A machine fills containers with a particular product. Assume the filling weights are normally distributed with a variance of 1.21 ounces^{2}. 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 
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 Zscore = 1.13 at pvalue = 0.13 and we
need to calculate machine's mean filling weight
Zscore can be calculated as
Zscore = (Xmean)/
So mean can be calculated as
Mean = X  Zscore*
= 24  (1.13*1.1) = 24 + 1.243 = 25.243
So its answer is C. i.e. 25.2430
Get Answers For Free
Most questions answered within 1 hours.