If you missed the simulating introduction to Continuous Variables and Normal Distributions it might be worth taking a few minutes to read.
Z Value
The functional form for a normal distribution is a bit complicated. It can also be difficult to compare two variables if their mean and or standard deviations are different, for example heights in centimeters and weights in kilograms, even if both variables can be described by a normal distribution. To get around both of these conflicts we can define a new variable:
(1)This variable gives a measure of how far the variable is from the mean $(x-\mu)$ then "normalizes" it by dividing by the standard deviation $(\sigma)$. This new variable gives us a way of comparing different variables. The z-value tells us how many standard deviations or "how many sigmas" the variable is from its respective mean.
When the distribution function is expressed in terms of the z-value it is sometimes called the "standard normal distribution." You've got to love the creativity!
(2)Areas Under the Curve
To calculate the probability that a variable is within a range we have to find the area under the curve… Hooray, calculus! It turns out that there is no indefinite integral of the function! However, smart folks have figured out how to do definite integrals, but they are a bit complex so the folks who have to work with normal distributions rely on tables, which you have in your formula booklet, or calculators.
An example of the table is shown in the collapsible box below. It's as close as I could find to the one the IB gives you…
Folks taking Exams in 2014 will not need use this table. They will only need to use their GDC.
These tables can be a bit scary, but you simply need to know how to read them.
- The left most column tells you how many sigmas above the the mean to 1 decimal place.
- The top row gives the second decimal place.
- The intersection of a row and column gives the probability.
For example, if we want to know the probability that a variable is no more than 0.51 sigmas above the mean we find select the 6th row down (corresponding to 0.5) and the 2nd column (corresponding to 0.01). The intersection of the 6th row and 2nd column is 0.6950. Which tells us that there is a 69.50% percent chance that a variable is less than 0.51 sigmas above the mean…
Notice that for 0.00 sigmas the probability is 0.5000. Thus showing that there is equal probability of being above or below the mean! So nice when stuff makes sense.
Using Ti-84 to Find Areas Under the Curve
A 6 minute video showing how to get area under the normal distribution given a range of z-values. It also covers the inverse, that is going from area to z-values. This is something you need to know how to do!
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"Simple" Examples
Example 1
Find $P(Z \leq 1.5)$
This problem essentially asks what is the probability that a variable is less than 1.5 sigmas above the mean. On the table of values find the row that corresponds to 1.5 and the row that column that corresponds to 0.00. Which gives a probability of 0.933.
Graphically this problem can be represented as such:
Example 2
Find $P(Z \geq 1.17)$
This problem essentially asks what is the probability that a variable is MORE than 1.17 sigmas above the mean. On the table of values find the row that corresponds to 1.1 and the row that column that corresponds to 0.07. Which gives a probability of 0.8790. However, this is the probability that the value is less than 1.17 sigmas above the mean. Since all the probabilities must sum to 1:
(3)Graphically this problem can be represented as such:
Example 3
Find $P( -1.16 \leq Z \leq 1.32)$
This example is a bit tougher… Graphically this problem can be represented as such:
This problem can be rewritten in the form below.
(4)The difficulty comes in our table of values does not allow us to directly calculate $P(Z\leq -1.16)$. However we can use the symmetry of the distribution.
(5)So we can say:
(6)IB Style Examples - In Progress
Example 4
a) First we have to find the z-value that corresponds to the 197 cm.
(7)Then using a GDC or table $P(Z>1) = 1-P(Z<1) = 0.159$
So 15.9% of the adults are taller than 197.
b) First we have to find how tall the 99th percentile is. Using a GDC (or table) we need an inverse normal distribution to get a z-value that corresponds to 99%.
(8)Then we can find the height:
(9)Then since the doorway needs to be 17 cm taller. So, the door needs to be 227 cm (3 S.F.).
Example 5
a) First we need to find the z-value that corresponds to $P(X>153)=0.705$. To do this we use the inverse normal (GDC or table).
This gives us $z=0.539$. Since the probability is greater than 0.5 this means that the z-value must be negative, i.e. 153 is less than the mean. Think about it. Draw a picture.
So we can say:
(11)From this $\mu = 155.7 cm$
b) Same old, same old… Find the z-value.
(12)Use a GDC or table… $P(Z>0.6)=1-P(Z<0.6)=0.274$
Example 6
a) We need to find $P(X<72)$. To do that we need the z-value.
(13)b) For this we need $P(90>X>72)$ which is the same as $P(1>Z>-1)$ because 90 is one standard deviation above the mean so its z-value is 1. You can use your GDC to quickly get an answer or…
(15)I'll let you guys do the shading. It looks very similar to the shading in example 3 above.
c) We need to find the z-value that corresponds to $P=0.04$. Use your GDC or table to find the inverse normal distribution. This is $z=-1.75$
Then using the definition of a z-value:
(16)From this $x=66$.
Want to add to or make a comment on these notes? Do it below.
Many errors in this article. Example 2 for the simple questions show P(Z>1.5) when the worked example is 1.17. Also, example 3 shows P(−1.16≥Z≤1.32) but then it changes correctly at the end to P(−1.16≤Z≤1.32).
Nonetheless, the article still helped me. Although, it'd be nice to have the worked examples for the harder questions since now I'd like help with those.
Thanks.
Thanks for the extra set of eyes. I fixed the two errors you mentioned.
I'll see if I can get an few more solutions up on the page.
-HS
I got solutions to the last 3 examples up. As always extra eyes are appreciated to find typos and errors.
-HS
For example 5
b) why did you use 153 for the mean.. I think it's supposed to be 155.7…
Thanks!
Question 6 - Part b. 72 is one standard deviation away, but 90 is not. 88 would be 1 standard deviation away.
Example 5:
Your answer for the mean is wrong in part a. I'm pretty sure you don't need to make it negative. I think the mean should be 150.3
Agree with Chelsea on Example 5:
The question states that the 0.705 is the probability for a boy being shorter than 153, or P(X<153)=0.705. You solved (and indeed wrote) the opposite: P(X>153)=0.705.
The price per kilogram of tomatoes, in euro, sold in various markets in a city is found to be normally distributed with a mean of 3.22 and a standard deviation of 0.84.
Find the price that is two standard deviations above the mean price.
Malthouse school opens at 08:00 every morning.
The daily arrival times of the 500 students at Malthouse school follow a normal distribution. The mean arrival time is 52 minutes after the school opens and the standard deviation is 5 minutes.
Find the probability that a student, chosen at random arrives at least 60 minutes after the school opens.