Question

The amount of time required to assemble a component on a factory assembly line is normally distributed with a mean of 3.1 minutes and a standard deviation of 0.6 minute. Find the probability that a randomly selected employee will take the given amount of time to assemble the component. (Round your answers to four decimal places.)

(a) more than 3.8 minutes

(b) between 1.8 and 2.5 minutes

Answer 1

Solution :

Given that ,

mean = = 3.1

standard deviation = = 0.6

a)

P(x > 3.8) = 1 - P(x < 3.8)

= 1 - P((x - ) / < (3.8 - 3.1) / 0.6)

= 1 - P(z < 1.17)

= 1 - 0.8790   

= 0.1210

Probability = 0.1210

b)

P(1.8 < x < 2.5) = P((1.8 - 3.1)/ 0.6) < (x - ) /  < (2.5 - 3.1) / 0.6) )

= P(-2.17 < z < -1)

= P(z < -1) - P(z < -2.17)

= 0.1587 - 0.0150

= 0.1437

Probability = 0.1437