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suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. most airlines charge a fee for baggage that weigh in excess of 50 pounds. determine what percent of airline passengers incur this fee.

Respuesta :

In most of the cases airlines charges a fees for baggage weigh in excess of 50 pounds. Then about 94.06% passengers incur this fees.

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

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 p-value, we get the probability that the value of the measure is greater than X.

Given That,

Mean (μ) = 45 pounds and Standard Deviation (σ) = 3.2

This means that μ = 46 , σ = 3.2

Since, Most airlines charges a fee for baggage that weigh in excess of 50 pound.

Then, As a proportion this is 1 subtracted by the P - value of Z when

X =50. So,

      Z = X - μ / σ

⇒ Z = 50 -45/ 3.2

⇒ Z = 5/3.2

⇒ Z = 1.56

when Z = 1.56 has a P- value of 0.9406

Now, 0.9406 × 100%

       = 94.06%

Therefor, 94.06% of airline passengers incur this fees.

Learn more about Standard Deviation:

https://brainly.com/question/16555520

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