Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 46 pounds and standard deviation 3.7 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee. (Round your answer to two decimal places.)

85.99% of airline passengers incur this fee.

Step-by-step explanation:

When the distribution is normal, we use 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.

Mean 46 pounds and standard deviation 3.7 pounds.

This means that

Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee.

As a proportion, this is 1 subtracted by the pvalue of Z when X = 50. So

has a pvalue of 0.8599

0.8599*100% = 85.99%

85.99% of airline passengers incur this fee.

1

step-by-step explanation:

it's the only multiple of both and the least

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet