What is the probability that simple random sample of 50 staff members provides a sample mean leave of 6.5 to 9.5 weeks?
More data needed. SD?
Can assume mean = (9.5-6.5)/2?
Z = (score-mean)/SEm
SEm = SD/√n
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability of the Z score.
To determine the probability, we need to make some assumptions and use the properties of the normal distribution. Here are the steps to calculate the probability:
Step 1: Determine the population mean (µ) and standard deviation (σ).
- If the population mean and standard deviation are known, use those values.
- If they are unknown, we can estimate them using the sample mean (x̄) and the standard deviation of the sample (s).
Step 2: Determine the sample size (n) and the desired sample mean range.
- In this case, the sample size is given as 50 staff members, and the desired sample mean range is 6.5 to 9.5 weeks.
Step 3: Calculate the standard error (SE), which is the standard deviation of the sample means.
- The standard error can be calculated using the formula: SE = σ / √n, where σ is the standard deviation of the population and n is the sample size.
Step 4: Convert the desired sample mean range to z-scores.
- Calculate the z-scores for both ends of the range using the formula: z = (x - µ) / SE, where x is the desired sample mean, µ is the population mean, and SE is the standard error.
Step 5: Determine the probability using the standard normal distribution table or a statistical software.
- Use the z-scores from step 4 to find the area under the standard normal distribution curve that corresponds to the desired range.
By following these steps and using the appropriate values for the population mean, standard deviation, sample size, and desired sample mean range, you can calculate the probability that a simple random sample of 50 staff members provides a sample mean within the specified range.