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An airliner carries 50 passengers and has doors with height of 72 in. Heights of men are normally distributed with mean of 69.0 in and standard deviation of 2.8

If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending
The probability is______

1 Answer

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Final answer:

Using the normal distribution, the probability that a randomly selected male passenger can fit through a door with a height of 72 inches without bending, given the men have an average height of 69.0 inches with a standard deviation of 2.8 inches, is approximately 85.77%.

Step-by-step explanation:

To determine the probability that a male passenger can fit through the door without bending, we utilize the normal distribution properties. Given the heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches, and the door height is 72 inches, we want to find the probability of a man being shorter than 72 inches.

Calculating the Probability

To do so, we calculate the Z-score using the formula:

Z = (X - μ) / σ

Where X is 72 inches (door height), μ is the mean height of men (69.0 inches), and σ is the standard deviation (2.8 inches). Plugging these values into the formula gives us:

Z = (72 - 69.0) / 2.8

Z = 3 / 2.8

Z ≈ 1.0714

Now, we look up the Z-score in a standard normal distribution table, or use a calculator to find the corresponding probability. A Z-score of 1.0714 corresponds to a probability of approximately 0.8577. This means there is an 85.77% chance that a randomly selected male passenger can fit through the door without bending.

User Abdullah Tellioglu
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