Question

(b) Find z such that 39% of all observations from a standard normal distribution are greater than z. (Enter your answer rounded to two decimal places.)

Answer 1

For 39% of all observations that are greater than z on the normal distribution table, the percentage of observations that are lesser than z is

100 - 39 = 61%

The proportion is 61/100 = 0.61

From the normal distribution table, the value of z corresponding to this proportion is 0.28

Thus, the required value of z is 0.28