Respuesta :
Using the normal distribution, it is found that there is a 0.0668 = 6.68% probability a randomly selected box has less than the 34 ounces of advertised weight.
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 = 34.15, \sigma = 0.1[/tex].
The probability a randomly selected box has less than the 34 ounces of advertised weight is the p-value of Z when X = 34, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34 - 34.15}{0.1}[/tex]
Z = -1.5
Z = -1.5 has a p-value of 0.0668.
0.0668 = 6.68% probability a randomly selected box has less than the 34 ounces of advertised weight.
More can be learned about the normal distribution at https://brainly.com/question/24537145
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