Download MATHEMATICS II (SHF 1124) CHAPTER 6 THE NORMAL

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 that 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
𝝈 = 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
𝝁

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 : THE 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:
𝝈=𝟏



𝝁=𝟎
The z value or z scores are the units marked on the horizontal axis of a standard
normal curve.
The continuos random variable Z is usually used to represent the standard normal
random variable, 𝑍~𝑁(0,1)
To standardize 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)
𝟎
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𝒛
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MATHEMATICS II (SHF 1124)
## When using the normal dist. table, pls take note that if area falls exactly
halfway between 2 z values, use the larger of the 2 values. Example area 0.45
falls halfway between 0.4495 & 0.4505... use z=1.65 rather than 1.64….. Sr.
Fizah 6/9/13
EXAMPLE:
If 𝑋~𝑁(4,9) find using standard normal table:
a) 𝑃(𝑋 ≥ 8)
0.0918
b) 𝑃(𝑋 ≤ 5)
0.6293
c) 𝑃(4 ≤ 𝑋 ≤ 6)
0.2486
d) 𝑃(𝑋 ≥ 3)
0.6293
e) 𝑃(𝑋 ≤ 2)
0.2514
f) 𝑃(1 ≤ 𝑋 ≤ 3)
0.2120
g) 𝑃(2 ≤ 𝑋 ≤ 5)
0.3779
h) 𝑃(3.5 < 𝑋 < 8.3)
0.4911
 Note
that
for
continous
random
variable,
𝑷(𝟑. 𝟓 < 𝑋 < 8.3) =
𝑷(𝟑. 𝟓 ≤ 𝑿 ≤ 𝟖. 𝟑)
i) the value of m if 𝑃(𝑋 ≥ 𝑚) = 0.1151
𝒎 = 𝟕. 𝟔
EXERCISE:
1. Compute the value of 𝑘 so that
i.
𝑃(𝑍 < 𝑘) = 0.3974
𝒌 = −𝟎. 𝟐𝟔
ii.
𝑃(−𝑘 < 𝑍 < 𝑘) = 0.8324
𝒌 = 𝟏. 𝟑𝟖
iii.
𝑃(𝑍 > 𝑘) = 0.3
𝒌 = 𝟎. 𝟓𝟐
iv.
𝑃(𝑍 < 𝑘) = 0.25
𝒌 = −𝟎. 𝟔𝟕
v.
𝑃(−1 < 𝑍 < 𝑘) = 0.6
𝒌 = 𝟎. 𝟕
vi.
𝑃(0.3 < 𝑍 < 𝑘) = 0.1
𝒌 = 𝟎. 𝟓𝟖
vii.
𝑃(𝑘 < 𝑍 < 1.75) = 0.88
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
DEPARTMENT OF MATHEMATICS
𝝁 = 𝟒𝟑. 𝟑
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MATHEMATICS II (SHF 1124)
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)
-2.15
1.62
0.0684
b)
0.3574
-0.44
1.92
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
1. The daily working hours of a taxi driver in Kuala Lumpur is normally distributed with
mean 12 hours and standard deviation 2 hours. Find the probability that a taxi driver
chosen at random in Kuala Lumpur drives
a) for less than 15 hours
0.9332
b) for more than 10 hours
0.8413
c) between 10 hours and 15 hours.
0.7745
d) less than 8 hours or more than14 hours
0.1815
2. Consider the intelligence quotient (I.Q) scores for people. I.Q. scores are normally
distributed with a mean of 100 and a standard deviation of 16. Find the probability that
a person selected at random will have an I.Q
(a) Between 100 and 115 0.3264
(b) greater than 90
0.7357
Find the 33rd percentile for I.Q. scores.
92.96
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
3. Ella found that the time taken for her to walk to her office from her house is normally
distributed with a mean of 45 minutes and a standard deviation of 6 minutes. If she
starts from her house at 8.10 a.m., find the probability that she will reach her office
before 9 a.m.
0.7967
4. To qualify for a police academy, candidates must score in the top 10% on a general
abilities test. The test has mean 200 and standard deviation of 20. Find the lowest
possible score to qualify. Assume that the test is normally distributed.
226
5. For a medical study, a researcher wishes to select people in the middle 60% of the
population based on blood pressure. If the mean systolic blood pressure is 120 and the
standard deviation is 8, find the upper and lower readings that would qualify people to
participate in the study.
𝟏𝟏𝟑. 𝟐𝟖 < 𝑋 < 126.72
6. In a normal distribution, find the standard deviation when mean is 100 and 2.68% of
the area lies to the right of 105.
2.59
7. The scores on a test have a mean of 100 and a standard deviation of 15. If a personnel
manager wishes to select from the top 75% of applicants who take the test, find the
cutoff score. Assume the variable is normally distributed.
89.95
8. An athletic association wants to sponsor footrace. The average time it takes to run the
course is 58.6 minutes, with a standard deviation of 4.3 minutes. If the association
decides to award certificates to the fastest 20% of the racers, what should the cutoff
time be? Assume the variable is normally distributed.
9.
54.99
The incomes of junior executives in a large corporation are normally distributed with a
standard deviation of $1200. A cutback is pending, at which time those who earn less
than $28000 will be discharged. If such a cut represents 10% of the junior executives,
what is the current mean salary of the group of junior executives?
29536
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
10. The average life of a certain type of small motor is 10 years with a standard deviation of
2 years. The manufacturer replaces free all motor that fail while under guarantee. If he
is willing to replace only 3% of the motor that fail, how long a guarantee should he
offer? Assume that the lives of the motors follow normal distribution.
6.24
6.4 : THE CENTRAL LIMIT THEOREM

