Answer to Question #173803 in Statistics and Probability for nada

Question

The mean and standard deviation of serum iron values which are normally distributed for a random sample of 50 healthy men are 120 and 15 micrograms respectively. Find 95% confidence limits within the population mean. (z=1.96)

Answer 1

Solution:

Given, $n=50,\overline{X}=120,\sigma=15$

For 95%, z=1.96

Now, confidence interval for $\mu=(\overline{X}-z(\sigma/\sqrt n), \overline{X}+z(\sigma/\sqrt n))$

$=(120-1.96(15/\sqrt {50}), 120+1.96(15/\sqrt {50})) \\\approx(120-4.157,120+4.157) \\=(115.843,124.157)$