Here's the problem statement...

A sample consisting of 48% men and 52% women is such that the height of men follows a normal distribution with mean 182cm and variance 7cm, and the height of women follows a normal distribution with mean 168cm and variance 5cm.

Given a person selected at random is between 174 and 175 cm tall what's the probability of that person being a women?

...I set the height of men as X~N(182, 7^2) and women as Y~N(168, 5^2). I calculated P(174 < Y < 175) / (P(174 < Y < 175) * P(174 < X < 175)) which gave me the wrong answer, 0.52. The answer is supposed to be 0.54.


Let the random variable $S$ denote the sex of a randomly selected individual. The possible values of $S$ are $m$, and $f$. Also,let $H$ denote the height of a randomly selected individual. The conditional probability in question is

$$P(S=f|174\le H\le 175)=\frac{P(S=f \cap 174\le H\le 175)}{P(174\le H\le 175)}.$$ First,

$$P(S=f \cap 174\le H\le 175)=P(174\le H\le 175| S=f)P(S=f).$$

Second, $$P(174\le H\le 175)=$$ $$=P(174\le H\le 175|S=m)P(S=m)+P(174\le H\le 175|S=f)P(S=f).$$

You have all the information to complete the calculation above.


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