Assume that the sample is taken from a large population and the correction factor can be ignored.
The average teacher's salary is $43152. Suppose that the distribution is normal with standard deviation equal to $7000. If we sample 100 teachers' salaries, what is the probability that the sample mean is less than $42249?
Use same process as described in your previous post.
To answer this question, we need to calculate the z-score and then use the z-table or a statistical software to find the corresponding probability.
The z-score is a measure of how many standard deviations the sample mean is from the population mean. It is calculated using the formula:
z = (sample mean - population mean) / (standard deviation / sqrt(sample size))
In this case:
sample mean = $42249
population mean = $43152
standard deviation = $7000
sample size = 100
z = ($42249 - $43152) / ($7000 / √100)
z ≈
(-903 / $700) * 10
z ≈ -1.29
Next, we can use the z-table (or a statistical software) to find the probability corresponding to a z-score of -1.29. The z-table provides the cumulative probability up to a given z-score.
Looking up the z-score of -1.29 in the z-table, we find that the cumulative probability is approximately 0.0985.
Therefore, the probability that the sample mean is less than $42249 is approximately 0.0985 or 9.85%.