A particular fruit's weights are normally distributed, with a mean of 657 grams and a standard deviation of 29 grams.

1. If you pick 23 fruit at random, what is the probability that their mean weight will be between 651 grams and 671 grams?

2. A population of values has a normal distribution with [tex]\mu = 101.2[/tex] and [tex]\sigma = 32.3[/tex]. You intend to draw a random sample of size [tex]n = 222[/tex].

a. Find the probability that a single randomly selected value is greater than 105.8. Enter your answer as a number accurate to 4 decimal places.

b. Find the probability that a sample of size [tex]n = 222[/tex] is randomly selected with a mean greater than 105.8. Enter your answer as a number accurate to 4 decimal places.

This problem involves statistics, specifically the application of Normal Distribution and the Central Limit Theorem. It includes calculating probabilities related to normally distributed weights of fruits and understanding how sample means approximate a normal distribution for large samples.

Explanation:

This is a Statistics related question specifically focusing on the application of Normal Distribution and the Central Limit Theorem. The first part of the problem deals with calculating the probability that the average weight of a randomly selected group of fruits is within a specific range. This can be computed by using the formulas of normal distribution where μ is the mean, σ is the standard deviation and n is the size of the randomly selected group. It uses the concept of the Central Limit Theorem which states that the distribution of sample means approximates a normal distribution as the sample size becomes larger.

The second part of the question involves using the formula for a single randomly selected value from a normally distributed population. Here, μ and σ are again the population mean and standard deviation respectively. We calculate the z-score to find the likelihood that a randomly selected fruit will be heavier than a certain weight. Finally, we do this again for a sample, only this time we use the standard error (σ/sqrt(n)), in place of the standard deviation in our calculations.

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