What is the probability that the mean weight of a random sample of 3 olives from a normally distributed population with a mean of 7.7 grams and a standard deviation of 0.2 grams is... What is the probability that the mean weight of a random sample of 3 olives from a normally distributed population with a mean of 7.7 grams and a standard deviation of 0.2 grams is greater than 8 grams? Read more

Steps to Solve

1. Identify Parameters of the Distribution The problem states that the weights of olives are normally distributed with a mean ($\mu$) of 7.7 grams and a standard deviation ($\sigma$) of 0.2 grams.

2. Determine the Sampling Distribution For a sample mean of size $n = 3$, the sampling distribution of the sample mean will have a mean $$\mu_{\bar{x}} = \mu = 7.7 \text{ grams}$$ and a standard deviation (standard error) given by $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{0.2}{\sqrt{3}} \approx 0.1155 \text{ grams}.$$

3. Standardize the Value To find the probability that the sample mean is greater than 8 grams, we first standardize this value using the z-score formula: $$ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} $$ Substituting the values: $$ z = \frac{8 - 7.7}{0.1155} \approx 2.59. $$

4. Find the Probability Now, we need to find the probability corresponding to $z = 2.59$. This can be found using a standard normal distribution table or calculator: $$ P(Z > 2.59) = 1 - P(Z \leq 2.59). $$ Using the standard normal table, we find that $P(Z \leq 2.59) \approx 0.9952$.

5. Calculate the Final Probability Thus, the final probability is: $$ P(Z > 2.59) = 1 - 0.9952 \approx 0.0048.$$