Normal distribution - that we love to hate! NZ curriculum, 2.12, Miss Fuller, Birkenhead College, BC, normally distributed....

1. Normal Distribution
Mock test
Friday Week 2!
/
Sample Assessments

2. God’s Mean
 This used to be called the ‘error curve’ – how far (say) people’s
heights deviated from ‘God’s mean’. We know now that data
occurring ‘naturally’ clusters about the mean, but spreads out
more and more thinly towards the extreme values.
 Pear orchard example
Miss Fuller made in “ Basketball player?
God’s image”

3. Standard Deviation
 If you don’t have a clue what SD really is, try this:
 Measure your thumbs 
 The SD is a measure of how far away from the mean your scores are.
 Our thumb ‘mean’ is 61.79 mm.
 On average, our 18 scores were 4.86 mm away from the mean, so
our SD is 4.86 mm.
 In the diagram below, 68% of the thumbs will be 1 SD away from the
mean, 95% of our thumbs will be 2 SDs away from the mean, and
99.7% will be 3 SDs away from the mean. Let’s see how close we get:

4. Standard Deviation
 How many ‘thumbs’ will we expect in each slice? Put yours in!
47.20 52.07 56.93 61.79 66.65 71.51 76.38
 68% of 19 is 12.969, 95.44% of 19 is 18.134 and 99.7% of 19 is 18.95
 Even with such a small sample, the standard deviation probs work!

5. Textbook references - Theta
Normal Distribution Curves Ex 24:01 1 – oral
Standard Deviation Ex 24:02 1 – read p 405
Q4 – scan gymnasts and disc
Normally Distributed data Ex 24:03 1 – DOIT. Need words.
Tables – p 418 dim. TC or GC 
The Standard Normal Distribution Ex 24:04 1 TC
Converting to the standard normal Ex 24:05
(and 06?)
2 use brains?
Contextual Standard Normal Distribution 24:07
24:08
2
Inverse Normal Distribution Ex 24:09 1
Applications using the ‘Inverse’ Normal z
Ex 24:11
value
2
Practice Assessment Work 8

6. The Normal Distribution- definitely!!
34%
0.5% 0.5%
2%
13.5% 13.5%
2%
34%
-3 -2 -1 mean +1 SD +2 +3

7. We now know that naturally occurring data
falls like this:
 68.26% of data
falls within 1 sd
either side of the
mean.
 It is _________ or
_________ that
data falls in this
region.
 95.44% of data
falls within 2 sd
either side of the
mean.
 It is ___________
 or ____________
 that data falls in
this region.
Green Thetad
 30.2 p.353
 99.74% of data
falls within 3 sd
either side of the
mean.
 It is _________
_____________
 that data falls in
this region.
If you become very familiar with these numbers you can estimate your answers – a great ‘mental check’!

8. We now know that naturally occurring data
falls like this:
 68.26% of data
falls within 1 sd
either side of the
mean.
 It is _________ or
_________ that
data falls in this
region.
 95.44% of data
falls within 2 sd
either side of the
mean.
 It is ___________
 or ____________
 that data falls in
this region.
Green Thetad
 30.2 p.353
 99.74% of data
falls within 3 sd
either side of the
mean.
 It is _________
_____________
 that data falls in
this region.
If you become very familiar with these numbers you can estimate your answers – a great ‘mental check’!

9. Just in case…

10. Just in case…

11. Standard Normal Distribution
 This is a perfect world called Zed
 The mean μ is always 0 and the
standard deviation σ is always 1
 We have tables to calculate the
probability that (say) a value is 1.5
or greater, or between 0 and 2.
 The curve is called ‘the normal
curve’ and the total area
underneath it is ONE. The area
under each half is 0.5.
 I will refer to this as ‘The Z World’
 Have a look at these examples:

12. The Normal Distribution- definitely!!
34%
0.5% 0.5%
2%
13.5% 13.5%
2%
34%
-3 -2 -1 mean +1 SD +2 +3

13. Practice
Find the probability that x is between…
1. 0 and 1.55
2. 0 and 0.54
3. 0 and 0.9
4. 0 and 0.04
5. 0 and 1.57
6. -1.4 and 0
7. -1.20 and 0
8. -1.3 and 1.3
9. 0.72 and 1.8
10. 1.8 and -0.05
Discuss differences column!!

