Question

The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time. Less than 19.5 hours?

Answer 1

Let x denotes the time taken to assemable cars mean µ = 20 hours and S.D σ = 2 hours

The standard normal variate

z = `(x - mu)/sigma`

= `(x - 20)/2`

P(less than 19.5 hours) = P(X < 19.5)

When x = 19.5

z = `19.5/2`

 = `(-0.5)/2`

= 0.25

P(X < 19.5) = P(Z < – 0.25)

= P(`-oo` < z < 0) – P(– 0.25 < z < 0)

= 0.5 – P(0 < z < 0.25)

= 0.5 – 0.0987

= 0.4013