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mojhsa [17]
3 years ago
7

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: _______%
Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0

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% .

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