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:
i believe its a
Step-by-step explanation:
Answer:
red is 50% blue is 25%
Step-by-step explanation:
red is half the circle so it would be 50% and blue is half of a half so 25%
It is the last one because if you go 3 left you have to change the sign to a positive (but it is still a negative) and the 2 remain the same sign so it’s h(x-3)-2
Answer:
104
Step-by-step explanation:
390/15=26
26 times 4 is 104