Round the answer to } 4 \text { decimal places) } \] c. Find the expected number of passengers who show up for the flight: \[ E[X]=142 \text { (Round the answer to the whole number) } \] d. Find the expected number of empty seats by subtracting the $E[X]$ from the plane capacity: \[ 14 \text { (Round the answer to the whole number) } \] 2. Assume that the airline sells 7 more ticket(s). Let $Y$ be the number of passengers who showed up for the flight. a. Describe the distribution of $Y$ : \[ Y \sim ? \vee(n=\square, p=\square) \] b. Find the probability that more passengers will show up than the plane can carry: \[ P(Y>\square)=\square \text { (Round the answer to } 4 \text { decimal places) } \]
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Solution

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Solution Steps

Step 1: Distribution of \( X \)

The number of passengers who showed up for the flight, denoted as \( X \), follows a binomial distribution given by:

\[ X \sim \text{Binomial}(n=158, p=0.91) \]

Step 2: Probability that the Flight is Not Full

To find the probability that not all passengers will show up, we calculate:

\[ P(X < 158) = 1 - P(X = 158) = 1.0 \]

This indicates that the probability of the flight being not full is:

\[ P(X < 158) = 1.0 \]

Step 3: Expected Number of Passengers

The expected number of passengers who show up for the flight is calculated as:

\[ E[X] = n \cdot p = 158 \cdot 0.91 = 144.0 \]

Step 4: Expected Number of Empty Seats

The expected number of empty seats can be found by subtracting the expected number of passengers from the plane's capacity:

\[ \text{Expected Empty Seats} = \text{Plane Capacity} - E[X] = 158 - 144 = 14.0 \]

Final Answer

  • Distribution of \( X \): \( X \sim \text{Binomial}(n=158, p=0.91) \)
  • Probability that the flight is not full: \( P(X < 158) = 1.0 \)
  • Expected number of passengers: \( E[X] = 144 \)
  • Expected number of empty seats: \( 14 \)

Thus, the final boxed answers are:

\[ \boxed{P(X < 158) = 1.0} \] \[ \boxed{E[X] = 144} \] \[ \boxed{\text{Expected Empty Seats} = 14} \]

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