Final answer:
The problem asks to calculate the probability that the sample mean cost of 36 schools is over $26,000, given a population mean of $26,489 and a standard deviation of $3204. By using the Central Limit Theorem and calculating the z-score, we can determine this probability utilizing standard normal distribution methods.
Step-by-step explanation:
The question involves finding the probability that the sample mean cost for 36 randomly selected four-year institutions is greater than $26,000, given that the population mean is $26,489 and the population standard deviation is $3204. This is a problem that can be solved using concepts from statistics, particularly the Central Limit Theorem (CLT). Because we have a sufficiently large sample size (n = 36), we can assume that the sampling distribution of the sample mean is approximately normally distributed, even if the population distribution is not normal.
First, we compute the standard error (SE) of the sample mean by dividing the population standard deviation by the square root of the sample size: SE = σ / √ n, which equals $3204 / √ 36.
Once we have the SE, we find the z-score for $26,000 by subtracting the population mean from $26,000 and dividing the result by the SE. The z-score tells us how many standard errors $26,000 is away from the population mean. Using the z-tables or standard normal distribution calculators, we can then find the probability associated with this z-score.
Last, since we are interested in the probability that the sample mean is greater than $26,000, we will look for the probability that z is greater than the calculated z-score. This probability will indicate how likely it is that the sample mean for the 36 schools is more than $26,000.