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I have the following task: The country has the same number of men and women. Normal height distributions for each sex are shown in the picture . The height of an individual person is 170 cm. Find the probability that this person is a woman.

I have an idea to simply write the normal distribution formulas for men and women, substituting the height 170 for x, and then find the desired probability as the ratio of the distribution value from 170 for men and the value of the sum of the distributions for men and women (also for x = 170). But I doubt the correctness of this approach a little, since it turns out that we divide the distributions by each other, and in the end we should get the probability

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  • $\begingroup$ How can an individual's height be exactly 170 cm? Height is a continuous distribution, so this probability is zero (for both genders). Perhaps you should re-phrase the question. $\endgroup$ Nov 22, 2021 at 12:41

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You use the frequency distributions to find the likelihood that woman is $170 cm$ and the probability that a man is $170 cm$

Then we use Bayes theorem to find the probability of a woman given the height is 170 cm

$\frac {P(170|woman)}{P(170|woman) + P(170|man)}$

There is a small shortcut here because the number of men and the number of women in the population is equal. If it were not we would need weighting factors for the sizes of the populations.

$\frac {P(170|woman)(\text{female pop})}{P(170|woman)(\text{female pop}) + P(170|man)(\text{male pop})}$

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  • $\begingroup$ Thank you very much for your reply! But wouldn't it be a mistake if instead of probabilities we substitute values calculated using the distribution formula? After all, the value f(x_0) corresponds to the probability P (X <x_0), and not P (X = x_0) $\endgroup$
    – Alex Scain
    Nov 22, 2021 at 10:49
  • $\begingroup$ Is your formula a frequency distribution or a cumulative distribution? $\endgroup$
    – user317176
    Nov 22, 2021 at 11:18
  • $\begingroup$ Hmm, I guess, this is formula of probability density: ibb.co/L6XgMxm $\endgroup$
    – Alex Scain
    Nov 22, 2021 at 11:28

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