A general overview of Normal distribution, Standard Normal Distribution, and relevant Examples to calculate.

1. NORMAL DISTRIBUTION
MMED ORTHOPAEDIC
MMED HAEMATOLOGY
MSC CARDIOVASCULAR NURSING
MPHARM CLINICAL PHARMACY
MDENT RESTORATIVE DENTISTRY
MSC NURSING MENTAL HEALTH

2. OBJECTIVES
At the end of the session, we should be able to:
• Define the normal distribution
• Explain properties of a normal distribution
• Calculate probabilities for the normal distribution

4. Probability
• Probability of an event is defined to be the proportion of times the event occurs in a
long series of random trials.
• Probabilities are proportions and so they take values between 0 and 1.
• A probability of 0 means that the event never occurs whereas the probability of 1
means the event certainly occurs.
• The sum of probabilities of all possible outcomes is 1.
• Example: in tossing a coin P(H) + P(T) = 1 and
• in tossing a die, P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1.

5. Normal distribution
• Also referred to as “ a bell curve” or Gaussian Curve
• Normal distribution is a hypothetical symmetrical distribution used to
make comparisons among scores or to make other kinds of statistical
decisions.
• The two main parameters are the mean and standard deviation.
• In a normal distribution most results are located in the middle and
few are spread on both sides.

6. CHARACTERISTICS OF NORMAL DISTRIBUTION
• It is a distribution of a continuous random variable
• It is a bell shaped curve
• The curve is symmetrical
• The curve is asymptotic (never touches the x-Axis)
• All measures of central tendency are equal and stable on the highest
peak ( Mean = Median=Mode)

7. The Normal Distribution: Graphically
Normal Curve is Symmetrical
Two halves identical
Mean, Median
and Mode are
equal.
Theoretically, curve
extends to - ∞
Theoretically, curve
extends to + ∞

8. THE EMPIRICAL RULE
• For normal distribution, almost all observed data will fall within three standard
deviations (denoted by σ) of the mean or average (denoted by µ).
• Total area under the curve (AUC)= 1
On the basis of mean µ and standard deviation s the area of normal curve is
distributed as
µ±1* s = 68.27% (µ-1s , µ+1s )
µ±2* s = 95.45% (µ-2s , µ+2s )
µ±3* s =99.73% (µ-3s , µ+3s )

9. • Approximately 68.26% of the data
is within 1-SD of the mean.
• Approximately 95.44% of the data
is within 2-SD of the mean.
• More than 99.72% of the data is
within 3-SD of the mean.

10. • μ and s define the shape of the curve
• changing µ shifts the whole curve to the left or right
• increasing s makes the curve flatter and more spread out
Parameters affecting normal curve
10

11. changing μ
11

13. Changing σ
13

14. Reference or spread limits
• Limits covering the 95% of the area are known as the 95% reference
or spread limits, and values contained in the interval are commonly
referred to as the NORMAL values
• 95% of the area under the normal curve lies in mean  1.96 SD
(usually approximated as mean  2 SD)

15. Example 1
Given that in a large population of health individuals, the mean systolic
blood pressure is 120 mmHg and that the standard deviation is 10
mmHg, obtain the approximate 95% reference range/limits for this
variable

16. Soln: For this question remember this curve!

17. Solution for Ex. 1 Z =
𝑋 − η
σ
1.96 =
𝑋 − 120
10
Z= 1.96
X=??
µ= 120
σ= 10
−1.96 =
𝑋 − 120
10
1.96(10) = X-120 -1.96(10) = X-120
X= 139.6 X= 100.4
Our 95% reference range is going to be 100.4 to 139.6

18. • Can also use the following formula to obtain the 95% reference range
Normal range = Xbar ± 2SD

19. Standard Normal Distribution
• Similar to the Normal distribution
• They are both unimodal, symmetrical and bell shaped curves.
• However, a normal distribution takes on any value as its mean and SD
• For the standard normal distribution, the mean is always “0” and the
standard deviation is always “1”
• By a change of units any normally distributed variable can be related
to the standard normal distribution

20. Standard Score (z-score)
 The standard score is the distance of the score from the mean
in terms of the standard deviation.
 It tells how many standard deviations the observed value lies
above or below the mean of the distribution.
 The standard score is useful in comparing observed values with
different distributions.

