S1: Chapter 9
The Normal Distribution
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
www.drfrostmaths.com
Last modified: 7th February 2016
Teacher Guidance
Possible lesson structure:
Lesson 1: Introduction and using z-tables
Go >
Lesson 2: Standardising, calculating probability of ranges
Lesson 3: Reverse z-table, calculating values given probabilities
Lesson 4: More on reverse z-tables: Quartiles, dealing with
𝑃(𝜇 < 𝑋 < 𝑎) and 𝑃 𝜇 − 𝑎 < 𝑋 < 𝜇 + 𝑎
Lesson 5 : Unknown 𝜇 and/or 𝜎
Lesson 6: Harder questions (including conditional probabilities)
Go >
Go >
Go >
What does it look like?
The following shows what the probability distribution might look like for a
random variable X, if X is the height of a randomly chosen person.
We expect this ‘bell-curve’ shape, where
we’re most likely to pick someone with a
height around the mean of 180cm, with
the probability diminishing symmetrically
either side of the mean.
p(x)
180cm
A variable with this kind of distribution is
said to have a normal distribution.
Height in cm (x)
For normal distributions we tend to draw
the 𝑦 axis at the mean for symmetry.
𝑝(𝑥)
What does it look like?
We can set the mean 𝜇 and the standard deviation 𝜎 of the Normal
Distribution.
Normal Distribution Q & A
Q1
For a Normal Distribution to be used, the
variable has to be:
continuous
?
Q2
With a discrete variable, all the probabilities
had to add up to 1.
For a continuous variable, similarly:
the area under the probability graph has
?
to be 1.
Q3
To find 𝑃 170 < 𝑋 < 190 , we could:
find the area between these values.
𝑝 𝑥
Q4
?
Would we ever want to find 𝑃 𝑋 = 200 say?
Since height is continuous, the probability someone
is ‘exactly’ 200cm is infinitesimally small. So not a
‘probability’ in the normal sense.
?
170cm
180cm 190cm
Height in cm (𝑥)
Notation
If a variable 𝑋 is ‘normally distributed’ (i.e. its probability function uses a
normal distribution), then we write:
...is distributed...
...using a Normal distribution with
mean 𝜇 and variance 𝜎 2
The random variable X...
𝑋~𝑁 𝜇, 𝜎
2
Example:
“𝑋 represents the height of a randomly chosen person,
with mean 160cm and standard deviation 10cm.”
2
𝑋~𝑁 160,10
?
Z value
! The Z value is the number of standard deviations a value is above the mean.
Example
p(x)
IQ, by definition, is normally
distributed for a given population. By
definition, 𝜇 = 100 and 𝜎 = 15
i.e. 𝑋~𝑁 100,152
IQ
Z
100
0
130
85
165
100
IQ (𝑥)
62.5
?
2 ?
-1 ?
4.333
?
-2.5 ?
Z table
Minimise
A z-table allows us to find the probability that the outcome will be less than a particular z value.
For IQ, 𝑷(𝒁 < 𝟐) would mean “the probability your IQ is less than 130”.
(You can find these values at the back of your textbook, and in a formula booklet in the exam.)
Expand
𝑃 𝑍 < 2 = 0.9772
?
𝑧=0 𝑧=1 𝑧=2
100 115 130
IQ (𝑥)
You may be wondering why we have to look up the values in a table, rather than
being able to calculate it directly. The reason is that calculating the area under
the graph involves integrating (see C2), but the probability function for the
normal distribution (which you won’t see here) cannot be integrated!
Use of the z-table
Suppose we’ve already worked out the z value, i.e. the number of standard deviations
above or below the mean.
1
2
𝑧
𝑧 = −0.3
𝑧 = −0.3
𝑧
𝑃 𝑍 > −0.3 = 𝑃 𝑍 < 0.3
= 0.6179?
𝑃 𝑍 < −0.3 = 1 − 𝑃 𝑍 < 0.3
= 0.3821?
3
4
This is clear by symmetry.
𝑧=1
𝑃 𝑍 >1 =1−𝑃 𝑍 <1
?
= 0.1587
𝑧
𝑧 = −2
𝑃 𝑍 < −2 = 1 − 𝑃 𝑍 < 2
= 0.0228?
