The engineer should consider a sample size of 325 electronic components to be 90% sure that the mean width will be within ± 0.3 mm.
To determine the required sample size, we can use the formula for the margin of error (ME) in the context of the normal distribution:
ME = (Z * σ) / √n
Where:
ME = Margin of error
Z = Z-score, which corresponds to the desired level of confidence (in this case, 90%)
σ = Standard deviation (3.3 mm in this problem)
n = Sample size
First, we need to find the Z-score corresponding to a 90% confidence level. This can be found using a Z-table or statistical software. The Z-score for a 90% confidence interval is approximately 1.645.
Now, we will rearrange the formula to solve for the sample size (n):
n = (Z * σ / ME)^2
We are given that the margin of error should be within ± 0.3 mm:
n = (1.645 * 3.3 / 0.3)^2
n ≈ (18.015)^2
n ≈ 324.54
Since we cannot have a fraction of a component, we round up to the nearest whole number to ensure the desired level of confidence:
n ≈ 325
The engineer should consider a sample size of 325 electronic components to be 90% sure that the mean width will be within ± 0.3 mm.