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jeka94
3 years ago
7

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.)

Mathematics
2 answers:
solong [7]3 years ago
6 0

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)

Alika [10]3 years ago
5 0

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

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