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?
Since the sample size is large (n=42), we can use the central limit theorem to assume that the sample means follow a normal distribution.
The mean of the sample means is the same as the population mean, which is $100. The standard deviation of the sample means (also called the standard error of the mean) is:
standard deviation / sqrt(sample size) = $20 / sqrt(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 can standardize the sample mean using the formula:
z = (sample mean - population mean) / standard error
z = ($90 - $100) / $3.08 ≈ -3.24
Using a standard normal distribution table or calculator, we can find the probability that z is greater than -3.24:
P(z > -3.24) ≈ 0.999
Therefore, the probability that the mean coffee expense of a randomly selected sample of 42 college students is greater than $90 is approximately 99.9%.
To solve this problem, we can use the Central Limit Theorem (CLT) since we don't have information about the distribution of the expenses.
The Central Limit Theorem states that for a large enough sample size (n > 30 in most cases), the distribution of sample means will be approximately normally distributed, regardless of the shape of the original population.
In this case, we want to find the probability that the mean coffee expense of a randomly selected sample of 42 college students is greater than $90.
Step 1: Find the mean of the sample. Since the mean monthly expenditure of college students on coffee is $100, the mean of our sample (denoted by x̄) will also be $100.
Step 2: Find the standard deviation of the sample. To find the standard deviation of the sample (s), we divide the standard deviation of the population ($20) by the square root of the sample size (42):
s = 20 / √42 ≈ 3.094
Step 3: Calculate the Z-score. The Z-score measures the number of standard deviations a data point is from the mean. In this case, we want to find the Z-score corresponding to an expense of $90. Using the formula:
Z = (x - μ) / σ
where x is the desired value, μ is the mean, and σ is the standard deviation, we have:
Z = (90 - 100) / 3.094 ≈ -3.231
Step 4: Find the probability. We want to find the probability that a randomly selected sample of 42 college students has a mean coffee expense greater than $90. We need to find the area under the standard normal distribution curve to the right of the calculated Z-score. We can use a Z-table or a statistical software to find the corresponding probability. In this case, the probability is approximately 0.9994.
Therefore, the probability that the mean coffee expense of a randomly selected sample of 42 college students is greater than $90 is approximately 0.9994, or 99.94%.