1. The standard error of the mean is approximately 12.65 minutes.
2. The likelihood the sample mean is greater than 320 minutes is 0.7952 or 79.52%.
3. The likelihood the sample mean is between 320 and 350 minutes is 0.7284 or 72.84%.
4. The likelihood the sample mean is greater than 350 minutes is 0.0566 or 5.66%.
1. Standard Error of the Mean:
The standard error of the mean (SEM) is calculated by dividing the population standard deviation by the square root of the sample size. In this case:
SEM = 80 minutes / √40 = 80 minutes / 6.325 ≈ 12.65 minutes
Therefore, the standard error of the mean is approximately 12.65 minutes.
2. Likelihood of Sample Mean Greater than 320 Minutes:
To calculate this, we need to find the z-score of the desired value (320 minutes) compared to the population mean (330 minutes) and then use a z-table or statistical software to find the corresponding probability.
z = (320 - 330) / SEM = -10 / 12.65 ≈ -0.79
Using a z-table or software, the probability of a z-score less than -0.79 is approximately 0.2048. So, the likelihood the sample mean is greater than 320 minutes is 1 - 0.2048 ≈ 0.7952 or 79.52%.
3. Likelihood of Sample Mean Between 320 and 350 Minutes:
We need to find the z-scores for both limits and then calculate the area between them under the standard normal curve.
z1 = (320 - 330) / SEM = -10 / 12.65 ≈ -0.79
z2 = (350 - 330) / SEM = 20 / 12.65 ≈ 1.58
Using a z-table or software, the area between z1 and z2 is approximately 0.9332 - 0.2048 = 0.7284. Therefore, the likelihood the sample mean is between 320 and 350 minutes is 0.7284 or 72.84%.
4. Likelihood of Sample Mean Greater than 350 Minutes:
Again, we need the z-score and corresponding probability:
z = (350 - 330) / SEM = 20 / 12.65 ≈ 1.58
From the z-table or software, the probability of a z-score greater than 1.58 is approximately 0.0566. Thus, the likelihood the sample mean is greater than 350 minutes is 0.0566 or 5.66%.
Question
According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers.
What is the standard error of the mean in this example?
What is the likelihood the sample mean is greater than 320 minutes?
What is the likelihood the sample mean is between 320 and 350 minutes?
What is the likelihood the sample mean is greater than 350 minutes?