Solve the problem. Round to the nearest tenth unless indicated otherwise.
The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. These weights could serve as limits used to identify which components should be rejected. Round to the nearest hundredth of a gram.
(1 point)
Responses
8.48 g and 8.52 g
8.30 g and 8.74 g
8.33 g and 8.67 g
8.46 g and 8.54 g
To find the weights that separate the top 3% and bottom 3% of the components, we need to find the z-scores that correspond to these percentiles.
For the top 3%:
z = (X - μ) / σ
z = (X - 8.5) / 0.09
Using a standard normal distribution table, we find that the z-score for the top 3% is approximately 1.8808.
Now we can solve for X:
1.8808 = (X - 8.5) / 0.09
1.8808 * 0.09 = X - 8.5
0.1693 = X - 8.5
X = 8.6693 g
For the bottom 3%:
z = (X - 8.5) / 0.09
Using a standard normal distribution table, we find that the z-score for the bottom 3% is approximately -1.8808.
Now we can solve for X:
-1.8808 = (X - 8.5) / 0.09
-1.8808 * 0.09 = X - 8.5
-0.1693 = X - 8.5
X = 8.3307 g
Therefore, the weights that separate the top 3% and bottom 3% of the components are 8.33 g and 8.67 g respectively.
Answer: 8.33 g and 8.67 g