Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation

Question

2 months ago

8

Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation

of 3.09 pounds. a) What is the probability that the weight of a bag will be greater than the maximum allowable weight of 50 pounds? Give your answer to four decimal places.

Mathematics

1 answer:

Zina [12.3K] · Answer 1 · 2026-03-07T08:58:11+03:00

Answer:

There is a probability of 24.51% that the weight of a bag exceeds the maximum permitted weight of 50 pounds.

Step-by-step explanation:

Problems dealing with normally distributed samples can be addressed using the z-score formula.

For a set with the mean $\mu$ and a standard deviation $\sigma$ , the z-score for a measure X is calculated by

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

Once the Z-score is determined, we consult the z-score table to find the related p-value for this score. The p-value signifies the likelihood that the measured value is less than X. Since all probabilities total 1, calculating 1 minus the p-value gives us the probability that the measure exceeds X.

For this case

Imagine the weights of passenger bags are normally distributed with a mean of 47.88 pounds and a standard deviation of 3.09 pounds, thus $\mu = 47.88, \sigma = 3.09$

What probability exists that a bag’s weight will surpass the maximum allowable of 50 pounds?

That translates to $P(X > 50)$

Thus

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

$Z = \frac{50 - 47.88}{3.09}$

$Z = 0.69$

$Z = 0.69$ has a p-value of 0.7549.

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Additionally, we have that

$P(X \leq 50) + P(X > 50) = 1$

$P(X > 50) = 1 - 0.7549 = 0.2451$

There is a probability of 24.51% that the weight of a bag will exceed the maximum allowable weight of 50 pounds.

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