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%.
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?
1 answer