The Ace AirLines has a policy of overbooking flights. The random variable x represents the number of passengers who cannot be boarded because there were more passengers that seats.

x P(x)
0 0.061
1 0.132
2 0.255
3 0.345
4 0.207

a)Is this a probability distribution? How do you know?
b)What is the mean number of passengers left at the gate?
c)What is the variance of the number of passengers left at the gate?
d) What is the standard deviation of the number of passengers left at the gate?

a) Is this a probability distribution? How do you know?

Yes, this is a probability distribution. We can tell because the sum of all the probabilities equals 1.

b) What is the mean number of passengers left at the gate?

To find the mean, we multiply each value of x by its corresponding probability and sum them up:

0*(0.061) + 1*(0.132) + 2*(0.255) + 3*(0.345) + 4*(0.207) = 1.988

So the mean number of passengers left at the gate is approximately 1.988.

c) What is the variance of the number of passengers left at the gate?

To find the variance, we calculate (x - mean)^2 for each value of x, multiply it by its corresponding probability, and sum them up:

(0 - 1.988)^2*(0.061) + (1 - 1.988)^2*(0.132) + (2 - 1.988)^2*(0.255) + (3 - 1.988)^2*(0.345) + (4 - 1.988)^2*(0.207) ≈ 1.177

So the variance of the number of passengers left at the gate is approximately 1.177.

d) What is the standard deviation of the number of passengers left at the gate?

To find the standard deviation, we take the square root of the variance:

√1.177 ≈ 1.085

So the standard deviation of the number of passengers left at the gate is approximately 1.085.

a) To determine if this is a probability distribution, we need to check two conditions:

1. The sum of all probabilities must equal 1.
2. All probabilities must be non-negative.

Calculating the sum of the probabilities:

0.061 + 0.132 + 0.255 + 0.345 + 0.207 = 1

Since the sum of the probabilities equals 1, the first condition is satisfied.

Checking that all probabilities are non-negative, we can see that all values given in the table are greater than or equal to 0. Therefore, the second condition is also satisfied.

Hence, this is a probability distribution.

b) The mean number of passengers left at the gate can be calculated by multiplying each value of x with their respective probabilities and then summing up the results.

Mean = (0 * 0.061) + (1 * 0.132) + (2 * 0.255) + (3 * 0.345) + (4 * 0.207) = 0 + 0.132 + 0.51 + 1.035 + 0.828 = 2.505

Therefore, the mean number of passengers left at the gate is 2.505.

c) The variance of the number of passengers left at the gate can be calculated using the formula:

Variance = (x^2 * P(x)) - mean^2

Variance = (0^2 * 0.061) + (1^2 * 0.132) + (2^2 * 0.255) + (3^2 * 0.345) + (4^2 * 0.207) - 2.505^2

Variance = 0 + 0.132 + 1.02 + 3.555 + 3.264 - 6.275025 = 1.696975

Therefore, the variance of the number of passengers left at the gate is approximately 1.696975.

d) The standard deviation can be calculated as the square root of the variance.

Standard Deviation = √(Variance)

Standard Deviation = √(1.696975) = 1.302

Therefore, the standard deviation of the number of passengers left at the gate is approximately 1.302.

a) To determine if this is a probability distribution, we need to check if the probabilities of all possible outcomes add up to 1. In other words, we need to verify that the sum of all the probabilities equals 1.

Let's calculate the sum of the probabilities:

0.061 + 0.132 + 0.255 + 0.345 + 0.207 = 1

Since the sum of the probabilities equals 1, we can conclude that this is indeed a probability distribution.

b) To find the mean number of passengers left at the gate, we need to multiply each outcome by its corresponding probability, and then sum up these products.

Mean (µ) = (0 * 0.061) + (1 * 0.132) + (2 * 0.255) + (3 * 0.345) + (4 * 0.207)

Mean (µ) = 0 + 0.132 + 0.51 + 1.035 + 0.828

Mean (µ) ≈ 2.505

Therefore, the mean number of passengers left at the gate is approximately 2.505.

c) To find the variance of the number of passengers left at the gate, we need to calculate the squared difference between each outcome and the mean, multiply it by its corresponding probability, and then sum up these products.

Variance (σ²) = [(0 - 2.505)² * 0.061] + [(1 - 2.505)² * 0.132] + [(2 - 2.505)² * 0.255] + [(3 - 2.505)² * 0.345] + [(4 - 2.505)² * 0.207]

Variance (σ²) ≈ (6.252025 * 0.061) + (2.272025 * 0.132) + (0.127525 * 0.255) + (0.373225 * 0.345) + (1.630525 * 0.207)

Variance (σ²) ≈ 0.382 + 0.300 + 0.033 + 0.129 + 0.337

Variance (σ²) ≈ 1.181

Therefore, the variance of the number of passengers left at the gate is approximately 1.181.

d) The standard deviation is the square root of the variance. So, to find the standard deviation, we take the square root of the previously calculated variance.

Standard Deviation (σ) ≈ √(1.181)

Standard Deviation (σ) ≈ 1.088

Therefore, the standard deviation of the number of passengers left at the gate is approximately 1.088.