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Lubov Fominskaja
21 day ago
9

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]21 day ago
5 0

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

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