prev next front |1 |2 |3 |4 |5 |6 |7 |8 |9 |10 |11 |12 |13 |14 |15 |16 |17 |18 |19 |20 |review

Exercise 3 Solution

 

In this question we need to determine the value of n (sample size). But this should be, so that only 10% of the sample mean exceeds μ = 70 by 2 or more. That is X – μ = 2.   From the normal distribution tables the value of z (which divides the area into the lower 90% and the upper 10%) is 1.28. By substituting this value of  z =1.28 in the formula below will yield the value of n.

 

 Z = X-μ

      σ /√n

 

1.28=72-70

        10/√n

 

Therefore, 1.28x10/√n= 2

 

              1.28 x10  = √n  

                   2

   6.40= √n

 

Thus, n= 6.40²

 

          = 40.96