7 THE NORMAL DISTRIBUTION
STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 276
Learning Outcomes
STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 277
At the end of this chapter you should be able to:
1. understand the properties of a normal random variable;
2. compute probabilities for a normal distribution using the standard normal
distribution tables and also R;
3. determine the distribution of sums of normal random variables;
4. model and solve problems using normal random variables;
5. be able to model and solve problems using a combination of continuous and
discrete random variables as appropriate.
Contents
STAT1520: Economic and Business Statistics M Firth, R N Khan and B A Turlach – 278
7.1 Introduction
7.2 Sums of Normal random variables
7.1 Introduction
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A very important continuous distribution.
Let the random variable X have a normal distribution with mean µ and variance
σ
2. We write X ∼ N(µ, σ2). Note that the variance is given in this description.
Let Y = aX + b. Then
E(Y) = aµ + b, Var(Y) = a2σ2,
and Y also has a normal distribution, Y ∼ N(aµ + b, a2σ2).
Standardisation
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In particular, if
Z =
X − µ
σ
=
1
σ
X −
µ
σ
(
a =
1
σ
, b = −
µ
σ
)
then E(Z) = 0 and Var(Z) = 1, so Z ∼ N(0, 1). We call Z the standard normal
distribution.
If Z ∼ N(0, 1) and X = µ + σZ, then X ∼ N(µ, σ2).
Probability density function — effect of changing the mean
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Below are sketched the pdf of the normal distribution for different values of the mean.
Note the effect of changing the mean.
−4 −2 0 2 4
0
.
0
0
.
1
0
.
2
0
.
3
0
.
4
Changing the mean of a normal distribution
x value
D
e
n
s
i
t
y
Means
mu=1
mu=−1
mu=0
Probability density function — effect of changing the variance
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Below are sketched the pdf of the normal distribution for different values of the
variance. Note the effect of changing the variance.
−6 −4 −2 0 2 4 6
0
.
0
0
.
2
0
.
4
0
.
6
0
.
8
Changing the standard deviations of a normal distribution
x value
D
e
n
s
i
t
y
Standard Deviations
sd=2
sd=0.5
sd=1
N(0, 1) distribution tables
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Tables list cumulative probabilities for the standard normal distribution. Tables in
exams will be the same as in the textbook, and also available online.
Normal Distribution Problems
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For X ∼ N(µ, σ2), there are usually two types of problems.
Given an interval for X, find the probability.
Given a probability, find the value(s) of x (the inverse problem).
The examples below illustrate the ideas.
Example 7.1
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Let Z ∼ N(0, 1). Determine the following.
(i) P(0 < Z < 1.0)
(ii) P(Z < 1.0)
Example 7.1 (ctd)
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(iii) P(Z > 1.0)
(iv) P(Z < −2.51)
Example 7.1 (ctd)
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(iv) P(Z > −2.51)
(vi) P(|Z| < 1.96)
Example 7.1 (ctd)
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(vii) P(−1.5 < Z < 2.7)
(viii) The value of z such that P(Z < z) = 0.9505
Example 7.1 (ctd)
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(ix) The value of z such that P(|Z| < z) = 0.95
Example 7.2
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Let X ∼ N(5, 16). Determine the following.
(i) P(X < 0)
Example 7.2 (ctd)
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(ii) P(X > 10)
(iii) P(−5 ≤ X ≤ 7)
Example 7.2 (ctd)
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(iv) The value of c such that P(|X − µ| < c) = 0.95
Example 7.3
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Let X ∼ N(µ, 0.25), and suppose P(X < 5.1) = 0.9772. Find the value of µ.
Example 7.4
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A machine fills bottles of soft drink to a mean volume of 210 mL with a standard
deviation of 10 mL. The label on the bottle specifies a volume of 200 mL. A bottle is
under-filled if it contains less than the labelled volume. Assume that the volumes of
the bottles are normally distributed.
(a) What percentage of bottles are under-filled?
Example 7.4 (ctd)
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(b) In order to reduce the percentage of under-filled bottles to 1% the company decides
to adjust the standard deviation of the volumes filled by the machine. What should
the standard deviation be reduced to?
7.2 SUM OF INDEPENDENT NORMAL RANDOM VARIABLES
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Let X ∼ N(µX, σ
2
X
), Y ∼ N(µY, σ
2
Y
), with X and Y independent. Then put W = X ± Y .
Then W ∼ N(µW, σ
2
W
), where
µW = µX ± µY
and σ2W = Var(X ± Y)
= Var(X) + Var(Y) = σ2X + σ
2
Y.
This result can be extended to a sum of several normal random variables.
Result
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Let X1, X2, · · · , Xn be independent normal random variables with mean E(Xi) = µi
and Var(Xi) = σ
2
i
, i = 1, 2, · · · , n. Put
Y =
n∑
i=1
Xi.
Then Y ∼ N(µY , σ
2
Y
), where
µY = E

n∑
i=1
Xi
 =
n∑
i=1
E (Xi) =
n∑
i=1
µi
and
σ
2
Y
= Var

n∑
i=1
Xi
 =
n∑
i=1
Var (Xi) =
n∑
i=1
σ
2
i .
7.2b SUM OF NON INDEPENDENT NORMAL RANDOM VARIABLES
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Let X ∼ N(µX, σ
2
X
), Y ∼ N(µY, σ
2
Y
), and put W = X ± Y .
Then
µW = µX ± µY
and σ2W = Var(X ± Y)
= Var(X) + Var(Y) ± 2Cov(X,Y).
Furthermore W ∼ N(µW, σ
2
W
) if X and Y are what we call bivariate normal.
While not every pair of normal random variables are bivariate normal, we will assume
that to be the case in STAT1520 without needing to mention it. So if you are working
on any problem with the sum of normal random variables, you may assume that the
sum is normally distributed.
Example 7.5
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Suppose the income of executives is normally distributed with the incomes of men and
women independent of each other. The means and standard deviations (in $10,000)
respectively are 20, 2.5 for men and 15, 2.0 for women.
(a) What is the probability that a randomly chosen female executive earns more than a
randomly chosen male executive?
Example 7.5 (ctd)
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Example 7.5 (ctd)
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(b) Find the probability that a randomly chosen female executive and a randomly
chosen male executive have a combined income of more than $400,000.
Example 7.6
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A machine makes washers with hole diameters that are normally distributed, with
mean 15.2 mm and variance 0.03 mm2. Another machine makes bolts with diameters
that are normally distributed, with mean 15.0 mm and variance 0.01 mm2.
(a) What is the probability that a randomly selected bolt with fit through a randomly
selected washer?
Example 7.6 (ctd)
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(b) What should be the mean diameter of the washer holes if 99% of the bolts are to
fit the washers?
Critical values of the standard normal distribution
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We often need the value of the Z-distribution corresponding to given tail
probabilities.
For convenience the commonly used values are listed below the standard normal
table.
They list the Z-values corresponding to either one-tail to two-tail probabilities.
Example Find the value of z such that
1. P(Z > z) = 0.01
2. P(Z < z = 0.05)
3. P(|Z| > z = 0.05)
4. P(|Z| > z = 0.01)