𝑧
Bro Tip: We can only
use the z-table
when:
a) The z value is
positive (i.e.
we’re on the
right half of the
graph)
b) We’re finding
the probability
to the left of this
z value.
Bro Tip: Either
changing the sign of
changing the
direction of the
inequality does “1 –”.
If we do both, they
cancel out.
Test Your Understanding
? = 𝟎. 𝟐𝟒𝟐𝟎
𝑃 𝑍 > 0.70 = 𝟏 − 𝑷 𝒁 < 𝟎. 𝟕𝟎
? = 𝟎. 𝟏𝟎𝟓𝟔
𝑃 𝑍 < −1.25 = 𝟏 − 𝑷 𝒁 < 𝟏. 𝟐𝟓
𝑃 𝑍 > −2.42 = 𝑷 𝒁 < 𝟐. 𝟒𝟐 =? 𝟎. 𝟗𝟗𝟐𝟐
Exercise 1
Determine the following probabilities, ensuring you show how you have manipulated your
probabilities (as per the previous examples).
1
2
3
4
5
6
7
8
9
10
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
< 1 = 𝟎. 𝟖𝟒𝟏𝟑 ?
> 2.1 = 1 − 𝑃 𝑍 < 2.1
? = 𝟎. 𝟎𝟏𝟕𝟗
> −0.3 = 𝑃 𝑍 < 0.3 =?𝟎. 𝟔𝟏𝟕𝟗
< 1.5 = 𝟎. 𝟗𝟑𝟑𝟐
?
< −0.7 = 1 − 𝑃 𝑍 < 0.7 = 𝟎.?𝟐𝟒𝟐𝟎
< −1.7 = 𝑃 𝑍 > 1.7 = 1 − 𝑃?𝑍 < 1.7 = 𝟎. 𝟎𝟒𝟒𝟔
> 0 = 𝟎. 𝟓
?
< −0.4 = 𝑃 𝑍 > 0.4 = 1 − 𝑃?𝑍 < 0.4 = 𝟎. 𝟑𝟒𝟒𝟔
> −2.4 = 𝑃 𝑍 < 2.4 = 𝟎. 𝟗𝟗𝟏𝟖
?
> 3 = 1 − 𝑃 𝑍 < 3 = 𝟎. 𝟎𝟎𝟏𝟑
?
‘Standardising’
We have seen that in order to look up a value in the 𝑧 table, we needed to first convert
our IQ into a 𝑧 value. We call this ‘standardising’ our variable, because we’re turning
our normally distributed variable 𝑋 into another one 𝑍 where the mean is 0 and the
standard deviation is 1.
IQ (X) world
Z world
(recall that 𝜇 = 100 and 𝜎 = 15)
𝑍=3
𝑋 = 145
?
By thinking of what calculation you did, can you therefore come up with a
formula for 𝑍 in terms of 𝑋, 𝜇 and 𝜎?
!𝑍 =
𝑋−𝜇
?
𝜎
Bro Side Note: 𝑍~𝑁 0,12 , but why? Well consider a z value of 3 for example. We understand that to
mean 3 standard deviations above the mean. But if 𝜇 = 0 and 𝜎 = 1, the 3 is 3 lots of 1 above 0!
Example
The heights in a population are normally distributed with mean 160cm and standard
deviation 10cm. Find the probability of a randomly chosen person having a height
less than 180cm.
Here’s how they’d expect you to lay out your working in an exam:
𝑋~𝑁 160, 102
?
𝑃 𝑋 < 180
180 − 160
=𝑃 𝑍< ?
10
= 𝑃 𝑍 <? 2
= 0.9772
?
No marks attached with
this, but good practice!
A statement of the
problem.
M1 for “attempt to
standardise”
Look up in z-table
Test Your Understanding
Edexcel S1 May 2012
𝑋~𝑁 162,7.52
𝑃 𝑋 > 150
150 − 162
=𝑃 𝑍>
= 𝑃 𝑍 > −1.6
7.5
= 𝑃 𝑍 < 1.6 = 0.9452
?
Edexcel S1 May 2013 (R)
𝑋~𝑁 150,102
𝑃 𝑋 < 145
145 − 150
=𝑃 𝑍<
= 𝑃 𝑍 < −0.5
10
= 1 − 𝑃 𝑍 < 0.5 = 0.3085
?
