Question

g Consider a brand of coffee. The weight of a pod of coffee this brand makes has mean 42.05 grams and standard deviation 0.025 grams. Using the central limit theorem and the normal distribution, what is the probability that the mean weight of 25 pods of coffee is less than 42.035 grams

Answer 1

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 $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• 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 $s = \frac{\sigma}{\sqrt{n}}$.

The parameters are given as follows:

$\mu = 42.05, \sigma = 0.025, n = 25, s = \frac{0.025}{\sqrt{25}} = 0.005$

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:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{42.035 - 42.05}{0.005}$

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 brainly.com/question/4079902

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