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
and standard deviation
, the zscore of a measure X is given by:

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:

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




The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Answer:
Step-by-step explanation:
If two ratios make a proportion then it can be shown as:
Let's verify the answer choices with this method
A. 1/4 and 5/25
- 1*25 = 4*5
- 25 = 20, incorrect, not a proportion
B. 1/4 and 7/24
- 1*24 = 4*7
- 24 = 28, incorrect, not a proportion
C. 1/4 and 10/48
- 1*48 = 4*10
- 48 = 40, incorrect, not a proportion
D. 1/4 and 9/36
- 1*36 = 4*9
- 36 = 36, correct, this is a proportion