Question

The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with a mean of 6.7 minutes and a standard deviation of 2.2 minutes. for a door selected at random, what is the probability the assembly time will be:
a.5 minutes or less?

b.10 minutes or more?

c.between 5 and 10 minutes?

Answer 1

We need to work out the z-value of each question to work out the probability

Question a)

We have
X = 5 minutes
The mean, μ = 6.7 minutes
Standard deviation, σ = 2.2 minutes
z-score = (X - μ) / σ = (5 - 6.7) / 2.2 = -0.77

We want to find the probability of z < -0.77 so we read the z-table for the value of z on the left of -0.77 we have the probability P(z< -0.77) = 0.2206

The table is attached in picture 1 below

The probability of assembly time less than 5 minutes is 0.2206 = 22.06%

Question b)

z-score = (10-5) / 2.2 = 2.27

Reading the z-table for P(z<2.27) = 0.9884
The table reading is shown in the second picture below

The probability the assembly time will be less than 10 minutes is 0.9884

We can use this information to find the probability of the assembly time will be more than 10 minutes  = 1 - 0.9884 = 0.0116 = 1.16%

Question c)

The value of z between 5 minutes and 10 minutes is

P(-0.77<z<2.27) = P(z<2.27) - P(z< -0.77)
P(-0.77<z<2.27) = 0.9884 - 0.2206 = 0.7678 = 76.78%