CHAPTER 5: NORMAL PROBABILITY DISTRIBUTIONS
Introduction
A continuous random variable X is a quantitative random variable that can assume an
uncountable number of values. Since for a continuous random variable all possible fractional
values of the variable cannot be listed, probabilities associated with the random variable are
determined by a function called probability density function. The function can be illustrated by a
curve, y = f (x). Probabilities are given by the area under the curve. The total area under a
probability density function must equal 1.0.
Normal Distribution
The probability density function f (x) of a normal random variable X depends on its mean, μ and
standard deviation, σ, where
,-∞<x<∞
f (x) =
A random variable X having a normal distribution with mean, μ and standard deviation, σ, can
also be written as X ~ N (μ, σ2). The normal distribution curve has the following properties:





It has a bell-shaped graph.
It is symmetrical about μ. Symmetrical means that each half of the distribution is a mirror
image of the other half.
It extends from - ∞ to +∞
The area under the curve yields the probabilities, so the total area under the curve is 1.
That is,
Because the distribution is symmetric, the area of the distribution on each side of the
mean is 0.5.
Finding probabilities
The probability that a random variable X is between a and b is written P (a < X < b)
Since X is continuous, P (a < X < b) =
. Since the normal function is so complex and
difficult to integrate, table of normal probabilities can be used instead.
The Standard Normal Distribution
The standard normal distribution is the normal distribution of variable Z with μ = 0 and σ = 1,
hence Z ~ N (0, 1). The density for Z is thus
f (z) =
,-∞<z<∞
1
Any value of X from a normally distributed population (X ~ N (μ, σ2)) can be transformed into
equivalent standard normal value Z by the formula Z =
A Z value restates the original value X in terms of the number of units of the standard deviation
by which the original value differs from the mean of the distribution. A negative value of Z
indicates that the original value X is below the value of the mean. If the value of X is more than
the mean, the Z value is positive.
Using standard normal tables
The standard normal table gives the area under the curve that is greater than a particular value
Z. This means that this area gives P (Z > z).
Example: Determine the probability or area for the portions of the normal distribution described.
(a) P (Z > 0.85)
= 0.1977
(b) P (Z > -1.25)
= 1 – 0.1056 = 0.8944
(c) P (Z ≤ 0.93)
= 1 – P (Z > 0.93)
= 1 – 0.1762
= 0.8238
(d) P (Z ≤ -1.21)
= 0.1131
(e) P (1.1 ≤ Z ≤ 2.38) = P (Z ≥ 1.1) – (P Z ≥ 2.38)
= 0.1357 – 0.0087
= 0.127
(f) P (-1.01 < Z < 0)
= P(Z > -1.01) – P(Z ≥ 0)
= 0.8438 – 0.5
= 0.3438
(g) P (-1.2 ≤ Z ≤ 1.61) = P (Z ≥ -1.2) – P (Z > 1.61)
= 0.8849 – 0.0537
= 0.8312
Example: Suppose X is a normal distribution with mean 70 and variance 4. Find:
(a) P (X ≥ 74)
(b) P (67 ≤ X ≤ 75)
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(c) P (71 < X ≤ 72)
(d) P (63 ≤ X ≤ 68)
X ~ N (70, 4). Standardize X using Z =
(a) P (X ≥ 74)
=P
= P (Z
2)
= 0.0228
(b) P (67 ≤ X ≤ 75)
=P
= P (-1.5
Z ≤ 2.5)
= P (Z ≥ -1.5) – P (Z > 2.5)
= 0.9332 + 0.0062 = 0.927
(c) P (71 < X ≤ 72)
=P
= P (0.5 < Z ≤ 1)
= P (Z > 0.5) – P (Z > 1)
= 0.3085 – 0.1587 = 0.1498
(d) P (63 ≤ X ≤ 68)
=P
= P (-3.5 ≤ Z ≤ -1)
= P (Z ≥ 3.5) – P (Z > 1)
= 0.9998 – 0.8413
= 0.1585
Example: The heights of students at a particular school are normally distributed with mean 168
cm and standard deviation 3 cm. Find the percentage of students who are:
(a) at least 172 cm
(b) less than 170 cm
(c) between 166 and 171 cm
Solution:
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X: The height of students; X ~ N (168, 32)
Standardize X using Z =
(a) P (X ≥ 172)
=P
= P (Z ≥ 1.33)
= 0.0918
Thus 9.18% of students are at least 172 cm.
(b) P (X < 170)
=P
= 1 – P (Z ≥ 0.67)
= 1 – 0.2514
= 0.7468
Thus, 74.86% of students are less than 170 cm.
(c) P (166 ≤ Z ≤ 171) = P
= P (-0.67 ≤ Z ≤ 1)
= 0.7486 – 0.1587
= 0.5899
Thus, 58.99% of students are between 166 and 171 cm.
Example: The time taken by 85 students to complete a test is normally distributed with a mean
of 50 minutes and a standard deviation of 3.5 minutes. Find the number of students who
complete the test:
(a) in more than 52 minutes.
(b) between 48 minutes and 55 minutes.
(c) in less than 62 minutes.
Solution:
X: The time taken to complete a test; X ~ N (50, 3.52)
Standardize X using Z =
4
(a) P (X > 52)
=P
= P (Z > 0.57)
= 0.2843
To find the number of students, multiply the probability by 85. Thus, 0.2583 x 85 ≈ 22 students
complete the test in more than 52 minutes.
(b) P (48 ≤ X ≤ 55)
=P
= P (-0.57 ≤ Z ≤ 1.43)
= P (Z ≥ -0.57) – P (Z > 1.43)
= 0.7157 – 0.0764
= 0.6393
Thus, 0.6393 x 150 ≈ 55 students complete the test between 48 minutes and 55 minutes.
(c) P (X < 62)
=P
= P (Z < 3.43)
= 1 – P (Z ≥ 3.43)
= 1 – 0.0003
= 0.9997
Thus, 0.9997 x 85 ≈ 85 students complete the test in less than 62 minutes.
Example: Suppose the scores of an examination are normally distributed with mean 75 and
standard deviation 10. The top 10.75% of the students receive As and the bottom 6.55% receive
Fs. Find:
(a) the minimum score to receive an A
(b) the minimum score to pass (not an F)
Solution:
X: Scores of an examination; X ~ N(75,102). Standardize X using Z =
(a) Suppose the minimum score to receive an A is a. Thus,
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P (X ≥ a)
= 0.1075
P
= 0.1075
From the table, P (Z ≥ 1.24) = 0.1075

