Question

3. In the November 1990 issue of Chemical Engineering Progress, a study discussed the percent purity of oxygen from a certain supplier. Assume that the mean was 99.61 with a standard deviation of 0.08. Assume that the distribution of percent purity was approximately normal.

a) What percentage of the purity values would you expect to be between 99.5 and 99.7?

b) What purity value would you expect to exceed 5% of the population?

Answer 1

µ = 99.61

sd = 0.08

a)

                                                

                                                 = P(-1.375 < Z < 1.125)

                                                 = P(Z < 1.125) - P(Z < -1.375)

                                                 = 0.8679 - 0.0846

                                                 = 0.7833

                                                 = 78.33%

b)

or, x = 99.61 + 1.645 * 0.08

or, x = 99.7416