A sample distribution (closely approximating a normal distribution) of a critical part dimension has an average of
11.85 inches and a sample standard deviation of 1.33 inches. The lower specification limit is 11.41 inches. The
upper specification limit is 12.83 inches. What percentage of parts will be produced below the lower specification
limit of 11.41 inches?
To find the percentage of parts that will be produced below the lower specification limit, we need to calculate the z-score for the lower specification limit and then find the corresponding percentage from the standard normal distribution.
The z-score is calculated using the formula:
z = (x - μ) / σ
Where:
x = the lower specification limit
μ = the average (mean) of the distribution
σ = the sample standard deviation
Substituting the values:
z = (11.41 - 11.85) / 1.33
z = -0.33 / 1.33
z ≈ -0.248
Now, we need to find the percentage of values below this z-score. We can use a standardized normal distribution table or a calculator to find this percentage.
Looking up the z-score of -0.248 in a standard normal distribution table, we find that the percentage of values below this z-score is approximately 0.4013.
Therefore, approximately 40.13% of parts will be produced below the lower specification limit of 11.41 inches.
To find the percentage of parts that will be produced below the lower specification limit of 11.41 inches, we can use the Z-score formula and the standard normal distribution.
The Z-score formula is given by:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the value we want to find the percentage for (in this case, the lower specification limit of 11.41 inches)
μ is the mean of the distribution (11.85 inches)
σ is the standard deviation of the distribution (1.33 inches)
First, we need to calculate the Z-score for 11.41 inches:
Z = (11.41 - 11.85) / 1.33
Z = -0.33 / 1.33
Z ≈ -0.2481
Next, we can use a Z-table or a statistical calculator to find the percentage associated with the Z-score of -0.2481. The Z-table will give us the area under the standard normal distribution curve up to a certain Z-score.
Looking up the Z-score of -0.2481 in the Z-table, we find that the area is approximately 0.4013.
Since the Z-table gives the area to the left of the Z-score, the percentage of parts produced below the lower specification limit of 11.41 inches is approximately:
Percentage = 0.4013 * 100 ≈ 40.13%
Therefore, approximately 40.13% of parts will be produced below the lower specification limit of 11.41 inches.