Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 47.88, \sigma = 3.09[/tex]
What is the probability that the weight of a bag will be greater than the maximum allowable weight of 50 pounds?
That is [tex]P(X > 50)[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{50 - 47.88}{3.09}[/tex]
[tex]Z = 0.69[/tex]
[tex]Z = 0.69[/tex] has a pvalue of 0.7549.
This means that [tex]P(X \leq 50) = 0.7549[/tex].
We also have that
[tex]P(X \leq 50) + P(X > 50) = 1[/tex]
[tex]P(X > 50) = 1 - 0.7549 = 0.2451[/tex]
There is a 24.51% probability that he weight of a bag will be greater than the maximum allowable weight of 50 pounds.