14. Practice – Answers!
Find the probability that x is between…
1. 0 and 1.55
2. 0 and 0.54
3. 0 and 0.9
4. 0 and 0.04
5. 0 and 1.57
6. -1.4 and 0
7. -1.20 and 0
8. -1.3 and 1.3
9. 0.72 and 1.8
10. 1.8 and -0.05
1. 0.4394
2. 0.2054
3. 0.3159
4. 0.0160
5. 0.4418
6. 0.4192
7. 0.3849
8. 2x0.4032 = 0.8064
9. 0.4641 - 0. 2642
10.0.4641 + 0.0199

15. Harder Practice
Find the probability that x is
between…
1. 0 and 0.231
2. 0 and 1.03
3. Find P( Z>2.135)
4. Find P( Z>2.135)
5. Find P( Z>-0.596)
6. Find P( Z>2.135)
7. Find P( Z<0.582)
8. Find P( Z<-1.452)
Sorry - no answers!

16. Mapping onto the ‘REAL’ World
0 1.2
 We can see that if Z was 1.2,
the probability of a point lying
between 0 and 1.2 is…
 That is, 38.49% or 0.3849 of our
sample lies between 0 and 1.2
 Can you see our problem when it
comes to ‘Real Life’?!!
 This information in our tables is
USELESS for data which is doesn’t have
a mean of 0 and a sd of 1 UNLESS… we
can MAP our real life data onto the
standard normal distribution.
 I will call this ‘the Real Life
X world’

17. Mapping - step by step
 Imagine this graph shows the X world,
where the mean (μ) is 5
and the SD(σ) is 2.
Stripes show
1 standard deviation.
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
 BUT in our maths Z world the mean is 0 and the SD is always 1.
 What must happen to map the graph above onto the Z world?
 The graph moves DOWN 5 to make the mean 0
 and then compresses so the SD is now 1.
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

18. How do we map the X world onto the Z?
 To go from X to Z (so we can use our tables) we...
 take off (subtract) the mean and divide by the standard deviation.
 And that gives us our formula:
Z=
푋 − μ
σ
Z(scor푒) =
(푋푠푐표푟푒) −푚푒푎푛
푠푡푑 푑푒푣′푛

19. From start to finish…
 The heights of BC students are normally distributed, with a
mean height of 1.7m and a standard deviation of 0.1 m.
 What percentage of students will be taller than 1.87m?
 Of the 700 students, how many will be shorter than 1.87m?
1. ALWAYS draw a diagram. Put info in and shade what you
need.
2. Estimate the answer and write it in.
3. This data is in ‘the X world’. Use the formula (next slide)
to translate it into the Z world (mean 0, sd = 1).

20. The Maths
4. Translate your X world to
the Z world.
5. Now we can use the
tables and look up 1.7.
Z=
푋 − μ
σ
P(X>1.87) = P(Z>1.87-1.7)
0.1
= P(Z>1.7)

21. The Answer
6. The answer from the
tables is 0.4554.
7. Look at your diagram and
adjust for your answer.
0.5 – 0.4554 = 0.0446
8. Look back at the original
estimate – how close?
9. Read the question and
give your real life
answer.
P(X>1.87) = P(Z>1.87-1.7)
Z=
푋 − μ
σ
0.1
= P(Z>1.7)
Prob that z is between 0
and 1.7 is… 0.4554
10. The probability that a
student is taller than 1.87m
is 4.46%
11. Have we finished? – nooo!

22.  For how many questions,
we multiply the
probability x the number.
 In this case we want to
know how many are
SHORTER than 1.87m.
 Our tables gave us 0.4554,
what else must we do?
 Add 0.5
 So 0.9554 x 700 = 668.78
 Give the maths and then
answer the question with
a ‘real life’ answer.
 There are about 669
students that are shorter
than 1.87m .
 Try the classic Christmas
tree problem, next.
‘How many’ questions

23. The Christmas Tree Problem
 The lengths of
Christmas tree
branches from a pine-tree
plantation can
be assumed to be
normally distributed
with a mean length of
1.8m and a standard
deviation of 20 cm.
 What percentage of
branches would measure
less than 2.07 m?
 If there were 80
branches, how many
would we expect to be
less than 2.07 m in
length?

24. Basics:
 Step 1: Write down important information.
 μ= 1.8 m sd(σ) = 0.2 m P(X< 2.07) = ?
 Step 2: Convert to Z score (sub mean, ÷ by sd):
P( Z < 2.07 – 1.8 )
0.2
 Step 3: ALWAYS draw a diagram – Doit!

