In order to be labeled overweight the weight of box should be of 34.15 ounces.
Given mean of 33 ounces, standard deviation=0.7 ounces. Production of packets of 32 ounces.
Because it is normally distributed samples are solved using the z score formula.
In this Z=X-meu/st
meu is sample mean and st is population standard deviation.
We have to find the z score and then p value from the z table.
meu =33 and st=0.7
Top 5% so X when Z has a p value of 1-0.05=0.95. So X when Z=1.645.
Z=(X-meu)/st
1.645=(X-33)/0.7
X-33=0.7*1.645
X-33=1.1515
X=1.1515+33
X=34.1515 ounces.
Hence to be labeled overweight the box must have 34.15 ounces weight.
Learn more about z test at https://brainly.com/question/14453510
#SPJ4