Question

The Mark VI monorail used at Disney World has doors with a height of 72in. Heights of men are normally distributed with a mean of 69.5in and a standard deviation of 2.4 in. What percentage of adult men can fit through the doors without bending?

Answer 1

SOLUTION:

From given data,

The Mark VI monorail used at Disney World has doors with a height of 72 in. Heights of men are normally distributed with a mean of 69.5 in and a standard deviation of 2.4 in.

Mean = $\mu$ = 69.5 in.

Standard deviation = $\sigma$ = 2.4 in

X = 72 in.

What percentage of adult men can fit through the doors without bending?

P(X < 72) = P((X-$\mu$) /$\sigma$  < (72-69.5) /2.4)

P(X < 72) = P( Z < 2.5 /2.4)

P(X < 72) = P( Z < 1.04)  

P(X < 72) = 0.85083

P(X < 72) = 0.8508

The percentage of adult men can fit through the doors without bending is 85.08%

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