Modular 11
Ch 7.1 to 7.2 Part I
Ch 7.1 Uniform and Normal Distribution
Objective A : Uniform Distribution
A1. Introduction
Recall: Discrete random variable probability distribution
Special case: Binomial distribution
Finding the probability of obtaining asuccess in nindependent trials
of a binomial experiment is calculated by plugging the value of a
into the binomial formula as shown below :
P( x  a)  n Ca p a (1  p) n  a
Continuous Random variable
For a continued random variable the probability of observing one
particular value is zero.
i.e. P( x  a )  0
Continuous Probability Distribution
We can only compute probability over an interval of values.
Since P( x  a)  0 and P ( x  b)  0 for a continuous random
variable,
P ( a  x  b)  P ( a  x  b)
To find probabilities for continuous random variables, we use
probability density functions.
Two common types of continuous random variable probability
distribution :
• Uniform distribution
• Normal distribution.
Ch 7.1 Uniform and Normal Distribution
Objective A : Uniform Distribution
A2. Uniform Distribution
1
ba
a
b
Note : The area under a probability density function is 1.
Area of rectangle = Height x Width
1 = Height x (b  a )
1
Height =
(b  a)
for a uniform distribution
Example 1 : A continuous random variable x is uniformly distributed
with 10  x  50 .
(a) Draw a graph of the uniform density function.
1
40
10 50
Area of rectangle = Height x Width
1 = Height x (b  a )
1
Height =
(b  a)
1
1

=
(50  10) 40
(b) What is P(20  x  30) ?
Area of rectangle = Height x Width

1
40
20
30
1
x (30  20)
40
1
x 10
40
1
  0.25
4

(c) What is P( x  15) ?
P( x  15)  P( x  15)
 P(10  x  15)
Area of rectangle = Height x Width

1
x5
40
1
  0.125
8

1
40
10
15
1
x (15  10)
40
Ch 7.1 Uniform and Normal Distribution
Objective A : Uniform Distribution
Objective B : Normal distribution
Ch 7.2 Applications of the Normal Distribution
Objective A : Area under the Standard Normal Distribution
Ch 7.1 Uniform and Normal Distribution
Objective B : Normal distribution – Bell-shaped Curve
Example 1: Graph of a normal curve is given.
Use the graph to identify the value of and .

  2
  2
  1   1
X
330 430 530 630 730
  530
  100
Example 2: The lives of refrigerator are normally distributed with
mean   14 years and standard deviation   2.5 years.
(a) Draw a normal curve and the parameters labeled.

  3
  2
  1   1
  2
  3
6.5 9 11.5 14 16.5 19 21.5
X
(b) Shade the region that represents the proportion of refrigerator
that lasts for more than 17 years.

17
6.5 9 11.5 14 16.5 19 21.5
X
(c) Suppose the area under the normal curve to the right x = 17 is 0.1151.
Provide two interpretations of this result.
Notation: P( x  17)  0.1151
The area under the normal curve for any interval of values of the
random variable x represent either:
– the proportions of the population with the characteristic
described by the interval of values.
11.51% of all refrigerators are kept for at least 17 years.
– the probability that a randomly selected individual from the
population will have the characteristic described by the interval
of values.
The probability that a randomly selected refrigerator will be kept
for at least 17 years is 11.51%.
Ch 7.1 Uniform and Normal Distribution
Objective A : Uniform Distribution
Objective B : Normal distribution
Ch 7.2 Applications of the Normal Distribution
Objective A : Area under the Standard Normal Distribution
Ch 7.2 Applications of the Normal Distribution
Objective A : Area under the Standard Normal Distribution

The standard normal distribution
– Bell shaped curve
–   0 and   1 .
 1
 3.5
2
 0
1
0 1
Negative Z
2
3.5 Z
Positive Z
The random variable for the standard normal distribution is Z .
Use the Z table (Table V) to find the area under the standard normal
distribution. Each value in the body of the table is a cumulative area
from the left up to a specific Z score.
Probability is the area under the curve over an interval.
The total area under the normal curve
is 1.
Z

0
Z
Under the standard normal distribution,
(a) what is the area to the right of   0 ? 0.5
(b) what is the area to the left of   0 ?
0.5
Example 1 : Determine the area under the standard normal curve.
(a) that lies to the left of -1.38.
From Table V
0.08
0.0838
 1.3
 1.38
Z
0
0.0838
P( Z  1.38)  0.0838
(b) that lies to the right of 0.56.
Table V only provides area to the left of Z = 0.56.
0.5
0.7123
1
From Table V
0.06
0.7123
Area under the whole
standard normal
distribution is 1.
0 0.56
Z
P( Z  0.56)  1  0.7123
 0.2877
(c) that lies in between 1.85 and 2.47.
0.9678
0.9932
0
1.85 2.47
Z
From Table V
2.47  0.9932
1.85  0.9678
P(1.85  Z  2.47)
 0.9932  0.9678
 0.0254