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Anvisha [2.4K]
3 years 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:
Leto [7]3 years ago
6 0

Answer:

There is a 24.51% probability that he weight of a bag will be greater than the maximum allowable weight of 50 pounds.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

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, so \mu = 47.88, \sigma = 3.09

What is the probability that the weight of a bag will be greater than the maximum allowable weight of 50 pounds?

That is P(X > 50)

So

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

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

Z = 0.69

Z = 0.69 has a pvalue of 0.7549.

This means that P(X \leq 50) = 0.7549.

We also have that

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

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

There is a 24.51% probability that he weight of a bag will be greater than the maximum allowable weight of 50 pounds.

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