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Answer :
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%
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Final answer:
To find the percent of women who have hemoglobin greater than 17 grams per deciliter, we can use the normal distribution curve with a mean of 14 and a standard deviation of 1. The area to the right of 17 can be found using a z-score.
Explanation:
To find the percent of women who have hemoglobin greater than 17 grams per deciliter, we need to find the area to the right of 17 on the normal distribution curve with a mean of 14 and a standard deviation of 1. We can use a z-score to find this area. The formula for the z-score is z = (x - mean) / standard deviation. Plugging in the values, we get z = (17 - 14) / 1 = 3.
Using a standard normal distribution table or a calculator, we can find that the area to the right of 3 on the standard normal distribution curve is approximately 0.0013, or 0.13%. Therefore, approximately 0.13% of women have hemoglobin greater than 17 grams per deciliter.
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