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If adult male heights are normally distributed with a mean of 180 cm and a standard deviation of 7 cm, how high should an aircraft lavatory door be to ensure that 99.9 percent of adult males will not have to stoop as they enter?

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toporc
From a standard normal distribution table, the value of z giving a cumulative probability of 0.999 is 3.08.
[tex]3.08=\frac{X-\mu}{\sigma}=\frac{X-180}{7}[/tex]
Solving for X we get X = 201.56.
The answer is 201.56 cm.
MrRoyal

The aircraft lavatory door should be 201.56 cm high to ensure that 99.9% of adult males will not have to stoop as they enter

How to determine the height of the door?

The p value is given as:

p = 99.9%

At p = 99.9%, the z value is:

z = 3.08

The z-score is calculated using:

[tex]z = \frac{x - \mu}{\sigma}[/tex]

Where:

Mean = 180

Standard deviation = 7

So, we have:

[tex]3.08 = \frac{x - 180}{7}[/tex]

Multiply both sides by 7

[tex]21.56 = x - 180[/tex]

Solve for x

x = 201.56

Hence, the height is 201.56 cm

Read more about normal distribution at:

https://brainly.com/question/4079902

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