Probabilities for Ranges
Again, let 𝑋 represent the IQ of a randomly chosen person, where 𝑋~𝑁 100,152
Thinking about the graph of the normal distribution, find:
𝑷 𝟗𝟔 < 𝑿 < 𝟏𝟏𝟐
This easiest way is to find 𝑃 𝑋 < 112
and ‘cut out’ the area corresponding to
𝑃 𝑋 < 96 :
𝑃 96 < 𝑋 < 112
= 𝑃 𝑋 < 112 − 𝑃 𝑋 < 96
112 − 100
96 − 100
=𝑃 𝑍<
−𝑃 𝑍 <
15
15
= 𝑃 𝑍 < 0.8 − 𝑃 𝑍 < −0.266
= 𝑃 𝑍 < 0.8 − 1 − 𝑃 𝑍 < 0.266
= 0.7881 − 1 − 0.6064 = 0.3945
?
z=0
96 100
IQ (𝑥)
112
! Probabilities of ranges:
𝑷 𝒂<𝑿<𝒃 =𝑷 𝑿<𝒃 −𝑷 𝑿<𝒂
Test Your Understanding
Let X represent the IQ of a randomly chosen person, where 𝑋~𝑁 100,152
Use your Z-table to find:
1 P(100 < X < 107.5) = P(Z < 0.5) – 0.5 = 0.1915
?
2 P(123 < X < 151) = P(1.53 < Z < 3.4) = P(Z < 3.4)
? – P(Z < 1.53) = 0.0627
3 P(70 < X < 90) = P(-2 < Z < -0.67) = (1 – 0.7486)
? – (1 – 0.9772) = 0.2286
Exercise 2
1
(On provided sheet)
[Jan 2013 Q4a] The length of time, L hours, that a
phone will work before it needs charging is normally
distributed with a mean of 100 hours and a standard
deviation of 15 hours. Find P(L > 127). = 0.0359
5
?
2
[Jan 2012 Q7a] A manufacturer fills jars with coffee. The
weight of coffee, W grams, in a jar can be modelled by a
normal distribution with mean 232 grams and standard
deviation 5 grams. Find P(W < 224). = 0.0548
?
6
?
3
[May 2011 Q4a] Past records show that the times, in
seconds, taken to run 100 m by children at a school can
be modelled by a normal distribution with a mean of
16.12 and a standard deviation of 1.60.
A child from the school is selected at random. Find the
probability that this child runs 100 m in less than 15 s.
= 0.2420
[Jan 2011 Q8a] The weight, X grams, of soup put in a tin
by machine A is normally distributed with a mean of
160 g and a standard deviation of 5 g. A tin is selected
at random. Find the probability that this tin contains
more than 168 g.
= 0.0548
?
[May 2009 Q8a,b] The lifetimes of bulbs used in a
lamp are normally distributed. A company X sells
bulbs with a mean lifetime of 850 hours and a
standard deviation of 50 hours.
(a) Find the probability of a bulb, from company
X, having a lifetime of less than 830 hours.
= 0.3446
(b) In a box of 500 bulbs, from company X, find
the expected number having a lifetime of less
than 830 hours.
= 172.3
?
?
?
4
[May 2010 Q7a] The distances travelled to work,
D km, by the employees at a large company are
normally distributed with D  N( 30, 82 ). Find the
probability that a randomly selected employee
has a journey to work of more than 20 km.
= 0.8944
7
[Jan 2009 Q6a] The random variable X has a
normal distribution with mean 30 and standard
deviation 5.Find P(X < 39).
= 0.9641
?
Exercise 2
(On provided sheet)
8 [May 2008 Q7a] A packing plant fills bags with cement.
The weight X kg of a bag of cement can be modelled by
a normal distribution with mean 50 kg and standard
deviation 2 kg. Find P(X > 53).
= 0.0668
?
