Question

The distribution of the amount of a certain brand of soda in 16 OZ bottles is approximately normal with a mean of 16.12 OZ and a standard deviation of 0.09 OZ. The percentage of the soda bottles that contain more than the 16 OZ advertised is: _______%

Answer 1

Answer: 90.82%

Step-by-step explanation:

Given : The distribution of the amount of a certain brand of soda in 16 OZ bottles is approximately normal .

Mean : $\mu=16.12\text{ OZ}$

Standard deviation: $\sigma=0.09\text{ OZ}$

Let X be the random variable that represents the amount of soda in bottles.

Formula for z-score : $z=\dfrac{x-\mu}{\sigma}$

Z-score for 16 oz: $z=\dfrac{16-16.12}{0.09}=-1.33$

Using the standard normal z-distribution table , the probability that the soda bottles that contain more than the 16 OZ is given by :_

$P(x>60)=P(z>-1.33)=1-P(x\leq-1.33)=1-0.0917591=0.9082409\approx0.9082\approx90.82\%$

Hence, the percentage of the soda bottles that contain more than the 16 OZ advertised is 90.82% .