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 about a 2.5% chance of this occurring. It is extremely unlikely for this to occur; the probability is very close to 0. There is about a 5% chance of this occurring. There is no way to evaluate this likelihood, since the sample size (n = 9) is too small. There is about a 16% 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 about a 2.5% chance of this occurring. It is extremely unlikely for this to occur; the probability is very close to 0. There is about a 5% chance of this occurring. There is no way to evaluate this likelihood, since the sample size (n = 9) is too small. There is about a 16% 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 of sample means will be equal to the population mean, and the standard deviation (known as the standard error) will be equal to 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 distribution of sample means is also 30 grams, and the standard error 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:
z = (X - μ) / SE
where X is the sample mean, μ is the population mean, and SE is the standard error.
Substituting the given values, we get:
z = (28 - 30) / 1 = -2
We can then look up this z-score in a standard normal distribution table, or use a calculator or software that can calculate the cumulative probability. The cumulative probability associated with a z-score of -2 is approximately 0.0228, or about 2.28%.
Therefore, there is about a 2.28% chance that the average weight of the bags in a randomly selected pack is 2 or more grams lighter than advertised. This is extremely unlikely to occur.
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