Question
The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 inches. Heights of men are normally distributed with a mean of 69 inches and a standard deviation of 2.8 inches.
b) assume that half of the 200 passengers are men, what doorway height satisfies the condition that there is a .95 probability that this height is greater than the mean height of 100 men?
b) assume that half of the 200 passengers are men, what doorway height satisfies the condition that there is a .95 probability that this height is greater than the mean height of 100 men?
Answers
anoynomous
a)
The value of z such that P(Z<z)= 0.95 is z = 1.6449
Solving for x:
z =1.6449=(x-69.0)/2.8
x = 1.6449*2.8+69 = 73.6
b)
Solving for xbar:
z =1.6449=(xbar-69.0)/(2.8/sqrt(100))
xbar = 1.6449*0.28+69 = 69.5
The value of z such that P(Z<z)= 0.95 is z = 1.6449
Solving for x:
z =1.6449=(x-69.0)/2.8
x = 1.6449*2.8+69 = 73.6
b)
Solving for xbar:
z =1.6449=(xbar-69.0)/(2.8/sqrt(100))
xbar = 1.6449*0.28+69 = 69.5
stupid but i got it
0.8577