Answer: 52
Step-by-step explanation:
Formula for sample size:-
[tex]n= (\dfrac{z_{\alpha/2\cdot \sigma}}{E})^2[/tex]
, where [tex]\sigma[/tex] = population standard deviation.
[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)
E= margin of error.
Given : tex]\sigma=\text{ 0.30 ounce}[/tex]
⇒Significance level for 99.6% confidence level :[tex]\alpha=1-0.996=0.004[/tex]
By using z-value table ,Two -tailed z-value for [tex]\alpha=0.01 [/tex]:
[tex]z_{\alpha/2}=z_{0.002}=2.878[/tex]
E= 0.12 ounces.
Minimum sample size will be :-
[tex]n= (\dfrac{2.878\cdot 0.30}{0.12})^2\\\\= (7.195)^2\\\\=51.768025\approx52[/tex]
Hence, the minimum sample size  for the number of packages the inspector must select = 52
