Question

The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with mean μ=6.7 minutes and standard deviation σ=2.2 minutes. For a door selected at random, what is the probability that the assembly-line time will be

a) 5 minutes or less?

b) 10 minutes or more?

c) between 5 and 8 minutes?

Please show work.

Answer 1

a) 0.2206

b) 0.668

c) 0.5018

Explanation:

Data provided in the question:

Mean, μ = 6.7 minutes

Standard deviation, σ = 2.2 minutes

Now,

z score = [ X - μ ] ÷ σ

For a) 5 minutes or less

z score = [ 5 - 6.7 ] ÷ 2.2

= - 0.773

From the normal curve table the P(z ≤ -0.773 ) = 0.2206

For b) 10 minutes or more

z score = [ 10 - 6.7 ] ÷ 2.2

= 1.5

From the normal curve table the P(z ≤ 1.5 ) = 0.668

c) between 5 and 8 minutes

P(greater than 5 min) = 1 - P(5 minutes or less)

= 1 - 0.2206 [From (a)]

= 0.7794

z score of 8 = [ 8 - 6.7 ] ÷ 2.2

= 0.59

From the normal curve table the P(z ≥ 0.59 ) = 0.2776

Therefore,

P( between 5 and 8 ) = P(greater than 5 min) + P(z ≥ 0.59 )

= 0.7794 - 0.2776

= 0.5018