9
The heights of raccoons in a Canadian town are
normally distributed with mean 30cm and standard
deviation 5cm. Determine the probability of a
randomly chose raccoon having a height between:
a) 28cm and 31cm
𝑷 𝟐𝟖 < 𝑿 < 𝟑𝟏 = 𝑷 −𝟎. 𝟒 < 𝒁 < 𝟎. 𝟐
= 𝟎. 𝟓𝟕𝟗𝟑 − 𝟏 − 𝟎. 𝟔𝟓𝟓𝟒
= 𝟎. 𝟐𝟑𝟒𝟕
b) 33cm and 37cm
𝑷 𝟑𝟑 < 𝑿 < 𝟑𝟕 = 𝑷 𝟎. 𝟔 < 𝒁 < 𝟏. 𝟐
= 𝟎. 𝟖𝟖𝟒𝟗 − 𝟎. 𝟕𝟐𝟓𝟕
= 𝟎. 𝟏𝟓𝟗𝟐
c) 25cm and 28cm
𝑷 𝟐𝟓 < 𝑿 < 𝟐𝟖 = 𝑷 −𝟏 < 𝒁 < −𝟎. 𝟒
= (𝟏 − 𝟎. 𝟔𝟓𝟓𝟒) − 𝟏 − 𝟎. 𝟖𝟒𝟏𝟑
= 𝟎. 𝟏𝟖𝟓𝟗
?
?
?
The reverse: Finding the z-value for a probability
Sometimes we’re given the probability, and need to find the 𝑧 value, so that we can
determine a missing value or the standard deviation/mean.
Just use the z-table backwards!
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑃
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
𝑍
<𝑧
<𝑧
<𝑧
<𝑧
>𝑧
<𝑧
<𝑧
<𝑧
<𝑧
<𝑧
>𝑧
>𝑧
>𝑧
= 0.87
= 0.92
= 0.7
= 0.85
= 0.23
= 0.1
= 0.95
= 0.7
= 0.46
= 0.01
= 0.86
= 0.975
= 0.43
→ 𝑧 = 1.13 ?
→ 𝑧 = 1.41 ?
→ 𝑧 = 0.5244
?
→ 𝑧 = 1.0364
?
→ 𝑃 𝑍 < 𝑧 = 0.77 ?
→ 𝑃 𝑍 < −𝑧 = 0.9 ?
→
→
→
→
→
→
→
𝑧 = 1.6449
?
𝑧 = 0.5244
?
𝑃 𝑍 < −𝑧 = 0.54?
𝑃 𝑍 < −𝑧 = 0.99 ?
𝑃 𝑍 < −𝑧 = 0.86 ?
𝑃 𝑍 < −𝑧 = 0.975?
𝑃 𝑍 < 𝑧 = 0.57 ?
For nice ‘round’ probabilities, we
have to look in the second z-table.
You’ll lose a mark otherwise.
𝑧 = 0.74
𝑧 = −1.2816
Bro Tip: Remember that either flipping the
inequality, or changing the sign of 𝑧 will cause
your probability to become 1 minus it.
→
→
→
→
→
𝑧 = −0.10
𝑧 = −2.3263
𝑧 = −1.08
𝑧 = −1.9600
𝑧 = 0.18
Dealing with two-ended inequalities
Bro Tip: The key is to use a
diagram (or otherwise) to convert
to a single-ended inequality.
Sometimes the range is two ended.
These can be one of two possible forms in an exam:
Find the value of 𝑎 such that
𝑃 0 < 𝑍 < 𝑎 = 0.3
Find the value of 𝑏 such that
𝑃 −𝑏 < 𝑍 < 𝑏 = 0.7
?
?
Suitable Diagram
0.3
0.7 Diagram
Suitable
0.15
𝑎
−𝑎
0.15
𝑎
This can be simplified to:
𝑷 𝒁< ?
𝒂 = 𝟎.?𝟖
This can be simplified to:
𝑷 𝒁< ?
𝒃 = 𝟎.?𝟖𝟓
Thus:
Thus:
𝒂 = 𝟎. 𝟖𝟒𝟏𝟔
?
𝒃 = 𝟏. 𝟎𝟑𝟔𝟒
?
Test Your Understanding
1
Find the value of 𝑒 such that 𝑃 −𝑒 < 𝑍 < 𝑒 = 0.9
𝑷 𝒁 < 𝒆 = 𝟎. 𝟗𝟓
?
𝒆 = 𝟏. 𝟔𝟒𝟒𝟗
2
Find the value of 𝑓 such that 𝑃 𝑓 < 𝑍 < 0 = 0.4
𝑷 𝒁 < 𝒇 = 𝟎. 𝟏
𝑷 𝒁 < −𝒇 = 𝟎. 𝟗
−𝒇 = 𝟏. 𝟐𝟖𝟏𝟔
?
