Respuesta :
Answer:
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Step-by-step explanation:
Problems of normally distributed samples are 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]
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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 175, \sigma = 22[/tex]
What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?
90th percentile
The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 175}{22}[/tex]
[tex]X - 175 = 22*1.28[/tex]
[tex]X = 203.16[/tex]
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
