0. The time required to instal
0. The time required to install a new aircraft engine isapproximately a normally distributed random variable with a mean of20 hours and a standard deviation of 1 hour. What is theprobability that the next installation takes
a. Between 20 and 21.5 hours?
b. Between 18 and 20 hours?
c. Over 23 hours?
d. At most 16.1 hours?
e. More than 18.3 hours?
Answer:
Solution :
Given that ,
mean = 
= 20
standard deviation = 
= 1
(a)
P(20 < x < 21.5) = P((20 – 20)/ 1) < (x – ) / <(21.5 – 20) / 1) )
= P(0 < z < 1.5)
= P(z < 1.5) – P(z < 0)
= 0.9332 – 0.5
= 0.4332
Probability = 0.4332
(b)
P(18 < x < 20) = P((18 – 20)/ 1) < (x – ) / <(20 – 20) / 1) )
= P(-2 < z < 0)
= P(z < 0) – P(z < -2)
= 0.5 – 0.0228
= 0.4772
Probability = 0.4772
(c)
P(x > 23) = 1 – P(x < 23)
= 1 – P((x – 
) / 
< (23 – 20) / 1)
= 1 – P(z < 3)
= 1 – 0.9987
= 0.0013
Probability = 0.0013
(d)
P(x 
16.1) = P((x – ) / (16.1 – 20) / 1)
= P(z -3.9)
= 0
Probability = 0
(e)
P(x > 18.3) = 1 – P(x < 18.3)
= 1 – P((x – ) / < (18.3 – 20) / 1)
= 1 – P(z < -1.7)
= 1 – 0.0446
= 0.9554
Probability = 0.9554
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