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The weight of males in Canada is normally distributed with mean $73.5\ kg$ and standard deviation $11.3\ kg$. The weight of females in Canada is normally distributed with mean $57.2\ kg$ and standard deviation of $8.8\ kg$. What is the probability that a couple (male+female) the combined weight is bigger than $150\ kg$?

Any ideas?

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  • $\begingroup$ The question conspicuously omits to mention anything about the correlation between weights of the two members of a couple, nor does it say whether they are jointly normally distributed. So it can't be answered without making additional assumptions, although one could get some bounds. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 29 '15 at 3:16
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To answer the question using the given data, we have to assume, quite unreasonably, that if $X$ is the weight of the male member of a randomly chosen couple, and $Y$ is the weight of the female member of a randomly chosen couple, then $X$ and $Y$ are independent. But let us be unreasonable.

Then the random variable $X+Y$ is normally distributed, with mean $130.7$, and variance the sum of the variances, that is, $(11.3)^2+(8.8)^2$. The rest is a standard calculation.

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  • $\begingroup$ wait, why use var and not sd? $\endgroup$ – Melissa Nov 29 '15 at 3:22
  • $\begingroup$ Because if $X$ and $Y$ are independent random variables, the variance of $X+Y$ is the sum of the variances of $X$ and $Y$. Usually, the standard deviation of $X+Y$ is not equal to the sum of the standard deviations. Of course, after computing the variance of $X+Y$ as in my answer, you will take the square root of $(11.3)^2+(8.8)^2$ to find the standard deviation of $X+Y$, and then calculate as you undoubtedly know how to do. $\endgroup$ – André Nicolas Nov 29 '15 at 3:27
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Nov 29 '15 at 3:35

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