In Exercises 21-26, use this information (based on data from the National Health
ID: 3155610 • Letter: I
Question
In Exercises 21-26, use this information (based on data from the National Health Survey): Men's bights are normally distributed with mean 69 0 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.6 in. and standard deviation 2.5 in. Doorway Height The Mark VI monorail used at Disney World and the Boeing 757-200 ER airliner have doors with a height of 72 in. What percentage of adult men can fit through the doors without bending? What percentage of adult women can fit through the doors without bending? Does the door design with a height of 72 in. appear to be adequate? Explain. What doorway height would allow 98% of adult men to fit without bending? Doorway Height The Gulfstream 100 is an executive jet that scats six, and it has a doorway height of 51.6 in. What percentage of adult men can fit through the door without bending? What percentage of adult women can fit through the door without bending? Does the door design with a height of 51.6 in. appear to be adequate? Why didn't the engineers design a larger door? What doorway height would allow 60% of men to fit without bending?Explanation / Answer
21)
A) THE DISTRIBUTION IS NORMALY DISTRIBUTED
THE FORMULA TO BE USED = Z = (X-MEAN)/STANDARD DEBIATION
FOR MEN
P(X<72) =
For x = 72, the z-value z = (72 - 69) / 2.8 = 1.07
Hence P(x < 72) = P(z < 1.07), now from the z table we will take the value of z score = 1.07
And that value will be the probability required.
= [area to the left of 1.07] = 0.8577
B) FOR WOMEN
P(X<72) =
For x = 72, the z-value z = (72 - 63.6) / 2.5 = 3.36
Hence P(x < 72) = P(z < 3.36), now from the z table we will take the value of z score = 3.36
And that value will be the probability required.
= [area to the left of 3.36] = 0.9996
C)YES IT APPEAR TO BE ADEQUATE BECAUSE FOR BOTH MEN AND WOMEN THE PROBABILITY IF MORE THEN SUFFICIENT THAT THEY CAN CROSS WITHOUT BENDING
D) Z SCORE FOR 98% = 2.33
FORMULA TO BE USED
X = MEAN+ZSCORE8STANDARD DEVIATION
THEREFORE
X = 69 +2.33*2.8
= 75.524