Final answer:
The probability that the mean cost for these 36 schools is less than $20,000 is practically 0. The probability that the mean cost is greater than $26,000 is about 0.8212, after converting the cost values to z-scores and using the standard normal distribution.
Step-by-step explanation:
The student is asking about finding the probability that the mean cost for 36 selected four-year institutions is either less than $20,000 or greater than $26,000 given a population mean cost of $26,489 and a standard deviation of $3,204.
First, we need to calculate the standard error of the mean, which is the standard deviation divided by the square root of the sample size. Since the sample size is 36, the standard error is:
Standard Error = Standard Deviation / sqrt(Sample Size)
Standard Error = $3,204 / sqrt(36) = $534
Next, we convert the cost values into z-scores using the formula:
Z = (X - Mean) / Standard Error
For a cost of $20,000:
Z = ($20,000 - $26,489) / $534
= -12.155 / 534
Z ≈ -12.08
For a cost of $26,000:
Z = ($26,000 - $26,489) / $534
= -489 / 534
Z ≈ -0.92
To find the probability, we look up these z-scores in a standard normal distribution table or use a statistical software. The probability for a z-score of -12.08 is practically 0, and for -0.92 is approximately 0.1788. Therefore, the probability that the mean cost for these 36 schools is less than $20,000 is practically 0 and the probability that the mean cost is greater than $26,000 is about 0.8212 (since 1 - 0.1788 = 0.8212).