25. Step 4: Estimate an answer…
We already have an idea of a
‘good’ answer now:
more than 0.5 + 0.34 = 0.84
About 0.84 or more is good 

26.  Step 5: Look up Z in the tables: Z = 1.35 gives us
0.4115
 Step 6: Adapt as required and answer the question:
 P(Z<1.35) = 0.5 + 0.4115 = 0.9115
 Step 7: Real life Answer - The probability that
branches measure less than 2.07 m is 0.
 If there were 80 branches, how many would we
expect to be less than 2.07 m in length?
 How many problems: _______ x _______ = ________
 Answer…?? Round for real life!

27.  Workbook – READ pages 34
and 35, or better still,
cover the model answers
and do each step,
uncovering and checking
as you go 
 Do p 37 – 42 
 Graphics Calculator ppl –
read p 36.
 Sky tower theta
– READ p 341
 Exercise 24.02 p 342
(10 mins only) then…
 Ex 24.03 p 348
 Ex 24.04 p 349 and 24.06
(the best and hardest)
p 352
 Expected value – 24.07
p354
Problems to try…
 Green theta
– READ p 341
 Exercise 30.2 p 353
(10 mins only) then…
 Ex 30.3 p 358
 Ex 30.4 p 359 (the best
and hardest – this
combines expected value)

28. Inverse - essential steps
1. ALWAYS draw a diagram!
2. After you have done that,
draw one for the ‘Z’
world. The probabilities
are the same AND the
tables ‘work’.
3. To find your missing z
point, look up the prob in
the BODY of the tables. –
GO LOW.
3. Adjust your answer for
the diagram – negative?
4. Then use the force-errr
formula to get your x
point.
5. Write your ‘real life’
answer with appropriate
rounding.

29. Mazda Man’s lightbulbs
 Light bulbs last for 200 hours
on average, with a standard
deviation of 40 hours and
they are normally
distributed.
 The Mazda Man wants to
create an ad that promises
that 80% of bulbs last for at
least a certain number of
hours.
 He is asking you to work out
the hours.
 Real world diagram!
 Z world diagram.
 Look up 0.3 in the BODY of the
tables – how close can you get?
 0.2995 – we still need 5 more –
differences – how close can you
get?
 3 – look up and 1 is your last
digit.
 The z number is… 0.841
 Adjust for your diagram
– positive or negative?

30. Mazda Light bulbs ctd
 Now do the maths
 In the diagram above we
HAVE z = -0.841 and we
need x.
 -0.841 = x – 200
40
 Multiply by 40 and add
200.
 X = 166.36
 Do a ‘sense check’ with
your problem – does this
sound right?
 Now apply to real life..
 80% of bulbs will last
longer than 166.36 hours
(yuk!)
 80% of bulbs will last
longer than 166 hours and
21.36 minutes (yuk!)
 I would advise the Mazda
man…
Z=
푋 − μ
σ

31. Lambsie-pies
 Newborn lamb weights are
normally distributed
about 1.5 kg with a
standard deviation of125
grams.
 What birth weight is
exceeded by 30% of
newborns?
 Real world diagram!
 Z world diagram.
 Look up 0.2 in the BODY of the
tables – how close can you get?
 0.1985 – we still need 15 more –
differences – how close can you
get?
 14 – look up and 4 is your last
digit.
 The z number is… 0.524
 Adjust for your diagram
– positive or negative?

32. Mazda Light bulbs ctd
 Now do the maths
 In the diagram above we
HAVE z = 0.524 and we
need x.
 0.524 = x – 1.5
0.125
X = 1.57 kg
Z=
푋 − μ
σ
 Do a ‘sense check’ with
your problem – does this
sound right?
 Now apply to real life..
 I expect that 30% of the
lambs will exceed the
birth weight of 1.57 kg

33. Inverse problems
 Green Theta p 366
 An army recruiting officer measures the length of the
feet of all new recruits before outfitting them with
Army-issue boots. He knows that these foot lengths are
normally distributed with a mean of 260mm and a
SD of 15 mm. 12% of the recruits are like Jack Reacher
and have feet so large that they do not fit any of the
boots. Find the maximum foot length to the nearest
mm which the army issue boots fit.
Workbook p 46 and 47

34. 2 4 6 8… you’re in the army now, son.
 Real world diagram!
 Z world diagram. Percentage??
 Look up ____ in the BODY of the
tables – how close can you get?
 The z number is… _____
 Adjust for your diagram
– positive or negative?
 Now do the maths
 In the Z diagram we HAVE
z = _____ and we need x.
 0._____ = x –…..
X = ________