Question

Suppose that the weights of passengers on a flight to Greenland on Frigid Aire Lines are normal with mean 175 pounds and standard deviation 22. What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

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:

$\mu = 175, \sigma = 22$

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

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

$1.28 = \frac{X - 175}{22}$

$X - 175 = 22*1.28$

$X = 203.16$

The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds