Question

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation is 1.6 mm, how many of these components should she consider to be 99% sure of knowing the mean will be within ±0.1±0.1 mm? a. 16 b. 4349 c. 1386 d. 42 e. 1699

Answer 1

Answer:

n=1705

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Assuming the X follows a normal distribution  

$X \sim N(\mu, \sigma=1.6)$  

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{1.6}{\sqrt{n}})$  

We know that the margin of error for a confidence interval is given by:  

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)  

The next step would be find the value of $\z_{\alpha/2}$, $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$  

Using the normal standard table, excel or a calculator we see that:  

$z_{\alpha/2}=\pm 2.58$  

If we solve for n from formula (1) we got:  

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$  

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

And we have everything to replace into the formula:  

$n=(\frac{2.58(1.6)}{0.1})^2 =1704.03$  

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=0.1$ mm is n=1705.