Question

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 51 lb and a standard deviation of 23 lb. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Round the answer to four decimal places.)

Answer 1

Answer:

The approximate probability that the total weight of their baggage will exceed the limit (6000 lb) is P=0.00005.

Step-by-step explanation:

To solve this problem, we have to think that we take a sample of 100 passengers, with a baggage weight that is a random variable normally distributed.

The parameters of the probability distribution of the sum of 100 normally distributed variables is:

$\mu_s=\sum_{i=1}^n\mu_i=n\mu_i=100*51=5100\\\\\sigma_s=\sqrt{\sum_{i=1}^n\sigma_i^2}=\sqrt{n\sigma_i^2}=\sqrt{n}\sigma_i=\sqrt{100}*23=10*23=230$

Now we can calculate the probability of $X_s>6000$.

$z=(X-\mu)/\sigma=(6000-5100)/230=900/230=3.913\\\\\\P(X>6000)=P(z>3.913)=0.00005$

Answer 2

The approximate probability that the total weight of their baggage will exceed the limit is; 2.8679 * 10⁻⁷

What is the Approximate Probability?

The limit we want to find the probability for is; 6000/100 = 60

We are given;

Sample mean; x' = 60

Population mean; μ = 51

standard deviation; σ = 18

Sample size; n = 100

Thus, z-score is;

z = (x' - μ)/(σ/√n)

z = (60 - 51)/(18/√100)

z = 5

From online p-value from z-score calculator, we have;

p = 2.8679 * 10⁻⁷

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