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!
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!