An airliner carries 200 passengers and has doors with a heig

An airliner carries 200 passengers and has doors with a height of 75 in. Heights of men are normally distributed with a mew of 69. 0 in and a standard deviation of 2. 8 in Complete parts (a) through (d). If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending. If half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 75 in.

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    75      
u = mean =    69      
          
s = standard deviation =    2.8      
          
Thus,          
          
z = (x - u) / s =    2.142857143      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2.142857143   ) =    0.983937714 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    75      
u = mean =    69      
n = sample size =    100      
s = standard deviation =    2.8      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    21.42857143      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   21.42857143   ) =    1 [ANSWER]