Respuesta :
Answer:
She must consider 3507 components to be 90% sure of knowing the mean will be within ± 0.1 mm.
Step-by-step explanation:
We are given that an engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation is 3.6 mm.
And she considers to be 90% sure of knowing the mean will be within ±0.1 mm.
As we know that the margin of error is given by the following formula;
The margin of error = [tex]Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }[/tex]
Here, [tex]\sigma[/tex] = standard deviation = 3.6 mm
n = sample size of components
[tex]\alpha[/tex] = level of significance = 1 - 0.90 = 0.10 or 10%
[tex]\frac{\alpha}{2} = \frac{0.10}{2}[/tex] = 0.05 or 5%
Now, the critical value of z at a 5% level of significance in the z table is given to us as 1.645.
So, the margin of error = [tex]Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }[/tex]
0.1 mm = [tex]1.645 \times \frac{3.6}{\sqrt{n} }[/tex]
[tex]\sqrt{n} = \frac{3.6\times 1.645}{0.1 }[/tex]
[tex]\sqrt{n}[/tex] = 59.22
n = [tex]59.22^{2}[/tex] = 3507.0084 ≈ 3507.
Hence, she must consider 3507 components to be 90% sure of knowing the mean will be within ± 0.1 mm.
