Question

1. airlines weigh luggage in order to safely load planes and to comply with weight restrictions. for one major airline, the distribution of passengers luggage weights is approximately normal with a mean of 37 pounds and a standard deviation of 4.8 pounds. let ( l ) = the weight of a randomly selected piece of luggage transported by this airline.

a. the airline charges an additional fee for luggage exceeding 50 pounds. find the probability that a randomly selected piece of luggage is subject to the additional fee.
b. the airline is considering lowering the weight limit so that more luggage is subject to the additional fee. the airline would like to set the limit so that 1% of the pieces of luggage are over the limit. what weight, to the nearest pound, should the airline set for the limit?

Explanation:

Step1: Calculate z-score for 50 lbs

$z = \frac{X - \mu}{\sigma} = \frac{50 - 37}{4.8} \approx 2.71$

Step2: Find upper tail probability

$P(L > 50) = P(z > 2.71) = 1 - P(z \leq 2.71)$
Using z-table, $P(z \leq 2.71) \approx 0.9966$, so $1 - 0.9966 = 0.0034$
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Step3: Find z for 99th percentile

For top 1% (99th percentile), z-score $\approx 2.33$

Step4: Solve for weight limit X

$X = \mu + z\sigma = 37 + (2.33 \times 4.8) \approx 37 + 11.18 = 48.18$
Round to nearest pound: 48

Answer:

a. 0.0034
b. 48 pounds