Final answer:
To find the probability for different sample mean costs, we can use the z-score formula. For a sample mean cost less than $25,000, the probability is approximately 0. For a sample mean cost greater than $26,000, the probability is approximately 0.352. And for a sample mean cost between $24,000 and $26,000, the probability is approximately 0.352.
Step-by-step explanation:
To find the probability that the sample mean cost for the 36 schools is less than $25,000, we need to calculate the z-score for this value. The formula to calculate the z-score is:
z = (x - mean) / (o / sqrt(n))
Substituting the given values:
z = (25000 - 26489) / (3204 / sqrt(36))
Simplifying:
z = -2489 / 534 = -4.66
Using a standard normal distribution table or calculator, we can find the probability associated with the z-score of -4.66. This probability is approximately 0.
Therefore, the probability that the sample mean cost for these 36 schools is less than $25,000 is approximately 0.
For part b, we need to calculate the z-score for $26,000 using the same formula. The z-score is:
z = (26000 - 26489) / (3204 / sqrt(36)) = -201 / 534 = -0.38
Using the standard normal distribution table or calculator, we find that the probability associated with a z-score of -0.38 is approximately 0.352.
Therefore, the probability that the sample mean cost for these 36 schools is greater than $26,000 is approximately 0.352.
For part c, we need to calculate the z-scores for $24,000 and $26,000. The z-scores are:
z1 = (24000 - 26489) / (3204 / sqrt(36)) = -2494 / 534 = -4.67
z2 = (26000 - 26489) / (3204 / sqrt(36)) = -201 / 534 = -0.38
Using the standard normal distribution table or calculator, we find that the probability associated with a z-score of -4.67 is approximately 0 and the probability associated with a z-score of -0.38 is approximately 0.352.
Therefore, the probability that the sample mean cost for these 36 schools is between $24,000 and $26,000 is approximately 0.352.