Suppose that the population of weights of airline passengers has mean 150 lbs. and standard deviation 25 lbs. A certain plane has a capacity of 7,800 lbs.
(a) What is the probability that a flight of 50 passengers will exceed capacity it?
(b) If we wish to reduce the chance of overload to 0.01, how much must the capacity be increased?
To answer these questions, we will use the normal distribution and the properties of the standard normal distribution. We can standardize the distribution of weights of airline passengers using the formula:
Z = (X - μ) / σ
Where:
Z is the standard score (also known as the z-score),
X is the value we want to standardize,
μ is the mean of the population,
and σ is the standard deviation of the population.
(a) To find the probability that a flight of 50 passengers will exceed capacity, we need to find the probability that the sum of weights of 50 passengers exceeds the capacity of 7,800 lbs.
Step 1: Calculate the mean and standard deviation of the sum of weights of 50 passengers.
The mean of the sum of weights is the sum of the means:
μ_sum = n * μ = 50 * 150 = 7,500 lbs
The standard deviation of the sum of weights is the sum of the standard deviations:
σ_sum = √(n * σ²) = √(50 * 25²) = 176.78 lbs
Step 2: Standardize the capacity using the formula mentioned earlier.
Z = (X - μ) / σ
Z = (7,800 - 7,500) / 176.78 = 1.703
Step 3: Look up the probability of exceeding the standardized value in the standard normal distribution table.
The table provides the area under the standard normal curve to the left of a given z-score. Since we want the probability of exceeding the capacity, we have to find the complement of that probability.
P(Z > 1.703) = 1 - P(Z < 1.703)
Using the standard normal distribution table or a calculator that provides the area to the left of the z-score, we find that P(Z < 1.703) = 0.9564. Therefore, the probability of exceeding the capacity is:
P(Z > 1.703) = 1 - 0.9564 = 0.0436
So, the probability that a flight of 50 passengers will exceed capacity is approximately 0.0436 or 4.36%.
(b) If we wish to reduce the chance of overload to 0.01, we need to find the new capacity that corresponds to that probability.
Step 1: Find the z-score that corresponds to a probability of 0.01 in the standard normal distribution.
P(Z > z) = 0.01
Using the standard normal distribution table or a calculator, we find that the z-score corresponding to P(Z > z) = 0.01 is approximately z = 2.33.
Step 2: Solve for X in the standardization formula by rearranging it:
Z = (X - μ) / σ
X - μ = Z * σ
X = Z * σ + μ
Substituting the known values:
X = 2.33 * 25 + 150
Step 3: Calculate the new capacity by rounding up to the nearest whole number:
X = 58.25 + 150 = 208.25 lbs
Therefore, to reduce the chance of overload to 0.01, the capacity must be increased to approximately 208 pounds.
To solve this problem, we will use the properties of a normal distribution. Let's break down the steps to find the solutions for each part:
(a) Let X represent the sum of the weights of the 50 passengers.
We know that X follows a normal distribution with mean (μ) 150 lbs and standard deviation (σ) 25 lbs.
The sum X of the weights of all 50 passengers is calculated by X = n * μ = 50 * 150 = 7,500 lbs.
Therefore, we need to find the probability that X exceeds the capacity, which is 7,800 lbs.
To find this probability, we'll use the z-score formula:
z = (X - μ) / σ
First, calculate the z-score:
z = (7,800 - 7,500) / 25 = 12
Now, we can find the probability by using a standard normal distribution table or a calculator to find the area (probability) to the right of z = 12. This probability represents the chance that the flight will exceed capacity.
Since z-scores above 3 are extremely rare and the area to the right of z = 3 is almost negligible, we can assume the probability is close to 0.
Therefore, the probability that a flight of 50 passengers will exceed capacity is approximately 0.
(b) To reduce the chance of overload to 0.01, we need to find the minimum capacity required.
We can solve this by finding the z-score that corresponds to the desired probability (0.01) and then calculating the minimum capacity needed.
First, we need to find the z-score corresponding to a probability of 0.01. For this, we can use a standard normal distribution table or a calculator to find the z-score that has an area of 0.01 to the right. Let's assume this z-score is z_critical.
Now, we can use the z-score formula to find the minimum capacity:
z_critical = (X - μ) / σ
Rearranging the formula to solve for X, we have:
X = z_critical * σ + μ
Substitute the values we have:
X = z_critical * 25 + 150
Calculate the minimum capacity (X) needed to reduce the chance of overload to 0.01 by substituting the value of z_critical:
X = z_critical * 25 + 150
Note: The standard deviation (σ) remains the same at 25 lbs.
Finally, calculate the minimum capacity needed by substituting the value of z_critical:
minimum capacity = z_critical * 25 + 150
Please make sure to determine the value of z_critical using a standard normal distribution table or a calculator.