Answer:
Option a) A 95% confidence interval for the mean diameter of the 120 bearings in the sample is
.
Explanation:
We are given the following information in the question:
Sample size, n = 120
Sample mean = 10 mm
Standard Deviation = 0.24 mm
Formula:
![\mu \pm z_(critical)(\sigma)/(√(n))](https://img.qammunity.org/2020/formulas/mathematics/high-school/zd6etee1rupgyy1dm7yi75g9lbvhed55e9.png)
![z_(critical)\text{ at}~\alpha_(0.05) = 1.96](https://img.qammunity.org/2020/formulas/mathematics/high-school/xlob5chz85089qlwh99zy010m8rebc84s3.png)
![10 \pm 1.96\displaystyle(0.24)/(√(120))](https://img.qammunity.org/2020/formulas/mathematics/college/rnt8mof8rqxtsrchh349kpc045iq2122jo.png)
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
.
We have to consider the factor of sampling of 120 ball bearings from a population of 10,000 ball bearings.