Question

A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 1.21 ounces². If 87% of the containers hold greater than 24 ounces, find the machine's mean filling weight (in ounces).

	a.	22.7570
	b.	None of the answers is correct
	c.	25.2430
	d.	22.6327
	e.	25.3673

Answer 1

Solution:
Given in the question
Filling weights are normally distributed with
Variance() = 1.21
So Standard deviation() = sqrt(Variance) = sqrt(1.21) = 1.1
Also given that 87% of the containers hold greater than 24 ounces that means
P(X>=24) = 0.87
So from Z table, we found Z-score = -1.13 at p-value = 0.13 and we need to calculate machine's mean filling weight
Z-score can be calculated as
Z-score = (X-mean)/
So mean can be calculated as
Mean = X - Z-score* = 24 - (-1.13*1.1) = 24 + 1.243 = 25.243
So its answer is C. i.e. 25.2430