posted by Jason .
A private medical clinic wants to estimate the true mean annual income of its patients. The clinic needs to be within $100 of the true mean. The clinic estimates that the true population standard deviation is around $2,100. If the confidence level is 95%, find the required sample size in order to meet the desired accuracy.
Use a formula to find sample size.
Here is one:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent the 95% confidence interval, sd = 2100, E = 100, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
Hope this helps.
A population is normally distributed with mean 26.8 and standard deviation 2.5. Find p( 26.8 < x < 29.3 ). (Round your answer to FOUR decimal places.)