21. To be able to find areas under the normal curve, observed
values must first be converted into standard scores

22. To change an observed value (x) into standard score use the equation:
Note:
A positive (+) z-score means that the score/observed value is above the mean.
A negative(-) z-score means that the score/observed value is below the mean.
Z =
𝑋 − µ
σ
Z =
𝑋 − xbar
𝑠
Where;
x= raw score/observed value
x̄ = mean
s= standard deviation
or

23. Converting to standard normal
distribution
µ = 0
σ = 1
µ = 158 cm
σ = 3 cm
x z
z = (x-µ)/ σ = (x-158) / 3

25. EXAMPLE 2
A study of blood pressure of Jangwani school girls gave a
distribution of systolic blood pressure (SBP) CLOSE TO NORMAL
with µ=105.8mmhg and s = 13.4mm Hg.
a) What percentage of girls would be expected to have SBP greater
than 120mm Hg

26. Solution
Step 1: Z =
𝑋−µ
σ
 Z =
120−150.8
13.4
 1.06
Step 2: Go to the table and get the corresponding value in the Z-score
Step 3: The value corresponds to 0.85543
Step 4: Subtract the 0.85543 from 1 (1-0.85543) = 0.14457
Step 5: Convert to percentage making it 14.5%
So about 14.5% of the girls would be expected to have systolic blood
pressure > 120mm Hg.

27. 0 1.06
The area under the
curve = 85.5%
Those above the
systolic of
120mmHg= 14.5%

28. b) What percentage of girls would be expected to have SBP below
120mmHg?
Answer:
• If 14.5% have SBP >120mmHg then
• 100-14.5= 85.5% will have SBP less <120mmHg.

29. C) What proportion of girls would be expected to have SBP between 85
and 120 mmHg.
Answer
SND1; Z =
85−105.8
13.4
= −1.55
SND2; Z =
120−105.8
13.4
= 1.06

30. The area to the left of SND 1.55 is 0.060571 and
The area to the right of the SND 1.06 is 0.85543
So the proportion with SBP between 85mmHg and
120mmHg= 0.85543-0.060571 = 0.794859
The proportion of girls between 85mmHg and 120mmHg is
79.49%
Continuation of solution…

31. Something to Note
• Varying Z-tables need different ways of tweaking around.
• Always remember to look at the title of your Z-table.
• Also look at the graph it represents

32. The varying Z-score tables

33. Exercise 1
The average fasting blood sugar level of adult male in some area
is 90mg% with SD of 5mg%. If fasting blood sugar levels are
normally distributed:
a) Would you regard fasting blood sugar of 95mg% as abnormal?
b) Is fasting blood sugar of 105mg% as abnormal?

34. Exercise 2
a) Find the z-score corresponding to a raw score of 132 from a normal
distribution with mean 100 and SD 15
b) A z-score of 1.7n was found from an observation coming from a
normal distribution with mean 14 and standard deviation 3. find
the raw score.

35. Exercise 3
Suppose the average length of stay in a chronic disease
hospital of a certain type of patient is 60 days with a
standard deviation of 15. If it is reasonable to assume an
approximately normal distribution of lengths of stay, find the
probability that a randomly selected patient from this group
will have a length of stay:
(a) Greater than 50 days.
(b) Less than 30 days.
(c) Between 30 and 50 days
(d) Greater than 90 days.

36. • Intuitive Biostatistics: A Nonmathematical Guide to Statistical Thinking by
Harvey Motulsky
• Principles of biostatistics by Marcello Pagano
• https://statisticsbyjim.com/basics/skewed-distribution/
• https://statisticsbyjim.com/basics/normal-distribution/
REFERENCES