Aquality inspector must verify whether a machine that packages snack foods is working correctly. the inspector will randomly select a sample of packages and weigh the amount of snack food in each. assume that the weights of food in packages filled by this machine have a standard deviation of 0.30 ounce. an estimate of the mean amount of snack food in each package must be reported with 99.6% confidence and a margin of error of no more than 0.12 ounces. what would be the minimum sample size for the number of packages the inspector must select?

52

Step-by-step explanation:

Formula for sample size:-

, where = population standard deviation.

= Two -tailed z-value for (significance level)

E= margin of error.

Given : tex]\sigma=\text{ 0.30 ounce}[/tex]

⇒Significance level for 99.6% confidence level :

By using z-value table ,Two -tailed z-value for :

E= 0.12 ounces.

Minimum sample size will be :-

Hence, the minimum sample size  for the number of packages the inspector must select = 52

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540

13cm

step-by-step explanation:

(apex)