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MATHEMATICS II (SHF 1124)
CHAPTER 6 THE NORMAL DISTRIBUTION
6.1 : PROPERTIES OF THE NORMAL DISTRIBUTION

In this chapter, the distribution of a continuous random variable will be discussed.
The most widely used are the normal and standard normal distributions.

Probability of a continuous random variable assumes within a certain interval is
given by the area under curve within the interval itself.
NORMAL DISTRIBUTION

A large number of phenomena in the world can be described or approximately
described by normal distribution. For example, the examination scores of students,
the weight of newborn babies, the length of a type of leaves and the amount of soda
in drinks.

A normal probability distribution is a bell-shaped curve with the following
characteristics:
a) The total area under the curve is 1.0.
b) The curve is symmetric about the mean 𝜇.
c) The 2 tails of the curve extend indefinitely.
d) The curve is described below

Normal distribution is identified by 2 parameters, mean 𝜇 and variance 𝜎 2 . Given
the values of 𝜇 and 𝜎 2 , the area under the curve can now be obtained.

Each different values of 𝜇 and 𝜎 2 give different normal distribution

The value of mean 𝜇 determines the center of the distribution whereas different
values of 𝜎 2 determines how much the observation is dispersed.

A normal random variable is a continuous random variable X that has a normal
distribution 𝑋~𝑁(𝜇, 𝜎 2 ).
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
6.2 : STANDARD NORMAL DISTRIBUTION

The normal distribution with mean 𝜇 = 0 and standard deviation 𝜎 = 1 is called the
standard normal distribution denoted by 𝑁(0,1) described below:

The Z value of Z scores are the units marked on the horizontal axis of standard
normal curve.
The continous random variable Z is usually used to represent the standard normal
random variable, 𝑍~𝑁(0,1)
To standardized a normal random variable X to a standard normal random variable
Z, the following formula is used:


𝒁=
𝑿−𝝁
𝝈
where 𝜇 = 𝑚𝑒𝑎𝑛
𝜎 = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛


The standard normal table can be used to find the area under the standard normal
curve (probability)
The table that will be used in this chapter is 𝑷(𝟎 ≤ 𝒁 ≤ 𝒛)
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
EXAMPLE:
If 𝑋~𝑁(4,9) find using standard normal table:
a)
b)
c)
d)
e)
f)
g)
h)
𝑃(𝑋 ≥ 8)
𝑃(𝑋 ≤ 5)
𝑃(4 ≤ 𝑋 ≤ 6)
𝑃(𝑋 ≥ 3)
𝑃(𝑋 ≤ 2)
𝑃(1 ≤ 𝑋 ≤ 3)
𝑃(2 ≤ 𝑋 ≤ 5)
𝑃(3.5 < 𝑋 < 8.3)
 Note that for continous random
𝑷(𝟑. 𝟓 ≤ 𝑿 ≤ 𝟖. 𝟑)
i) the value of m if 𝑃(𝑋 ≥ 𝑚) = 0.1151
variable,
0.0918
0.6293
0.2486
0.6293
0.2514
0.2120
0.3779
0.4911
𝑷(𝟑. 𝟓 < 𝑋 < 8.3) =
𝒎 = 𝟕. 𝟔
EXERCISE:
1. Compute the value of 𝑘 so that
i. 𝑃(𝑍 < 𝑘) = 0.3974
ii. 𝑃(−𝑘 < 𝑍 < 𝑘) = 0.8324
iii. 𝑃(𝑍 > 𝑘) = 0.3
iv.
𝑃(𝑍 < 𝑘) = 0.25
v.
𝑃(−1 < 𝑍 < 𝑘) = 0.6
vi.
𝑃(0.3 < 𝑍 < 𝑘) = 0.1
vii. 𝑃(𝑘 < 𝑍 < 1.75) = 0.88
𝒌 = −𝟎. 𝟐𝟔
𝒌 = 𝟏. 𝟑𝟖
𝒌 = 𝟎. 𝟓𝟐
𝒌 = −𝟎. 𝟔𝟕
𝒌 = 𝟎. 𝟕
𝒌 = 𝟎. 𝟓𝟖
𝒌 = −𝟏. 𝟒𝟏
2. Let 𝑋 be a normal random variable with mean 𝜇 and standard deviation 𝜎.
a)
If 𝑃(𝑋 ≥ 50) = 0.2514 , 𝜎 = 10 find 𝜇
𝝁 = 𝟒𝟑. 𝟑
b)
If 𝑃(25 < 𝑋 < 30) = 0.3944 , 𝜇 = 25, find 𝜎
𝝈=𝟒
3. Find the area of the curve below by using standard normal table
a)
Normal curve
b)
Normal curve
0.0684
0.3574
4. 𝑋~𝑁(𝜇 , 𝜎 2 ). If 𝑃(𝑋 > 12) = 0.3 and 𝑃(𝑋 < 6) = 0.4, find the value of 𝜇 and 𝜎.
𝝁 = 𝟕. 𝟗𝟓 , 𝝈 = 𝟕. 𝟕𝟗
6.3 : APPLICATIONS OF THE NORMAL DISTRIBUTION
Refer hondout given to the students
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
6.4 : THE CENTRAL LIMIT THEOREM



