Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 194 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 36 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 22 passengers. What is the approximate probability that the total weight of the passengers exceeds 4668 pounds? (Round your answer to four decimal places.)

Answer 1

Answer:

the approximate probability that the total weight of the passengers exceeds 4668 pounds is 0.0089

Step-by-step explanation:

Using the central limit Theorem

z = (x - u )/Q÷ sqrt n

Where z is the z-score

x is the average weight = 4668/22 = 212.2

u is the passengers average = 194 pounds

Q is standard deviation = 36 pounds

Therefore

z= (212.2 - 194) / 36 ÷ sqrt22

= 18.2/ 36÷ 4.69

= 18.2/7.68

z=2.3698 aprx 2.37

Find the value of 2.37 z-score on the z-table to determine the approximate probability

P(z>2.37) = 0.0089 (from the z-table)

Answer 2

Answer:

Probability = 0.0089

Step-by-step explanation:

From the question ;

Mean (μ) = 194 pounds

Standard deviation(σ) = 36 pounds

Minimum total weight = 4688

Passengers carried = 22

Thus average weight = 4668/22 = 212.18

The z-value is given by;

Z = (x - μ) / ((σ)/√n)

Z= (212.18 - 194)/(36/√22)

Z = 18.18/7.675 = 2.37

Using the table I attached ;

P(z>2.37) = 0.0089