Respuesta :
Answer:
The sample size required is 865.
Step-by-step explanation:
The (1 - α)% confidence interval for population mean is:
[tex]CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
The margin of error of this interval is:
[tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
It is provided that all values of breakdown voltage are between 40 and 70.
So, the minimum value of breakdown voltage is, Min. = 40 and the maximum value of breakdown voltage is, Max. = 70.
Assume that the population standard of the distribution of breakdown voltage is known.
The standard deviation is:
[tex]\sigma=\frac{Range}{4}=\frac{Max.-Min.}{4}=\frac{70-40}{2}=15[/tex]
The margin of error is, MOE = 1 kV.
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Compute the sample size as follows:
[tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}\\1=1.96\times \frac{15}{\sqrt{n}}\\n=(\frac{1.96\times 15}{1})^{2}\\n=864.36\\n\approx865[/tex]
Thus, the sample size required is 865.
