A machine fills 16-ounce soda cans with a mean of 16.06 ounces of soda and a standard deviation of 0.03 . Approximately what percentage of the cans will actually contain less than the advertised less than the advertised 16 ounces of soda?
A) 2.5%
B) 5%
C) 10%
D) 12.5%
E) 15%

Question

A machine fills 16-ounce soda cans with a mean of  16.06  ounces of soda and a standard deviation of  0.03 . Approximately what percentage of the cans will actually contain less than the advertised less than the advertised 16 ounces of soda?
A)  2.5% 
B)  5% 
C)  10% 
D)  12.5% 
E)  15%

Quizplus · Accepted Answer

Using the Z-score formula, $Z = \frac{X - \mu}{\sigma}$, where $X = 16$ ounces, $\mu = 16.06$ ounces, and $\sigma = 0.03$ ounces, we find $Z = \frac{16 - 16.06}{0.03} = -2$. Looking up a Z-score of -2 in a standard normal distribution table gives us an area of approximately 2.5%, indicating that about 2.5% of the cans will contain less than 16 ounces of soda.

A Machine Fills 16-Ounce Soda Cans with a Mean of