Answer:
Option a) A 95% confidence interval for the mean diameter of the 120 bearings in the sample is [tex]10 \pm 1.96\displaystyle\frac{0.24}{\sqrt{120}}[/tex].
Step-by-step explanation:
We are given the following information in the question:
Sample size, n = 120
Sample mean = 10 mm
Standard Deviation = 0.24 mm
Formula:
[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
[tex]10 \pm 1.96\displaystyle\frac{0.24}{\sqrt{120}}[/tex]
Hence, the correct interpretation for the confidence interval is given by option a).
A 95% confidence interval for the mean diameter of the 120 bearings in the sample is [tex]10 \pm 1.96\displaystyle\frac{0.24}{\sqrt{120}}[/tex].
We have to consider the factor of sampling of 120 ball bearings from a population of 10,000 ball bearings.