The central limit theorem is used to solve problems involving sample means for
large samples

For random samples from a population which is not necessarily normally
distributed, the distribution of the sample means, 𝑥̅ , is approximately normal when
the sample size n is large.

As n increases the approximation to a normal distribution improves.

The distribution of the means has reducing standard deviation as 𝑛 increases but 𝜇𝑥̅
is constant and equal to the population mean, 𝜇.

As sample size 𝑛 increases:
𝑠𝑥̅
𝑠𝑥̅ decreases as 𝑠𝑥̅ =
𝑠𝑥̅
𝜎
√𝑛
𝑠𝑥̅
and 𝜇𝑥̅ = 𝜇 always.
Remember with the Central Limit Theorem we are looking the distributions of the
sample means ̅𝑥, not at the distribution of individuals scores.

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
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
Properties:
mean of sample mean, 𝑋̅ will be the same as population mean 𝜇
i.
𝝁𝑿̅ = 𝝁
ii.
𝜎 = population standard deviation
𝜎𝑋̅ = sample mean standard deviation
𝑠𝑥̅ =

𝜎
√𝑛
Using the Central Limit Theorem
As the distribution of the sample means approaches the normal distribution,
the standard normal variable, z, will be given by:
𝑧=
𝑥̅ − 𝜇𝑥̅
𝑠𝑥̅
𝑤ℎ𝑒𝑟𝑒 𝜇𝑥̅ = 𝜇
𝑎𝑛𝑑 𝑠𝑥̅ =
𝜎
√𝑛
{𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝐿𝑖𝑚𝑖𝑡 𝑇ℎ𝑒𝑜𝑟𝑒𝑚}
𝑆𝑜,
𝒛=
̅−𝝁
𝒙
𝝈
√𝒏
Note: In problems involving sample means this formula enables us to use the
standard normal distribution to calculate probabilities
where
̅ = 𝐬𝐚𝐦𝐩𝐥𝐞 𝐦𝐞𝐚𝐧
𝑿
𝝁 = 𝐦𝐞𝐚𝐧 𝐨𝐟 𝐬𝐚𝐦𝐩𝐥𝐞 𝐦𝐞𝐚𝐧=population mean
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
6.4 : THE CENTRAL LIMIT THEOREM
The probability distribution of ̅
X is called its sampling distribution.
A sampling distribution of sample means is a distribution obtained by using the means
computed from random samples of a specific size taken from a population.
Consider a small population: {0, 2, 4, 6, 8}
Possible samples of size 2 taken with replacement and mean of each sample :
Sample
0,0
0,2
0,4
0,6
0,8
2,0
2,2
2,4
2,6
2,8
4,0
4,2
4,4
Mean
0
1
2
3
4
1
2
3
4
5
2
3
4
Sample
4,6
4,8
6,0
6,2
6,4
6,6
6,8
8,0
8,2
8,4
8,6
8,8
Mean
5
6
3
4
5
6
7
4
5
6
7
8
Sampling distribution for the sample means :
̅
X
̅̅̅
P(X)
0
1/25
= 0.04
1
2/25
=0.08
2
3/25
= 0.12
3
4/25
= 0.16
4
5/25
= 0.20
5
4/25
= 0.16
6
3/25
= 0.12
7
2/25
=0.08
8
1/25
= 0.04
The shape of the histogram for the distribution above appears to be approximately normal.
The mean of the sample means, 𝜇𝑋̅ =
The population mean, 𝜇 =
0+1+2+⋯+8
0+2+4+6+8
5
25
=4
=4
Hence, 𝜇𝑋̅ = 𝜇
The standard deviation of the sample means, 𝜎𝑋̅ is
𝜎𝑋̅ = √
(0−4)2 +⋯+(8−4)2
25
= 2 is the same as the population standard deviation divided by
√2.
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
In summary if all possible samples of size n are taken with replacement from the same
population :
1. the mean of the sample means = population mean (𝜇𝑋̅ = 𝜇)
𝜎
2. 𝜎𝑋̅ =
√𝑛
3. the shape of the distribution is explained by the central limit theorem as follows:
The Central Limit Theorem – As the sample size n increases without limit, the shape
of the distribution of the sample means will approach a normal distribution with
mean 𝜇 and standard deviation
𝜎
√𝑛
.
The standard normal variable, z, will be given by: 𝑧 =
̅ −μ
X
σ
√n
It is important to remember that in using the Central Limit Theorem,
a) when the original variable is normally distributed, the distribution of the sample
means will be normally distributed, for any n.
b) when the original variable is not normally distributed, n must be quite large
(n ≥ 30) before normal distribution gives a satisfactory approximation. The
larger n is, the better the approximation will be.
##### (Note : if original variable is not normal & n < 30… not in our syllabus )
EXAMPLE :
1. The average cholesterol content of a certain brand of eggs is 215 milligrams and the
standard deviation is 15 milligrams. Assume the variable is normally distributed.
a) If a single egg is selected, find the probability that the cholesterol content will be
more than 220 milligrams.
0.3707
b) If a sample of 25 eggs is selected, find the probability that the mean of the
sample will be larger than 220 milligrams.
0.0475
