Commercial airliners have a maximum allowable weight of passenger checked baggage. It is known that the distribution of individual passengers checked-in baggage weight has a mean of 42 pounds and a standard deviation of 25 pounds. Suppose that a particular airplane, with a capacity of 125 passengers, can take at most 6,000 pounds of checked baggage. Assuming the plane has no empty seats, what is the approximate probability that the baggage limit will not be exceeded?

Choose one answer.

a. 0.9791

b. 0.9832

c. 0.9963

d. 1.0000

e. none of these

here is a page that will calculate it for you without tables.

6000/125 = 48 pounds

so you want the prob that the weight will be below 48

so enter the mean as 42
the standard dev as 25
and click the below after entering 48

I get .5948

To calculate the probability that the baggage limit will not be exceeded, we can use the Central Limit Theorem.

The Central Limit Theorem states that, for a large number of independent samples, the distribution of the sample means will be approximately normally distributed, regardless of the shape of the original population.

Given that the mean baggage weight is 42 pounds and the standard deviation is 25 pounds, we can calculate the mean of the sample means (μx̄) and the standard deviation of the sample means (σx̄) as follows:

μx̄ = μ = 42 pounds
σx̄ = σ / sqrt(n) = 25 pounds / sqrt(125) ≈ 2.232 pounds

Next, we need to find the z-score corresponding to the baggage limit of 6,000 pounds. The z-score formula is given by:

z = (x - μx̄) / σx̄

where x is the value we want to find the z-score for.

z = (6000 - 42) / 2.232 ≈ 2673.75

Using a standard normal distribution table or calculator, we can find the cumulative probability associated with a z-score of 2673.75 (z > 2673.75). This probability will give us the approximate probability that the baggage limit will not be exceeded.

The value of the cumulative probability is practically 1. Therefore, the approximate probability that the baggage limit will not be exceeded is:

1 - 1 = 0 or 0%.

Thus, the correct answer is d. 1.0000.

To solve this problem, we need to first calculate the total weight capacity of the airplane.

Given:
Mean baggage weight (μ) = 42 pounds
Standard deviation (σ) = 25 pounds
Capacity of the airplane (n) = 125 passengers
Maximum allowable weight of checked baggage = 6,000 pounds

Since the mean and standard distribution of individual passenger baggage weight are given, we can assume that the distribution of the total checked baggage weight will be approximately normal.

To find the total weight capacity, we multiply the number of passengers by the mean baggage weight:
Total weight capacity (μn) = μ * n = 42 * 125 = 5,250 pounds.

Now, we need to calculate the probability that the baggage limit will not be exceeded.

The probability distribution of the sum of independent normally distributed variables is also a normal distribution. The mean of the sum is the sum of the individual means, and the variance of the sum is the sum of the individual variances.

To calculate the standard deviation (σn) of the total baggage weight (n passengers), we use the formula:
σn = σ / √n = 25 / √125 ≈ 2.23 pounds.

Next, we calculate the z-score (standard score) for the limit of 6,000 pounds:
z = (6,000 - 5,250) / 2.23 ≈ 335.03

To determine the probability that the baggage limit will not be exceeded, we need to find the area under the normal curve to the left of the z-score. This represents the probability that the total weight of checked baggage will be less than or equal to 6,000 pounds.

Using a standard normal distribution table or a statistical calculator, we find that the probability corresponding to a z-score of 335.03 is essentially 1.0000.

Therefore, the approximate probability that the baggage limit will not be exceeded is 1.0000, which corresponds to option d.