The central limit theorem is used to solve problems involving sample means for
large samples
A sampling distribution of sample means
Distribution using the means computed from all possible random samples of
a specific size taken from a population
Sampling error
The difference between the sample measure and the corresponding
population measure due to the fact that the sample is not a perfect
presentation of the population
Properties:
i. mean of sample mean, 𝑋̅ will be the same as population mean 𝜇
𝝁𝑿̅ = 𝝁
ii.
𝜎 = population standard deviation
𝜎𝑋̅ = sample mean standard deviation
𝝈𝑿̅ =

𝝈
√𝒏
The central limit theorem
As sample size n increasing, the shape of distribution of 𝜇𝑋̅ taken with
replacement from 𝜇 and 𝜎 will approach normal distribution where
𝝁𝑿̅ = 𝝁
𝝈𝑿̅ =
𝝈
√𝒏
The Z-values :
̅ −𝝁
𝑿
𝒁= 𝝈
⁄ 𝒏
√
where
̅ = 𝐬𝐚𝐦𝐩𝐥𝐞 𝐦𝐞𝐚𝐧
𝑿
𝝁 = 𝐦𝐞𝐚𝐧 𝐨𝐟 𝐬𝐚𝐦𝐩𝐥𝐞 𝐦𝐞𝐚𝐧
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
6.5 : THE NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION



A binomial distribution with parameters 𝑛 and 𝑝 can be approximated by a normal
distribution with mean 𝜇 = 𝑛𝑝 and standard deviation 𝜎 = √𝑛𝑝𝑞.
The normal approximation to the binomial distribution is good if 𝑛 is large and 𝑝
close to 0.5.
The condition to use normal approximation is when :
𝒏𝒑 ≥ 𝟓 and 𝟎. 𝟏 ≤ 𝒑 ≤ 𝟎. 𝟗


Note that 𝑋~𝐵(𝑛, 𝑝) is a discrete random variable whereas 𝑋~𝑁(𝜇 , 𝜎 2 ) is
continuous.
Before using the normal approximation, a correction for continuity must be carried
out because a continuous distribution is used as an approximation for a discrete
distribution
A correction for continuity
a)
b)
c)
d)
e)
f)
g)
𝑷(𝑿 ≥ 𝒂) = 𝑷(𝑿 > 𝒂 − 𝟎. 𝟓)
𝑷(𝑿 > 𝒂) = 𝑷(𝑿 > 𝒂 + 𝟎. 𝟓)
𝑷(𝑿 ≤ 𝒂) = 𝑷(𝑿 < 𝒂 + 𝟎. 𝟓)
𝑷(𝑿 < 𝒂) = 𝑷(𝑿 < 𝒂 − 𝟎. 𝟓)
𝑷(𝒂 ≤ 𝑿 ≤ 𝒅) = 𝑷(𝒂 − 𝟎. 𝟓 < 𝑿 < 𝒅 + 𝟎. 𝟓)
𝑷(𝒂 < 𝑿 < 𝒅) = 𝑷(𝒂 + 𝟎. 𝟓 < 𝑿 < 𝒅 − 𝟎. 𝟓)
𝑷(𝑿 = 𝒂) = 𝑷(𝒛 − 𝟎. 𝟓 < 𝑿 < 𝒂 + 𝟎. 𝟓)
Guideline to use normal approximation to binomial distribution
1.
2.
3.
4.
5.
Check for 𝒏𝒑 ≥ 𝟓
Find 𝜇 and 𝜎 where 𝝁 = 𝒏𝒑 and 𝝈 = √𝒏𝒑𝒒
Write the problem in probability notation
Use continuity correction (show area under the curve)
Find z-values from standard normal table
EXAMPLE:
1. 80% of the books in a cupboard are academic books. If there are 50 books in the
cupboard, find the probability that 40 books taken at random from the cupboard are
academic books, by using
a) binomial distribution,
b) the normal approximation to the binomial distribution.
DEPARTMENT OF MATHEMATICS
0.1398
0.1428
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