2. During the last week of the semester, students at a certain college spend on the average
4.2 hours using the school’s computer terminals with a standard deviation of 1.8 hours.
For a random sample of 36 students at that college, find the probabilities that the
average time spent at the computer terminals during the last week of the semester is
a) at least 4.8 hours
0.0228
DEPARTMENT OF MATHEMATICS
b) between 4.1 and 4.5 hours
0.4706
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MATHEMATICS II (SHF 1124)
3. Assume that the mean systolic blood pressure of normal adults is 120 mm Hg and the
standard deviation is 5.6. Assume the variable is normally distributed.
a) If an individual is selected, find the probability that the individual’s pressure will be
between 120 and 121.8 mm Hg.
0.1255
b) If a sample of 30 adults is randomly selected, find the probability that the sample
mean will be between 120 and 121.8 mm Hg.
0.4608
c) Why is the answer for part (a) so much smaller than (b)?
4. Kindergarten children have heights that are approximately normally distributed about
a mean of 39 in. and a standard deviation of 2 in. A random sample of size 25 is taken
and the mean is calculated.
a) What is the probability that this mean value will be between 38.5 and 40.0 inches?
0.8882
b) Within what limits would the middle 90% of the sampling distribution of sample
̅ < 39.33
means for samples of size 100 fall?
𝟑𝟖. 𝟔𝟕 < 𝑿
5. In redesigning jet ejection seats to better accommodate woman as pilots, it is found that
women’s weights are normally distributed with a mean of 143 lb and a standard
deviation of 29 lb. If a woman is randomly selected,
a) What is the probability that she weighs between 143 lb and 201 lb?
0.4772
b) Find the weight separating the bottom 10% from the top 90%.
105.88
c) Find the maximum and minimum weights which include in the middle 30% of the
women weights.
𝟏𝟑𝟏. 𝟔𝟗 < 𝑋 < 𝟏𝟓𝟒. 𝟑𝟏
If a sample of 20 women is selected, find the probability that the mean of the sample
will be less than 150 lb.
0.8599
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) 𝑷(𝑿 ≥ 𝒂) = 𝑷(𝑿 > 𝑎 − 0.5)
b) 𝑷(𝑿 > 𝑎) = 𝑷(𝑿 > 𝑎 + 0.5)
c) 𝑷(𝑿 ≤ 𝒂) = 𝑷(𝑿 < 𝑎 + 0.5)
d) 𝑷(𝑿 < 𝑎) = 𝑷(𝑿 < 𝑎 − 0.5)
e) 𝑷(𝒂 ≤ 𝑿 ≤ 𝒅) = 𝑷(𝒂 − 𝟎. 𝟓 < 𝑿 < 𝑑 + 0.5)
f) 𝑷(𝒂 < 𝑿 < 𝑑) = 𝑷(𝒂 + 𝟎. 𝟓 < 𝑿 < 𝑑 − 0.5)
g) 𝑷(𝑿 = 𝒂) = 𝑷(𝒂 − 𝟎. 𝟓 < 𝑿 < 𝑎 + 0.5)
Guideline to use normal approximation to binomial distribution
1. Check for 𝒏𝒑 ≥ 𝟓 𝒂𝒏𝒅 𝟎. 𝟏 ≤ 𝒑 ≤ 𝟎. 𝟗
2. Find 𝜇 and 𝜎 where 𝝁 = 𝒏𝒑 and 𝝈 = √𝒏𝒑𝒒
3. Write the problem in probability notation
4. Use continuity correction (show area under the curve)
5. Find z-values from standard normal table
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
EXAMPLE:
1. Use the normal approximation to the binomial to find the probabilities for the
specific values of X:
a) n = 50, p = 0.8, X = 44
0.0516
b) n = 50, p = 0.6, X ≤ 40
0.9988
2. 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,
0.1398
b) the normal approximation to the binomial distribution.
0.1428
3. According to recent surveys, 60% of households have personal computers. If a
random sample of 180 households is selected, find the probability that more than 60
but fewer than 100 have a personal computer.
0.0984
4. Seventy-eight percent of U.S homes have a telephone answering device. In a random
sample of 250 homes, find the probability that fewer than 50 do not have telephone
answering device.
0.2005
5. A survey on the preference of 1000 consumers found 180 consumers preferred
frozen local food ‘Pau Kaya’. If the percentage of consumers that prefer Pau Kaya is
actually 20%, find the approximate probability of observing 180 or less in a sample
of 1000 that prefer Pau Kaya.
0.0618
6. An airline finds that 5% of the persons making reservation on a certain flight will
not show up for the flight. If the airline sells 180 tickets for a flight with only 175
seats, what is the probability that a seat will be available for every person holding a
reservation and planning to fly?
0.9382
DEPARTMENT OF MATHEMATICS
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MATHEMATICS II (SHF 1124)
6.6 : THE NORMAL APPROXIMATION TO THE POISSON DISTRIBUTION