𝒇 = −𝟏. 𝟐𝟖𝟏𝟔
We’ll come back to this later…
Where we are so far
“IQ has the distribution 𝑋~𝑁 100,152
An IQ of 130 corresponds with what
z-value?”
𝑿−𝝁
𝟏𝟑𝟎−𝟏𝟎𝟎
𝒁 = 𝝈 = ?𝟏𝟓 = 𝟐
 Understand what a Z-value means and
the formula to calculate it.
 Calculate a probability of being above or
below a particular value.

“Find the probability that I have an IQ
above 115.”
𝑷 𝑿 > 𝟏𝟏𝟓
= 𝑷 𝒁 > 𝟏 =?
𝟏−𝑷 𝒁<𝟏
= 𝟎. 𝟏𝟓𝟖𝟕
Calculate a z-value corresponding to a
probability.
𝑃
𝑃
𝑃
𝑃
𝑍
𝑍
𝑍
𝑍
<𝑧
>𝑧
<𝑧
>𝑧
= 0.58
= 0.22
= 0.15
= 0.95
𝒛 = 𝟎. 𝟐𝟎 ?
𝒛 = 𝟎. 𝟕𝟕 ?
𝒛 = −𝟏. 𝟎𝟑𝟔𝟒
?
𝒛 = −𝟏. 𝟔𝟒𝟒𝟗
?
COMING
NOW!
Calculate a value corresponding with a
probability.
“Find the IQ corresponding with the
bottom 30% of the population.”
COMING
SOON!
Calculate a missing value of 𝜇 and/or 𝜎.
Solve more complex problems (e.g. involving conditional probabilities)
Retrieving the original value
Again, let X represent the IQ of a randomly chosen person, where 𝑋~𝑁 100,152
What IQ corresponds to the bottom 78% of the population?
State the problem in probabilistic terms.
𝑃 𝑋 < 𝑥? = 0.78
𝑃 𝑍 < 𝑧? = 0.78
‘Standardise’.
0.78
z = 0.77
Bro Tip: Draw a diagram
for these types of
questions if it helps.
𝑧 =?
0.77
Identify z value
(as we did in the previous lesson).
Use z formula to find 𝑥
𝑥 − 100
= 0.77
15 ?
𝑥 = 100 + 0.77 × 15 = 111.55
Retrieving the original value
Again, let X represent the IQ of a randomly chosen person, where 𝑋~𝑁 100,152
What IQ corresponds to the bottom 90% of the population?
𝑃 𝑋<?
𝑥 = 0.9
State the problem in probabilistic terms.
‘Standardise’
𝑃 𝑍 < 𝑧? = 0.9
0.9
Identify z value
z = 1.2816
z = 1.2816
?
Use z formula to find 𝑥
𝑥 − 100
= 1.2816
15 ?
𝑥 = 119.224
Retrieving the original value
Again, let X represent the IQ of a randomly chosen person, where 𝑋~𝑁 100,152
What IQ corresponds to the bottom 30% of the population?
State the problem in probabilistic terms.
𝑃 𝑋 < 𝑥? = 0.3
𝑃 𝑍 < 𝑧? = 0.3
𝑃 𝑍 < −𝑧 = 0.7
−𝑧 = 0.5244
?
𝑧 = −0.5244
‘Standardise’
Identify z value
Use z formula to find 𝑥
𝑥 − 100
? 15
𝑥 = 92.143
−0.5244 =
Retrieving the original value
Again, let X represent the IQ of a randomly chosen person, where 𝑋~𝑁 100,152
What IQ does 80% of the population have a value more than?
State the problem in probabilistic terms.
𝑃 𝑋 > 𝑥? = 0.8
‘Standardise’
𝑃 𝑍 > 𝑧? = 0.8
𝑃 𝑍 < −𝑧 = 0.8
−𝑧 = 0.8416
?
𝑧 = −0.8416
Identify z value
Convert back into an IQ.
𝑥 − 100
? 15
𝑥 = 87.376
−0.8416 =
Test Your Understanding
[May 2008 Q7b] A packing plant fills bags with cement. The weight X kg of a bag
of cement can be modelled by a normal distribution with mean 50 kg and standard
deviation 2 kg.
Find the weight that is exceeded by 99% of the bags. Remember:
(5)
1.
2.
?
3.
4.
State your problem in probabilistic terms.
