Answer to Question #171932 in Statistics and Probability for Bikram Chaudhary

Question

The average outstanding balance of loan issued by a bank varies from month to month. From past experience it is known that the amounts are normally distributed with a standard deviation of Rs. 5000. The bank wishes to estimate the average by drawing a random sample such that the probability is 0.95 that the mean of the sample will not deviate by more than Rs. 600 from the universe mean. What should be the sample size?

Answer 1

The critical value for $\alpha=0.05$  is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}},\bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

Given $\sigma=5000$

$$z_c\times\dfrac{\sigma}{\sqrt{n}}\leq300$$

$$n\geq(\dfrac{z_c\times\sigma}{300})^2$$

$$n\geq(\dfrac{1.96\times5000}{300})^2$$

$$n\geq1068$$