# $\mu$of a normal distribution

We have a normal distribution with $\sigma = 18.6$ A line $732$ divides the area to the left of $740$ in 2 parts, such that the area of the part to the left is $1.5$ times as big as the area to the right. Get $\mu$.

This is the answer in the correction model:

Calculator with graphing ability:

$y_1 = 2 . normalcdf (-10^99, 732, X, 18.6)$

$y_2 = 3 . normalcdf (732, 740, X, 18.6)$

Calculate the intersection and that's the answer ($\mu = 746,4$)

My question is, shouldn't the 2 and the 3 be switched, since it says that the left part is 1.5 times bigger than the right part, yet here the left part is multiplied by 2 and the right part by 3..

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We can work backwards and see that the answer of $746.4$ does not work for reasonable interpretations of the problem.
Let $X$ be normal, mean $746.4$, standard deviation $18.6$.
Then $\Pr(X\le 740)\approx 0.4$ and $\Pr(X\le 732)\approx 0.25$. So the point $732$ divides the region to the left of $740$ in approximately the ratio $5:3$. This is a fair distance from the sought-for $1.5:1$.