= 1.24

= 87.4
(b) Suppose the minimum score to pass is f. Thus,
P (X ≤ f)
= 0.0655
P
= 0.0655
P (Z ≤
)
= 0.0655
P (Z ≥
)
= 0.0655
From the table, P (Z ≥ 1.51) = 0.0655
 –
)
= 1.51
 f
= 59.9
Therefore, a student must score more than 59.9 in order to pass the examination.
The normal approximation to the Binomial Distribution
Given a random variable X ~ b (n, p), if n ≥ 30 and both np ≥ 5 and nq ≥ 5, then X ~ N (np, npq)
approximately.
Continuity Corrections
A continuity correction is needed when a continuous curve is being used to approximate
discrete probability distributions. 0.5 is added or subtracted as a continuous correction factor
according to the form of the probability statement as follows:
(a) P (X = x)
P (x – 0.5 < X < x + 0.5)
(b) P (X ≤ x)
P (X < x + 0.5)
(c) P (X < x)
P (X ≤ x – 0.5)
(d) P (X ≥ x)
P (X > x – 0.5)
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(e) P (X > x)
P (X ≥ x + 0.5)
Example: A fair coin is tossed 100 times. Find the probability that tail occurs:
(a) less than 45 times
(b) exactly 60 times
(c) between 48 and 53 times
Solution:
Let X: Number of times a tail occurs when a fair coin is tossed 100 times.
X ~ B(100, 0.5) => X ~ N(50, 25)
Standardized X using Z =
(a) P (X < 45)
= P (X ≤ 45 – 0.5)
= P (X ≤ 44.5)
=P
= P (Z ≤ - 1.1)
= 1 – 0.8643
= 0.1357
(b) P (X = 60)
= P (60 – 0.5 < X < 60 + 0.5)
= P (59.5 < X < 60.5)
=P
= P (1.9 < Z < 2.1)
= P (Z > 1.91) – P (Z < 2.1)
= 0.0287 – 0.0179
= 0.0108
(c) P (48 < X < 53)
= P (48 + 0.5 ≤ X ≤ 53 – 0.5)
= P (48.5 ≤ X ≤ 52.5)
=P
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= P (-0.3 ≤ X ≤ 0.5)
= P (Z ≥ -0.3) – P (Z > 0.5)
= 0.6179 – 0.3085
= 0.3094
The normal approximation to the Poisson distribution
The mean and the variance for the normal probability distribution that is used to approximate
Poisson probabilities is μ = λ and σ2 = λ
Given a random variable X ~ Po (λ), if λ is relatively large, then X ~ N (λ, λ)
Example: At a supermarket an average of 1 customer per two minutes arrive at a checkout
stand. Find the probability that more than 70 customers arrive at the stand during a particular
interval of 2 ¼ hour?
Solution:
Let X: Number of customers arriving at a checkout stand in a 2-minute interval
X ~ Po (1)
In a 2 ¼ hr interval, λ = ½ x 135 = 67.5
Let Y: Number of customers arriving at a checkout stand in a 2 ¼ hr interval
Y ~ Po (67.5) ⇒ Y ~ N (67.5, 67.5). Standardize Y using Z =
Thus P (Y > 70)
= P (Y ≥ 70 + 0.5)
= P (Y ≥ 70.5)
=P
= P (Z ≥ 0.37)
= 0.3557
Example: A telephone operator receives an average of 2 calls per minute. What is the
probability that the operator receives at most 45 calls in 25 minutes?
Solution:
Let X: Number of calls in one minute; X ~ Po (2)
In 25 minutes, λ = 2/1 x 25 = 50
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Let Y: Number of calls in 25 minutes; Y ~ Po (50) ⇒ Y ~ N (50, 50)
Standardize Y using Z =
Thus, P (Y ≤ 45)
= P (Y < 45 + 0.5)
= P (Y < 45.5)
=P
= P (Z ≤ -0.64)
= 1 – P (Z > - 0.64)
= 1 – 0.7389
= 0.2611
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