Question

ASK YOUR TEACHER An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable x with a mean value of 49 lb and a standard deviation of 19 lb. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit

Answer 1

0%

Explanation:

so i guess its asking what is the chance of a passenger going over the limit so i guses it would be the aproximent decimal .

so if 6000 divided by 100 is 60 the chance of them excedding the limit is 0% chance because the answer is even so it dosent have a decimal so each passenger wont exceed

Answer 2

0% approximate probability that the total weight of their baggage will exceed the limit

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sum of a sample of size n, we have that the mean is
`\mu*n` and the standard deviation is
`\sigma √(n)`

In this problem, we have that:

`n = 100, \mu = 49*100 = 4900, s = 19√(100) = 190`

If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit

This is 1 subtracted by the pvalue of Z when X = 6000. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (6000 - 4900)/(190)`

`Z = 5.79`

`Z = 5.79` has a pvalue of 1

1 - 1 = 0

0% approximate probability that the total weight of their baggage will exceed the limit