Answer:
a) 135 minutes
b) 15
c) 0.5
d) 0.3707
Step-by-step explanation:
A) What is the distribution's mean
mean(m) = (120 + 150)/2 = 270/2 = 135 minutes
B) What is the distribution's standard deviation
standard deviation(s) [tex]= \sqrt{\frac{(120-135)^{2}+(150-135)^{2} }{2} }[/tex]
s = 15
C) What is the probability that a flight is less than 135 minutes
firstly calculate z score
z score = (x - m)/s = (135 - 135)/15 = 0
P(x<135) = z(0) = 0.5
D) What is the probability that a flight is more than 140 minutes?
z score = (x - m)/s = (140 - 135)/15 = 0.33
P(x>140) = 1 - P(x<140) = 1 - 0.6293 = 0.3707
