Standard Normal Distribution Question

qwert said:

Hey guys, I got a little stuck with this question here:

The random variable X has a normal distribution with mean =m and variance =4. Given that P(X<5) =P(X>3k),
(i) Show that m=(5+3k)/2

I've been given the solution but I'm unsure of some of the workings. I hope someone can me with it. Thanks!

Solution:
X~N(m, 2^2)
Given P(X<5) =P(X>3k),
P(Z<(5-m)/2) = P(Z>(3k-m)/2)
(5-m)/2 = - (3k - m)/2 [This is the part im unsure of- why must we multiply (3k-m)/2 with a negative sign?]
5-m=m-3k
m=(5+3k)/2

Thank you very much!

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Ishuda said:

For a symmetric distribution about the mean [which the normal distribution is], the probability that x is greater than p plus the mean is the same as the probability that -x plus the mean is less than p. That is, for a non-negative x less than or equal to the mean, the area under the curve of the probability distribution function from -∞ to -(x-m) [the probability that Z is less than the distance of -x to the mean] is the same as the area under the curve of the probability distribution function from x-m to +∞ [the probability that Z is greater than the distance from the mean to x]. So P(Z>(3k-m)/2) = P(Z<-(3k-m)/2)

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