Answer:
0.13%
Step-by-step explanation:
We dont care about the men distribution, so we can focus on women distribution of hemoglobin in grams per deciliter. Lets call Y the random variable that states the hemoglobin in grams per deciliter for a woman. The exercise tells us that Y has distribution N(14,1) We want Y greater than 17, so we need to calculate P(Y > 17). We can standarize Y by substracting its expected value 14 and by dividing by its standard deviation 1.
The variable [tex] Z = \frac{Y - 14}{1} = Y-14 [/tex] is a Standard normal random variable, and its cummulated distribution function, Ф, is tabulated. The values of Ф can be found on the attached file.
P(Y > 17) = P(Y-14 > 17-14) = P(Z > 3) = 1 - P(Z ≤ 3) = 1 - Ф(3) = 1 - 0.9987 = 0.0013
Thus, the percent of women with greater hemoglobin than 17 grams per deciliter is 100*0.0013 = 0.13%