# Probability of a point from one normal distribution being higher than a point from another independent normal distribution

Given two independent normal distributions:

Distribution 1: Mean $$= 23.95$$, SD $$= 7.44$$

Distribution 2: Mean $$= 16.29$$, SD $$= 7.79$$

How often on average will a point from Distribution 2 be greater than a point from Distribution 1?

I apologize for any nomenclature that is incorrect.

### Progress

I know that the distribution of $$Z=Y-X$$ is normal, as well as its mean and variance.

• Hint: Since the random variables are normal and independent, $Y-X$ is a normal random variable with mean $\mu_Y-\mu_X$ and variance $\sigma_Y^2+\sigma_X^2$. Do you know how to determine $P\{Y-X > 0\} = P\{Y > X\}$ from this knowledge about $Y-X$? Dec 16, 2014 at 20:54
• I see what you did with the inequality I think. So in this case the mean is 7.66 and the SD is 10.77? So Given this knowledge how do I determine how likely P{Y > X}. Would it help if I explained my practical reasoning for calculating this since I am such a stat. novice? Dec 16, 2014 at 21:26
• Write $Z=Y-X$. What do you know about the distribution of $Z$? You know its mean and variance but what else? (Hint: read the adjectives applied to the object of the verb is in the main clause of the first sentence in my previous comment). Then, read the second sentence of my previous comment, applying it to $Z$ about which you know, I hope, a lot by this time. Dec 16, 2014 at 21:40
• I know that the distribution of Z is normal, as well as its mean and variance. So the probablity of Z > 0 is equal to the probability of Y > X. I calculate this to be 0.238469339. Is this correct? Thanks so much Dilip for your help by the way. Dec 16, 2014 at 22:13

Let $X_1 \sim N(23.95,7.44), X_2 \sim N(16.29,7.79)$,

then, as mentioned by Dilip Sarwate in the comments, $X_1 - X_2 \sim N(7.66,{\sqrt{7.44^2 + 7.79^2}})$

Thus the problem becomes $$\mathbb{P}(X_1<X_2) = \mathbb{P}(X<0),\ \ where \ \ X = X_1 - X_2$$

This can be calculated directly or reverting to the standard $N(0,1)$ distribution as follows:

\begin{align} \mathbb{P}( X < 0 ) = \mathbb{P}(\frac{X-7.66}{{\sqrt{7.44^2 + 7.79^2}}}<\frac{-7.66}{\sqrt{7.44^2 + 7.79^2}}) \end{align}

Noting that $$\frac{X-7.66}{{\sqrt{7.44^2 + 7.79^2}}} = Z \sim N(0,1)$$

So the required probability is, approximately,

\begin{align} \mathbb{P}(Z<-0.7110976012) &= \mathbb{P}( Z > 0.7110976012) \\ &= 1 -\mathbb{P}(Z<0.7110976012) \\ &\approx 1 - 0.7611\\ &= 0.2389 \end{align}

The numerical value found by looking up 0.71 in a standard normal table.

• Cool, thanks a lot for the help Daniel. So if I wanted to determine the probablity of X2 exceeding X1 by more than zero, I would just plug in a different number for X, I think. For example if I wanted to know the probability of X2 exceeding X1 by more than 7, I would plug in -7 for X correct? If I do this, my new z-score is -1.36 and therefore the P of X2 > X1 + 7 = .0869. Is this correct? Dec 17, 2014 at 13:17
• @RobbieTraylor That's right. $X_1 + 7 < X_2 \Leftrightarrow X_1 - X_2 + 7 < 0$ and we know that $X_1 - X_2 + 7 \sim N(7.66+7,{\sqrt{7.44^2 + 7.79^2}})$. Doing the same calculations as above yields the Z-score and probability you mention. Dec 17, 2014 at 13:53
• I'm really greatful to everyone who helped. I've got another problem about adding normal distributions, but I think it should probably have its own question. Dec 17, 2014 at 18:58