Standardise. Manipulate if necessary so
that your probability is above 0.5 and
you’re finding 𝑃 𝑍 < 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔 .
Use your z-table backwards to find the zvalue.
𝑥−𝜇
Use 𝑧 =
to find your value of 𝑥.
𝜎
[May 2011 Q4b] Past records show that the times, in seconds, taken to run 100 m by
children at a school can be modelled by a normal distribution with a mean of 16.12 and
a standard deviation of 1.60.
On sports day the school awards certificates to the fastest 30% of the children in the
100 m race. Estimate, to 2 decimal places, the slowest time taken to run 100 m for
which a child will be awarded a certificate.
(4)
?
Additional Practice (Outside of Class)
On the planet Frostopolis, the mean height of a
Frongal is 1.57m and the standard deviation 0.2m.
Determine:
a) The height for which 65% of Frongals have a
height less than.
𝑃 𝑋 < 𝑥 = 0.65
𝑥 − 1.57
𝑃 𝑍<
= 0.65
0.2
𝑥 − 1.57
= 0.39
0.2
𝑥 = 1.648 𝑚
b) The height for which 40% of Frongals have a
height more than.
?
c) The height for which 23% of Frongals have a
height less than.
𝑃 𝑋 < 𝑥 = 0.23
𝑃 𝑍 < 𝑧 = 0.23
𝑃 𝑍 < −𝑧 = 0.77
𝑧 = −0.74
𝑥 − 1.57
= −0.74
0.2
𝑥 = 1.422 𝑚
?
𝑃 𝑋 > 𝑥 = 0.4
𝑃 𝑍 > 𝑧 = 0.4
𝑃 𝑍 < 𝑧 = 0.6
𝑥 − 1.57
= 0.2533
0.2
𝑥 = 1.62 𝑚
?
Remember:
1. State your problem in probabilistic terms.
2. Standardise. Manipulate if necessary so that your probability is above 0.5 and you’re finding 𝑃 𝑍 < 𝑧 .
3. Use your z-table backwards to find the z value.
𝑥−𝜇
4. Use 𝑧 =
to find your value of 𝑥.
𝜎
Exercise 3a
(On provided sheet)
1 [Jan 09 Q6b-d] The random variable X has a
normal distribution with mean 30 and
standard deviation 5.
(b) Find the value of d such that
P(X < d) = 0.1151.
(4)
𝒅 = 𝟐𝟒
?
(c) Find the value of e such that
P(X > e) = 0.1151.
(2)
𝒆 = 𝟑𝟔
?
(d) Find P(d < X < e).
(2)
𝟎. 𝟕𝟔𝟗𝟖
?
2
[Jan 2013 Q4b] The length of time, L hours,
that a phone will work before it needs
charging is normally distributed with a mean
of 100 hours and a standard deviation of 15
hours. Find the value of d such that
P(L < d) = 0.10. (3)
𝒅 = 𝟖𝟎. 𝟕𝟕𝟔
?
3 [Jan 2011 Q8b] The weight, X
grams, of soup put in a tin by
machine A is normally distributed
with a mean of 160 g and a
standard deviation of 5 g. The
weight stated on the tin is w
grams. Find w such that P(X < w) =
0.01. (3)
𝒘 = 𝟏𝟒𝟖.
? 𝟑𝟕
2
4 [Ex9C Q6] Given 𝑌~𝑁 30, 5 ,
find 𝑎 such that 𝑃 𝑌 > 𝑎 = 0.30
𝒂 = 𝟑𝟐. 𝟔
?
[Ex9C Q7] Given 𝑋~𝑁 15, 32 ,
5 find 𝑎 such that 𝑃 𝑋 > 𝑎 = 0.15
𝒂 = 𝟏𝟖. 𝟏
?
Exercise 3a
(On provided sheet)
6 [Jan 2007 Q7b] The measure of intelligence,
8 [Solomon Paper E Q3c] The random
variable 𝑋 is normally distributed with a
IQ, of a group of students is assumed to be
mean of 42 and a variance of 18. Find
Normally distributed with mean 100 and
the value of 𝑥 such that
standard deviation 15.
𝑃 𝑋 ≤ 𝑥 = 0.95
The probability that a randomly selected
𝒙 = 𝟒𝟗. 𝟎
student as an IQ of at least 100 + k is 0.2090.
?
