Question

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation is 3.6 mm, how many of these components should she consider to be 90% sure of knowing the mean will be within ± 0.1 ±0.1 mm?

Answer 1

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 =  $Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }$  

Here, $\sigma$ = standard deviation = 3.6 mm

         n = sample size of components

         $\alpha$ = level of significance = 1 - 0.90 = 0.10 or 10%

         $\frac{\alpha}{2} = \frac{0.10}{2}$ = 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 =  $Z_(_\frac{\alpha}{2}_) \times \frac{\sigma}{\sqrt{n} }$  

                  0.1 mm        =  $1.645 \times \frac{3.6}{\sqrt{n} }$

                    $\sqrt{n} = \frac{3.6\times 1.645}{0.1 }$

                    $\sqrt{n}$ = 59.22

                     n = $59.22^{2}$ = 3507.0084 ≈ 3507.

Hence, she must consider 3507 components to be 90% sure of knowing the mean will be within ± 0.1 mm.