Question

According to a 2011 publication, the average monthly expenditure of college students on coffee is $100. Given a standard deviation of $20, but with no confirmation of a normal distribution, what is the probability that the mean coffee expense of a randomly selected sample of 42 college students is greater than $90?

Answer 1

To answer this question, we can use the Central Limit Theorem, which states that the distribution of sample means tends to be approximately normal, regardless of the shape of the population distribution, as long as the sample size is large enough.

In this case, the sample size is 42, which is considered large enough for the Central Limit Theorem to apply. We can approximate the distribution of the sample mean using a normal distribution.

The mean of the sample means will still be $100, as it is the same as the population mean. However, the standard deviation of the sample means, also known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size. In this case, the standard error is $20 / √42 ≈ $3.08.

To find the probability that the mean coffee expense of a randomly selected sample of 42 college students is greater than $90, we need to calculate the z-score and then find the corresponding probability using a standard normal distribution table or calculator.

The z-score is calculated as follows:
z = (sample mean - population mean) / standard error
= ($90 - $100) / $3.08
≈ -3.25

Using the z-score of -3.25, we can find the corresponding probability from the standard normal distribution table or calculator. The probability will be the area to the right of the z-score, which represents the probability that the mean coffee expense is greater than $90.

Note that since the z-score is very negative, the probability will be very close to 1.

Answer 2

The Central Limit Theorem states that, given a sufficiently large sample size, the sampling distribution of the mean for a variable will approximate a normal distribution regardless of the underlying distribution of the variable. In this case, we have a sample size of 42, which is large enough for the Central Limit Theorem to apply.

The mean of the sampling distribution of the mean is equal to the population mean, which is $100. The standard deviation of the sampling distribution of the mean is equal to the population standard deviation divided by the square root of the sample size, which is $20 / sqrt(42) ≈ $3.08.

We can standardize to find the z-score for a sample mean of $90: z = ($90 - $100) / $3.08 ≈ -3.25. Using a z-table, we find that the probability of getting a z-score less than -3.25 is approximately 0.0006. Therefore, the probability that the mean coffee expense of a randomly selected sample of 42 college students is greater than $90 is approximately 1 - 0.0006 = 0.9994.