Suppose that for a particular airline 25 of all flight are l

Suppose that, for a particular airline, 25% of all flight are late. The airline randomly selects 50 of their flights. Let X = the number of late flights among those selected Find the E[X] Find the standard deviation of X

Solution

a)

E(x) = np = 50*0.25 = 12.5 [ANSWER]

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B)

s(x) = sqrt(np(1-p)) = sqrt(50*0.25*(1-0.25)) = 3.061862178 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    9.5      
u = mean =    12.5      
          
s = standard deviation =    3.061862178      
          
Thus,          
          
z = (x - u) / s =    -0.979795897      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.979795897   ) =    0.836406561 [ANSWER]