A manufacturer makes bags of popcorn and bags of potato chips. The average weight of a bag of popcorn is supposed to be 3.05 ounces with an allowable deviation of 0.02 ounces. The average weight of a bag of potato chips is supposed to be 5.07 ounces with an allowable deviation of 0.04 ounces. A factory worker randomly selects a bag of popcorn from the assembly line and it has a weight of 3.02 ounces. Then the worker randomly selects a bag of potato chips from the assembly line and it has a weight of 5.03 ounces. Which description closely matches the findings on the assembly line

Step-by-step explanation:

Hello!

You have two variables of interest.

X₁: Weight of a popcorn bag.

It's mean is μ= 3.05 ounces and it's standard deviation δ= 0.02 ounces.

X₂: Weight of a potato chips bag.

With mean μ= 5.07 ounces and standard deviation δ= 0.04 ounces.

A bag of popcorn is randomly selected and its weight is X₁= 3.02 and a bag of potato chips is randomly selected with weight X₂= 5.03.

Since these two values are from completely different distributions, you cannot compare them, but if you convert these values to their equivalent Z value. For this, you will subtract the mean of their distribution and dive them by their standard deviation.

Z= (X-μ)/δ ~N(0;1)

Bag of popcorn: Z=(3.02-3.05)/0.02= -1.5

The selected bag of popcorn is 1.5δ away from the mean.

I hope it helps!

d.192

step-by-step explanation:

12*16=192

answer: a

step-by-step explanation: