Using the normal distribution, it is found that approximately 79.67% of the bags weigh 3.00 ounces or more.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 3.05, \sigma = 0.06[/tex]
The proportion of the bags that weigh 3.00 ounces or more is one subtracted by the p-value of Z when X = 3, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 3.05}{0.06}[/tex]
Z = -0.83
Z = -0.83 has a p-value of 0.2033.
1 - 0.2033 = 0.7967 = 79.67%.
79.67% of the bags weigh 3.00 ounces or more.
More can be learned about the normal distribution at https://brainly.com/question/27643290
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