The concept of normal approximation to the binomial distribution can be extended
to the Poisson distribution.
If X has a Poisson distribution with the parameter 𝜆, the 𝐸(𝑋) = 𝜆 and 𝑉𝑎𝑟(𝑥) = 𝜆.
For large values of 𝜆, that is 𝜆 > 30 , 𝑋 is approximately normal with mean 𝜆 and
variance 𝜆.
EXAMPLE:
1. Let 𝑋~𝑃𝑜(60). Using normal approximation, find the following probabilities.
a) 𝑃(𝑋 ≤ 50)
𝟎. 𝟏𝟎𝟗𝟑
b) 𝑃(36 ≤ 𝑋 ≤ 54)
𝟎. 𝟐𝟑𝟖𝟖
c) 𝑃(𝑋 > 52)
𝟎. 𝟖𝟑𝟒
2. The number of absentees per day in a kindergarten follows a Poisson distribution
with an average of 10 absentees in a day. Find the probability that there are at least
50 absentees per week. (Assume that classes are conducted 7 days a week).
𝟎. 𝟗𝟗𝟐𝟗
3. The number of printing errors on any page of a certain book has a Poisson
distribution with mean 0.4.
a) Find the probability that the total number of errors in the first 10 pages is more
than 3.
0.5665
b) The book has 250 pages. If there are more than 110 errors then the publishers will
have the book corrected and reprinted. Use a suitable approximation to find the
probability of this happening.
0.1469
4. When pond water is examined under the microscope there are nasty bugs to be
seen. The concentration of bugs in the water may be assumed to have a Poisson
distribution with an average concentration of three per milliliter.
a) Find the probability that a random sample of 5 milliliters of pond water contains
exactly 14 bugs.
0.1025
b) Using a suitable approximation, calculate the probability that the number of bugs to
be found in a random sample of 200 milliliters of pond water is between 560 and
620 inclusive.
0.750
DEPARTMENT OF MATHEMATICS
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