Statistics as level cie math question ?please help

It is given that Y ∼ N(33, 21). Find the value of a given that P(33 − a < Y < 33 + a) = 0.5.?

Please explain thoroughly. I do not get it , any kind of help would be great

First of all, draw a nice sketch of the normal distribution for Y, marking on the mean.

You should notice that the region $33-a < Y < 33+a$ is actually symmetrical about the mean, which means that the probabilities $P(33-a < Y < 33)$ and $P(33<Y<33+a)$ are indeed the same. This means we only need to focus on one of them if we can.
Indeed we can because know that $P(33-a < Y < 33+a) = P(33-a < Y < 33) + P(33 < Y < a+33) = 2P(33 < Y < 33+a) = 0.5$

This implies that we have $P(33 < Y < 33+a) = 0.25$. We can also rewrite the LHS by saying that $P(33 < Y < 33+a) = P(Y<33+a) - P(Y < 33)$ hence we have $P(Y<33+a) - 0.5 = 0.25$ hence $P(Y < 33+a) = 0.75$

From here it should be straight-forward to determine $a$ such that $P(Y < 33+a) = 0.75$ holds.

Thank you very much , I really appreciate all that , I get it , but the only question is that is p(y<33) equal to 0.5? Thank u again , really helpful

Thank you very much , I really appreciate all that , I get it , but the only question is that is p(y<33) equal to 0.5? Thank u again , really helpful

Because 33 is the mean, and in normal distribution the mean=mode=median.

I just meant where did u get the 0.5 from

I just meant where did u get the 0.5 from

First of all, draw a nice sketch of the normal distribution for Y, marking on the mean.

You should notice that the region $33-a < Y < 33+a$ is actually symmetrical about the mean, which means that the probabilities $P(33-a < Y < 33)$ and $P(33<Y<33+a)$ are indeed the same.

This means we only need to focus on one of them if we can.
Indeed we can because know that



This implies that we have $P(33 < Y < 33+a) = 0.25$.

We can also rewrite the LHS by saying that $P(33 < Y < 33+a) = P(Y<33+a) - P(Y < 33)$ hence we have $P(Y<33+a) - 0.5 = 0.25$ hence $P(Y < 33+a) = 0.75$

From here it should be straight-forward to determine $a$ such that $P(Y < 33+a) = 0.75$ holds.