SOLUTION: the mark vi monorail used at disney world has doors with a height of 72 inches. heights of men normally distributed with a mean of 69.4 inches and a standard deviation of 2.2 inche

Algebra.Com
Question 1091258: the mark vi monorail used at disney world has doors with a height of 72 inches. heights of men normally distributed with a mean of 69.4 inches and a standard deviation of 2.2 inches
A. what percentage of adult men can fit through the doors without bending?
B. what doorway height would allow 99% of adult men to fit without bending?

Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
z=(x-mean)/sd
want z for a mean of 69.4
percent of men over 72 inches
z=(72-69.4)/2.2=1.182
z <+1.182 is the probability desired,of 0.8814 or 88.1%
for 99%, need the value of z (0.99), which is +2.328
multiply that by the sd and get 5.1216. That is the difference between the mean and the height of the door.
The door height needs to be 69.4+5.1 or 74.5 inches.

RELATED QUESTIONS

the mark vi monorail used at disney world has doors with a height of 72 inches. heights... (answered by Boreal)
Men's heights are normally distributed with a mean of 69.0 inches and a standard... (answered by ikleyn)
6.4 11/An airliner carries 150 passengers and has doors with a height of 78 in. Heights... (answered by Boreal)
ASSUME THAT HEIGHTS OF MEN ARE NORMALLY DISTRIBUTED WITH A MEAN OF 68.4 INCHES AND A... (answered by solver91311)
If a random sample of eight 18-year-old men is selected, what is the probability that the (answered by VFBundy)
Men’s heights are normally distributed with a mean of 72 inches and a standard... (answered by Boreal)
Men’s heights are normally distributed with a mean of 71.8 inches and a standard... (answered by Boreal)
The sitting height (from seat to top of head) of drivers must be considered in the design (answered by stanbon)
Suppose the heights of men are normally distributed with mean, μ= 70 inches, and... (answered by Boreal)