If the standard deviation for waiting times in a line is 5 minutes, the standard deviation for the averages of 30 randomly chosen waiting times is _____ minutes.
SEm = standard error(deviation) of the mean
SEm = SD/√n
0.9
To find the standard deviation for the averages of 30 randomly chosen waiting times, we need to divide the standard deviation of the individual waiting times by the square root of the sample size.
Given that the standard deviation for the waiting times in a line is 5 minutes, we can use the formula:
Standard deviation for averages = (Standard deviation for individual waiting times) / √(Sample size)
Substituting the given values, we have:
Standard deviation for averages = 5 / √30
Calculating this expression, we find:
Standard deviation for averages ≈ 0.9129
Therefore, the standard deviation for the averages of 30 randomly chosen waiting times is approximately 0.9129 minutes.
To find the standard deviation for the averages of 30 randomly chosen waiting times, you need to apply the concept of the Central Limit Theorem. The Central Limit Theorem states that for a large enough sample size, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution.
In this case, you are given the standard deviation of the waiting times in a line, which is 5 minutes. However, you do not have information about the population mean. Nevertheless, as long as the sample size is sufficiently large (typically, >30), you can assume that the distribution of the sample means will be normally distributed.
To find the standard deviation for the averages of 30 randomly chosen waiting times, you can use the formula:
Standard Deviation of Sample Means = Standard Deviation of Population / Square Root of Sample Size
From the given information, we know that the standard deviation of the waiting times in a line is 5 minutes and the sample size is 30.
Plugging these values into the formula:
Standard Deviation of Sample Means = 5 minutes / √30
Using a calculator, the standard deviation of the averages of 30 randomly chosen waiting times can be calculated as approximately 0.91 minutes.