Answer:
The sample size required is 865.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean is:
![CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The margin of error of this interval is:
![MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
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:
![\sigma=\frac{Range}{4}=\frac{Max.-Min.}{4}=\frac{70-40}{2}=15](https://tex.z-dn.net/?f=%5Csigma%3D%5Cfrac%7BRange%7D%7B4%7D%3D%5Cfrac%7BMax.-Min.%7D%7B4%7D%3D%5Cfrac%7B70-40%7D%7B2%7D%3D15)
The margin of error is, MOE = 1 kV.
The critical value of <em>z</em> for 95% confidence level is:
![z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3Dz_%7B0.05%2F2%7D%3Dz_%7B0.025%7D%3D1.96)
Compute the sample size as follows:
![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](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C1%3D1.96%5Ctimes%20%5Cfrac%7B15%7D%7B%5Csqrt%7Bn%7D%7D%5C%5Cn%3D%28%5Cfrac%7B1.96%5Ctimes%2015%7D%7B1%7D%29%5E%7B2%7D%5C%5Cn%3D864.36%5C%5Cn%5Capprox865)
Thus, the sample size required is 865.