Find, to the nearest integer, the value of k. (6)
2
𝒌 =?𝟏𝟐
9 [Ex9C Q9] Given 𝑌~𝑁 100,15 . Find
the value of 𝑎 and 𝑏 such that:
[June
2005
Q6b,c]
A
scientist
found
that
the
7
a) 𝑃 𝑌 < 𝑎 = 0.40
time taken, M minutes, to carry out an
experiment can be modelled by a normal
random variable with mean 155 minutes and
standard deviation 3.5 minutes.
(b)P(150 ≤ M ≤ 157),
(4)
𝟎. 𝟔𝟑𝟗𝟑
?
(c) the value of m, to 1 decimal place, such
that P(M ≤ m) = 0.30.
(4)
𝒎 = 𝟏𝟓𝟑.
? 𝟐
? 𝟔
𝒂 = 𝟕𝟎.
b) 𝑃 𝑋 > 𝑏 = 0.6915
?𝟖
𝒃 = 𝟖𝟎.
c) 𝑃 𝑏 < 𝑋 < 𝑎
= 𝟎. 𝟎𝟑𝟔𝟒
?
Harder reverse probability questions
Recap: Find the value 𝑎 such that 𝑃 −𝑎 < 𝑍 < 𝑎 = 0.7
𝑷 𝒁 < 𝒂 = 𝟎. 𝟖𝟓
?
𝒂 = 𝟏. 𝟎𝟑𝟔𝟒
Q: If 𝑋~𝑁 30,52 , find the value of 𝑘 such that 𝑃 30 < 𝑋 < 𝑘 = 0.2
Observe that the 30 at the bottom end of the inequality
is the mean. Thus:
𝑷 𝑿 < 𝒌 = 𝟎. 𝟕
𝒌 − 𝟑𝟎 ?
= 𝟎. 𝟓𝟐𝟒𝟒
𝟓
𝒌 = 𝟑𝟐. 𝟔𝟐𝟐
Quartiles and Percentiles
Using IQ: 𝑋~𝑁 100, 152
Find:
The Lower Quartile 𝑄1
?𝟏 = 𝟎.?𝟐𝟓
𝑷 𝑿<𝑸
𝑷 𝒁 < 𝒛 = 𝟎. 𝟐𝟓
𝑷 𝒁 < −𝒛 = 𝟎. 𝟕𝟓
𝑸𝟏 − 𝟏𝟎𝟎
−𝟎. 𝟔𝟕 = ?
𝟏𝟓
𝑸𝟏 = 𝟖𝟗. 𝟗𝟓
The 70th percentile 𝑃70
i.e. 25% of people have an IQ less
than the lower quartile.
The Upper Quartile 𝑄3
𝑷 𝑿 < 𝑸𝟑 = 𝟎. 𝟕𝟓
𝑷 𝒁 < 𝒛 = 𝟎. 𝟕𝟓
𝑸𝟑 − 𝟏𝟎𝟎
𝟎. 𝟔𝟕 = ?
𝟏𝟓
𝑸𝟑 = 𝟏𝟏𝟎. 𝟎𝟓
𝑷 𝑿 < 𝑷𝟕𝟎 = 𝟎. 𝟕
𝑷𝟕𝟎 − 𝟏𝟎𝟎
𝟎. 𝟓𝟐𝟒𝟒 = ?
𝟏𝟓
𝑸𝟏 = 𝟏𝟎𝟕. 𝟖𝟔𝟔
Test Your Understanding
Edexcel S1 May 2012
?
Edexcel S1 May 2010
a)
b)
c)
P(D > 20) = P(Z > -1.25)
= P(Z < 1.25) = 0.8944
P(Z < z) = 0.75
-> z = 0.67
Q3 = 30 + (0.67 x 8) = 35.36
Q1 = 30 – (0.67 x 8) = 24.64
?
Exercise 3b
(On provided sheet)
Finding missing 𝜇 and 𝜎
The random variable 𝑋~𝑁 𝜇, 32
Given that P(X > 20) = 0.20, find the
value of 𝜇.
𝑃(𝑋 < 20) = 0.8
𝑃 𝑍 < 𝑧 = 0.8 (Standardising)
𝑧 = 0.8416
?
20 − 𝜇
= 0.8416
3
𝜇 = 17.5
The random variable 𝑋~𝑁 50, 𝜎 2
Given that P(X < 46) = 0.2119, find the
value of 𝜎.
