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Using the normal distribution, it is found that there is a 0.0013 = 0.13% probability that the mean weight of 25 pods of coffee is less than 42.035 grams.
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.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
The parameters are given as follows:
[tex]\mu = 42.05, \sigma = 0.025, n = 25, s = \frac{0.025}{\sqrt{25}} = 0.005[/tex]
The probability that the mean weight of 25 pods of coffee is less than 42.035 grams is the p-value of Z when X = 42.035, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{42.035 - 42.05}{0.005}[/tex]
Z = -3
Z = -3 has a p-value of 0.0013.
0.0013 = 0.13% probability that the mean weight of 25 pods of coffee is less than 42.035 grams.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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