Conditional probability and the normal distribution

Until fairly recently, you could throw a handkerchief over the variety of normal distribution questions you might expect to see in an EdExcel S1 exam. It would be one or more of:

  • given a mean and a standard deviation, work out the probability that the random variable is larger or smaller than a given observation (very common)
  • given a probability (as well as $\mu$ and $\sigma$, work out the observation (fairly common))
  • given two observations and two probabilities, work out the mean and standard deviation (rather rare)

You might get something involving the symmetry of the graph, or requiring you to know that the mean, median and mode are identical, but -- for the most part -- the normal distribution questions were pretty predictable and routine.

I presume someone at Ofqual spotted this, too, as another sort of question has crept into play recently, combining the normal distribution with conditional probability -- in most of my students' books, a sort of anti-Rastamouse affair: making a bad thing worse.

The thing is, once you know what to look out for, conditional probability with normal distributions isn't especially complicated, as long as you remember the golden rule of conditional probability1 : The probability of one event given another is (the probability of both events happening) divided by (the probability of the thing you're given).

For example, the probability of a given email in my inbox being spam, given that it starts 'Dear Flying', would need me to work out:

  • The probability of an email starting "Dear Flying" -- of the 100 or so emails I've had in the last few days, four of them have started "Dear Flying", so let's call it 4%.
  • The probability of an email starting "Dear Flying" and being spam -- well, three of them were trying to sell me SEO services and got put in the spam bucket quicker than you can say 'front page of Google', and one was someone trying to be funny. So 3% of the emails started badly and were spam

Now to do the sum: the probability of an email being spam given that it starts "Dear Flying" is the probability of both things being true (3%) divided by the probability of the 'given' bit being true (4%), making a conditional probability of $\frac 34$. That is to say, three quarters of the emails that started "Dear Flying" were spam.

When you're doing it with normal distribution, it works just the same way.

So we have an example to work with, let's take a variable, $X$, which is normally distributed with a mean of 100 and a standard deviation of 10.

One form of conditional question is to ask directly: what is $P( X > 120 \mid X > 90 )$?

Let's try it: we need to work out the probability of $X$ being greater than 90. First, work out the $z$-score: how many standard deviations is the value above the mean? It's one standard deviation below, so $z = -1$. Looking in the big table, $\Phi(1.00) = 0.8413$. Looking at the picture you've no doubt draw, you notice that the probability has to be bigger than a half, so we don't need to take it away from one; the bottom of our fraction is 0.8413.

As for the top, we want to know what the probability of $X$ being greater than both 90 and 120. Well, if it's bigger than 120, it's automatically bigger than 90, so we just need $P( X > 120)$. For that, the $z$-score is 2, and $\Phi(2.00) = 0.9772$. This time, the probability is under a half, so the top of the fraction is 0.0228.

Throw it together: the probability is $\frac{0.0228}{0.8413} \approx 0.0271$.

A second version is wordier, and might involve setting up the distribution as something 'real-life': let's say it's the number of guests expected at a wedding. The venue can only hold 110 people, so it's a disaster if more than than show up; similarly, if fewer than 95 people come along, the wedding's a disaster because the happy couple look like Billy No Mateses. You notice that the mother of the bride is throwing a hissy fit about the wedding being a disaster -- what's the probability that she's upset because too many people came along?

The first thing to do is to translate this into something you can work out: you're given that the wedding is a disaster (either too many or too few showed up). What was the probability of the two cases? The probability of too few guests involves a $z$-score of -0.5; $\Phi(0.50) = 0.6915$, but we want the less-than-half version, which is 0.3085. The probability of two many involves a $z$-score of 1; $\Phi(1.00)=0.8413$, and again, we want the less-than-half version; the probability is 0.1587.

So, the probability of (disaster and too many) is 0.1587; the probability of (any disaster) is 0.4672; the probability of there being too many guests, given that the wedding is a disaster, is $\frac{0.1587}{0.4672} \approx 0.3397$.

* Edited 2015-05-27 to fix LaTeX and typos.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Bayes's Theorem, if you like []

Share

5 comments on “Conditional probability and the normal distribution

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter