Suppose that a candy company packages a bag of jelly beans whose weight is supposed to be 30 grams, but in fact, the weight varies from bag to bag according to a normal distribution with mean μ = 30 grams and standard deviation σ = 3 grams.If the company sells the jelly beans in packs of 9 bags, what can we say about the likelihood that the average weight of the bags in a randomly selected pack is 2 or more grams lighter than advertised? There is no way to evaluate this likelihood, since the sample size (n = 9) is too small. There is about a 5% chance of this occurring. It is extremely unlikely for this to occur; the probability is very close to 0. There is about a 16% chance of this occurring. There is about a 2.5% chance of this occurring.
Question
Suppose that a candy company packages a bag of jelly beans whose weight is supposed to be 30 grams, but in fact, the weight varies from bag to bag according to a normal distribution with mean μ = 30 grams and standard deviation σ = 3 grams.If the company sells the jelly beans in packs of 9 bags, what can we say about the likelihood that the average weight of the bags in a randomly selected pack is 2 or more grams lighter than advertised? There is no way to evaluate this likelihood, since the sample size (n = 9) is too small. There is about a 5% chance of this occurring. It is extremely unlikely for this to occur; the probability is very close to 0. There is about a 16% chance of this occurring. There is about a 2.5% chance of this occurring.
Solution
To answer this question, we first need to understand the concept of the distribution of sample means. According to the Central Limit Theorem, if we take a large number of random samples from a population, the distribution of the sample means will approximate a normal distribution, regardless of the shape of the population distribution. The mean of this distribution will be equal to the population mean, and the standard deviation (known as the standard error) will be the population standard deviation divided by the square root of the sample size.
In this case, the population mean (μ) is 30 grams, the population standard deviation (σ) is 3 grams, and the sample size (n) is 9. Therefore, the mean of the sample means is also 30 grams, and the standard error (σ/√n) is 3/√9 = 1 gram.
We are asked to find the probability that the average weight of the bags in a randomly selected pack is 2 or more grams lighter than advertised, i.e., less than 28 grams. To find this, we need to calculate the z-score, which is the number of standard errors that a particular sample mean is away from the population mean. The z-score is given by the formula (X - μ) / (σ/√n), where X is the sample mean.
Substituting the given values, we get a z-score of (28 - 30) / 1 = -2. This means that the sample mean of 28 grams is 2 standard errors below the population mean.
Finally, we need to find the probability associated with this z-score. Looking up a z-score of -2 in a standard normal distribution table, we find that the probability is approximately 0.0228, or about 2.28%. Therefore, there is about a 2.28% chance of a randomly selected pack of 9 bags having an average weight that is 2 or more grams lighter than advertised.
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