The mean of nonresidential buildings is 30 years and the standard deviation of the ages of the buildings is 5 years. Determine the probability that a random sample of a group of 50 nonresidential buildings will have a mean age of less than 27.5 years.
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To determine the probability that a random sample of 50 nonresidential buildings will have a mean age of less than 27.5 years, we can use the Central Limit Theorem. According to the Central Limit Theorem, as the sample size increases, the sampling distribution of the mean approaches a normal distribution, even if the population distribution is not normal.
First, let's calculate the standard deviation of the sampling distribution of the mean, also known as the standard error. Since the standard deviation of the ages of the buildings is 5 years and the sample size is 50, the standard error can be calculated as follows:
Standard Error = Standard Deviation / Square Root of Sample Size
= 5 / sqrt(50)
= 5 / 7.071
≈ 0.707
Next, we need to calculate the z-score, which measures the distance between the sample mean and the population mean in terms of standard errors. The z-score can be calculated as follows:
z = (Sample Mean - Population Mean) / Standard Error
= (27.5 - 30) / 0.707
≈ -3.536
Now, we can use a standard normal distribution table or calculator to find the probability corresponding to the z-score. Since we are interested in the probability of a mean age less than 27.5 years, we need to find the area to the left of the z-score (-3.536).
Using a standard normal distribution table or calculator, the probability corresponding to a z-score of -3.536 is approximately 0.0002.
Therefore, the probability that a random sample of 50 nonresidential buildings will have a mean age of less than 27.5 years is approximately 0.0002, or 0.02%.