𝑃 𝑋 < 46 = 0.2119
𝑃 𝑍 < 𝑧 = 0.2119
𝑃 𝑍 < −𝑧 = 0.7881
𝑧 = −0.8
46 − 50
?= −0.8
𝜎
𝜎=5
If your standard deviation is negative,
you know you’ve done something
wrong!
Finding missing 𝜇 and 𝜎
The random variable 𝑋~𝑁 𝜇, 𝜎 2
Given that P(X > 35) = 0.025 and P(X < 15) = 0.1469, find the value of 𝜇 and the
value of 𝜎
First deal with P(X > 35) = 0.025
Next deal with P(X < 15) = 0.1469
P(X < 35) = 1 – 0.025 = 0.975
If P(Z < z) = 0.975, then z = 1.96.
𝑃 𝑋 < 15 = 0.1469
𝑃 𝑍 < 𝑧 = 0.1469
𝑃 𝑍 < −𝑧 = 0.8531
𝑧 = −1.05
35 − 𝜇?
= 1.96
𝜎
𝜇 + 1.96𝜎 = 35
?
15 − 𝜇
= −1.05
𝜎
𝜇 − 1.05𝜎 = 15
We now have two simultaneous equations. Solving gives:
𝜇 = 22.0
𝜎 =?6.64
Test your understanding
For the weights of a population of squirrels, which are
normally distributed, Q1 = 0.55kg and Q3 = 0.73kg.
Find the standard deviation of their weights.
 = 0.114

 = 0.124

 =
0.134
 = 0.144

Due to symmetry,  = (0.55 + 0.73)/2 = 0.64kg
If P(Z < z) = 0.75, then z = 0.67.
0.64 + 0.67 = 0.73
 = 0.134
Only 10% of maths teachers live more than 80 years. Triple that
number live less than 75 years. Given that life expectancy of maths
teachers is normally distributed, calculate the standard deviation and
mean life expectancy.
𝑷 𝑿 < 𝟖𝟎 = 𝟎. 𝟗
 = 76.15

 = 76.25

 = 76.35

 =
76.45
 =
2.77
 =
2.78
 =79
 =
2.80
𝑷 𝒁 < 𝒛 = 𝟎. 𝟗
𝒛 = 𝟏. 𝟐𝟖𝟏𝟔
𝟖𝟎 − 𝝁
= 𝟏. 𝟐𝟖𝟏𝟔
𝝈
𝝁 + 𝟏. 𝟐𝟖𝟏𝟔𝝈 = 𝟖𝟎
Similarly
 – 0.5244 = 75
Exam Questions
Edexcel S1 May 2013 (R)
?
Edexcel S1 Jan 2011
?
Edexcel S1 Jan 2002
a) z-value for 0.975 is 1.96. By symmetry, 235 is
1.96 standard deviations below mean.
So 𝜇 − 1.96𝜎 = 235. The result follows.
b) P(Z < z) = 0.85. So z = 1.04
𝜇 + 1.04𝜎 = 286
c) Solving, 𝜇 = 268.32, 𝜎 = 17
d) If 0.683 in the middle, (0.683/2)+0.5=0.8415
prob below value above mean. Thus z = 1.
So values are 154.8 – 2.22 and 154.8 + 2.22
?
Summary
A
B
A normal distribution is good for modelling
data which:
tails off symmetrically about some mean/
has a bell-curve like ?
distribution.
A z-value is:
The number of standard deviations above
?
the mean.
C
If a random variable X is normally
distributed with mean 50 and standard
deviation 2, we would write:
𝑿~𝑵 ?𝟓𝟎, 𝟐𝟐
D
A z-table is:
A cumulative distribution function for a
normal distribution with
? mean 0 and
standard deviation 1.
P(IQ < 115) = 0.8413
?
E
We need to use the second z-table
whenever:
we’re looking up the z value for
certain ‘round’ probabilities.
?
F
P(a < X < b) = P(X < b) –?P(X < a)
G
We can treat quartiles and
percentiles as probabilities.
For IQ, what is the 85th
percentile?
100 + (1.04 x?15) = 115.6
H
We can form simultaneous
equations to find the mean and
standard deviation, given known
values with their probabilities.
Mixed Questions
Edexcel S1 June 2001