Question

0. The time required to install a new aircraft engine is approximately a normally distributed random variable with a mean of 20 hours and a standard deviation of 1 hour. What is the probability 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 1

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