To answer these questions, we will use the properties of the sampling distribution of the sample mean, which is characterized by the Central Limit Theorem (CLT). The CLT states that, under certain conditions, the distribution of sample means will be approximately normal, regardless of the shape of the population distribution.
(a) The distribution of the sample mean on-campus housing cost for a random sample of 50 students will also be approximately normally distributed. The mean of this distribution will be the same as the population mean, μ = $1553, and the standard deviation of the distribution of sample means, also known as the standard error, can be calculated using the formula:
Standard Error (SE) = Population Standard Deviation (σ) / Square Root of Sample Size (n)
SE = 329 / √50 ≈ 46.462
So, the distribution of the sample mean on-campus housing cost for a random sample of 50 students is normally distributed with a mean of $1553 and a standard deviation of approximately $46.462.
(b) To calculate the probability that a randomly selected on-campus student spends more than $1200 per semester, we need to find the z-score corresponding to this value and then use the z-table to find the corresponding probability.
Z-score = (X - μ) / σ
Z-score = (1200 - 1553) / 329 ≈ -1.056
Using the z-table, we can find the probability associated with a z-score of -1.056. This probability represents the area under the standard normal distribution curve to the left of -1.056. To find the probability to the right of -1.056 (corresponding to spending more than $1200), we subtract this probability from 1.
P(X > 1200) = 1 - P(Z < -1.056)
Look up the probability value for -1.056 in the z-table (or use a calculator) to find P(Z < -1.056), and then subtract it from 1 to get the final probability.
(c) To calculate the probability that a random sample of 50 students spends an average of less than $1500, we need to find the z-score for this value and use the z-table to find the corresponding probability.
Z-score = (X - μ) / (σ / √n)
Z-score = (1500 - 1553) / (329 / √50) ≈ -0.638
Using the z-table, locate the probability associated with a z-score of -0.638. This probability represents the area under the standard normal distribution curve to the left of -0.638.
P(XÌ„ < 1500) = P(Z < -0.638)
Look up the probability value for -0.638 in the z-table to find the final probability.
(d) To calculate the probability that a random sample of 50 students spends an average of $1450 to $1600 per semester, we need to find the z-scores for these values and use the z-table to find the corresponding probabilities.
Z-score for $1450: (1450 - 1553) / (329 / √50) ≈ -2.548
Z-score for $1600: (1600 - 1553) / (329 / √50) ≈ 1.433
The probability can be calculated as the difference between the areas under the standard normal distribution curve corresponding to these two z-scores.
P(1450 < XÌ„ < 1600) = P(-2.548 < Z < 1.433)
Using the z-table, find the probabilities associated with -2.548 and 1.433 and then subtract the smaller probability from the larger probability to get the final probability.
Note: These calculations assume that the population distribution of on-campus housing costs for a semester follows a normal distribution and that the sample is a simple random sample.