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1 answer:
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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:
dy/dx at x=6 is 0.334
Step-by-step explanation:
The center difference method requires that the values of the function are given in equal intervals which is the case, and allows one to find the value for x = 6 using those of the function for x = 5.5 and for x = 6.5 as follows: