Mathematics, 14.04.2020 22:31 coolman5999alt

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

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Answer from: Aliciaonfleek

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$ .

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 $\sigma \sqrt{n}$

In this problem, we have that:

$n = 100, \mu = 49*100 = 4900, s = 19\sqrt{100} = 190$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{6000 - 4900}{190}$

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

Answer from: Quest

x= -3

step-by-step explanation:

step 1: simplify brackets

4 - x - 1 = 6

step 2: simpilfy

3 - x = 6

step 3: subtract 3 from both sides

-x = 6 - 3

step 4: simpilfy 6 - 3

-x = 3

step 5: multiply both sides by -1

x = -3

Answer from: Quest

the answer will be

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ASK YOUR TEACHER An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Supp...

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