An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation of all

Question

2 months ago

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An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation of all

components is 2.2 mm, how many of these components should she consider to be 80% sure of know the mean will be within 0.3 mm?

Mathematics

1 answer:

Leona [12.6K] · Answer 1 · 2026-04-11T16:54:05+03:00

Answer:

$n=(\frac{0.539(2.2)}{0.3})^2 =15.62 \approx 16$

In this scenario, the solution for n is determined to be 16 when rounded up to the closest integer

Step-by-step explanation:

The following information has been provided for this case:

$\sigma = 2.2$ symbolizes the population standard deviation

$Confidence =0.8$

$ME = 0.3$ signifies the preferred margin of error

The formula for the margin of error of the actual mean is given as:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

In this case, the margin of error (ME) is 0.03, and to find n, we rearrange equation (a) to get:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

With a confidence level of 80%, and the significance level being $\alpha=0.2$ and $\alpha/2 =0.1$ , the critical value is $z_{\alpha/2}=0.539$ . Plugging in these values into formula (b) results in:

$n=(\frac{0.539(2.2)}{0.3})^2 =15.62 \approx 16$

Thus, the solution for this situation results in n=16, rounded to the nearest whole number