Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 1

Chapter 2: Random Variables (Cont’d)

Section 2.2: Continuous Random Variables

Problem (1): A random variable X takes values between 4 and 6 with a probability

density function:

𝑓(𝑥) =1

𝑥 𝑙𝑛(1.5) 𝑓𝑜𝑟 4 ≤ 𝑥 ≤ 6

𝑓(𝑥) = 0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

(a) Make a sketch of the probability density function.

(b) Check that the total area under the probability density function is

equal to 1.

(c) What is P(4.5 ≤ X ≤ 5.5)?

(d) Construct and sketch the cumulative distribution function.

(problem 2.2.2 in textbook)

Solution:

𝒙

𝒇(𝒙)

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 2

𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑃𝐷𝐹 = ∫ 𝑓(𝑥)6

4

= ∫1

𝑥 𝑙𝑛(1.5)

6

4

= [ln(𝑥)

𝑙𝑛(1.5)]

4

6

= [ln(6)

𝑙𝑛(1.5)−

ln(4)

𝑙𝑛(1.5)] = 1.0 (𝑂𝑘)

≤ ≤ =

𝑃(4.5 ≤ 𝑋 ≤ 5.5) = ∫ 𝑓(𝑥)5.5

4.5

= ∫1

𝑥 𝑙𝑛(1.5)

5.5

4.5

= [𝑙𝑛(𝑥)

𝑙𝑛(1.5)]

4.5

5.5

= [ln(5.5)

𝑙𝑛(1.5)−

ln(4.5)

𝑙𝑛(1.5)] = 0.4949

𝐹(𝑥) = ∫ 𝑓(𝑥)𝑥

4

= ∫1

𝑦 𝑙𝑛(1.5)

𝑥

4

𝑑𝑦 = [ln(𝑦)

𝑙𝑛(1.5)]

4

𝑋

= [ln(𝑋)

𝑙𝑛(1.5)−

ln(4)

𝑙𝑛(1.5)]

𝐹(𝑥) =ln(𝑥) − ln(4)

ln(1.5) 𝑓𝑜𝑟 4 ≤ 𝑥 ≤ 6

𝒙

𝑭(𝒙)

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 3

Problem (2): A random variable X takes values between –2 and 3 with a probability

density function:

15

64+

𝑥

64 − 2 ≤ 𝑥 ≤ 0

𝑓(𝑥) = 3

8+ 𝑐𝑥 0 ≤ 𝑥 ≤ 3

0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

(a) Find the value of c and sketch the probability density function.

(b) What is P(−1 ≤ X ≤ 1)?

(c) Construct and sketch the cumulative distribution function.

(problem 2.2.3 in textbook)

Solution:

𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑃𝐷𝐹 = ∫ 𝑓(𝑥)0

−2

+ ∫ 𝑓(𝑥)3

0

1.0 = ∫ (15

64+

𝑥

64)

0

−2

+ ∫ (3

8+ 𝑐𝑥)

3

0

1.0 = [15

64𝑥 +

𝑥2

128]

−2

0

+ [3

8𝑥 +

𝑐𝑥2

2]

0

3

1.0 = [(15

64(0) +

(0)2

128) − (

15

64(−2) +

(−2)2

128)]

+ [(3

8(3) +

𝑐 × (3)2

2) − (

3

8(0) +

𝑐 × (0)2

2)]

𝑐 = −1

8

15

64+

𝑥

64 − 2 ≤ 𝑥 ≤ 0

𝑓(𝑥) = 3

8−

𝑥

8 0 ≤ 𝑥 ≤ 3

0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 4

𝒙

𝒇(𝒙)

− ≤ ≤

𝑃(−1 ≤ 𝑋 ≤ 1) = ∫ (15

64+

𝑥

64)

0

−1

+ ∫ (3

8−

𝑥

8)

1

0

=69

128

∫ (15

64+

𝑦

64)

𝑥

−2

=15

64𝑥 +

𝑥2

128+

7

16 − 2 ≤ 𝑥 ≤ 0

𝐹(𝑥) = 7

16+ ∫ (

3

8−

𝑦

8)

𝑥

0

= 7

16+

3

8𝑥 −

𝑥2

16 0 ≤ 𝑥 ≤ 3

0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

𝒙

𝑭(𝒙)

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 5

Problem (3): A random variable X takes values between 0 and 4 with a cumulative

distribution function:

𝐹(𝑥) =𝑥2

16 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 4

(a) Sketch the cumulative distribution function.

(b) What is P(X ≤ 2)?

(c) What is P(1 ≤ X ≤ 3)?

(d) Construct and sketch the probability density function.

(problem 2.2.4 in textbook)

Solution:

𝒙

𝑭(𝒙)

≤ = ≤

𝐹(𝑋 = 2) =22

16= 0.25

≤ ≤ = = ≤ ≤

𝑃(1 ≤ 𝑋 ≤ 3) = 𝐹(𝑋 = 3) − 𝐹(𝑋 = 1) =32

16−

12

16= 0.50

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 6

𝑓(𝑥) =𝑑𝐹(𝑥)

𝑑𝑥=

𝑑(𝑥2/16)

𝑑𝑥=

2𝑥

16=

𝑥

8 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 4

𝒙

𝒇(𝒙)

Problem (4):

A random variable X takes values between 0 and ∞ with a cumulative

distribution function:

𝐹(𝑥) = 𝐴 + 𝐵𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

(a) Find the values of A and B and sketch the cumulative distribution

function.

(b) What is P(2 ≤ X ≤ 3)?

(c) Construct and sketch the probability density function.

(problem 2.2.5 in textbook)

Solution:

𝐹(∞) = 1.0

𝐴 + 𝐵 𝑒−𝑥 = 1.0

𝐴 = 1.0

𝐹(0) = 0.0

𝐴 + 𝐵 𝑒−𝑥 = 0.0

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 7

1 + 𝐵 = 0.0

𝐵 = −1.0

𝐹(𝑥) = 1 − 𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

𝒙

𝑭(𝒙)

≤ ≤ = = ≤ ≤

𝑃(2 ≤ 𝑋 ≤ 3) = 𝐹(𝑋 = 3) − 𝐹(𝑋 = 2)

= (1 − 𝑒−3) − (1 − 𝑒−2)

= 0.9502 − 0.8647 = 0.0855

𝑓(𝑥) =𝑑𝐹(𝑥)

𝑑𝑥=

𝑑(1 − 𝑒−𝑥)

𝑑𝑥= 𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

𝑓(𝑥) = 𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

𝒙

𝒇(𝒙)

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 8

Problem (5): A random variable X takes values between 0 and ∞ with a cumulative

distribution function:

𝐹(𝑥) = 𝐴 + 𝐵𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

(a) Find the value of both A and B.

(b) What is the probability that the random variable X takes a value

between 1 and 7.

(c) Find the probability density function.

(Question 4 (7 points): in Midterm Exam 2010)

Solution:

𝐹(∞) = 1.0

𝐴 + 𝐵 𝑒−𝑥 = 1.0

𝐴 = 1.0

𝐹(0) = 0.0

𝐴 + 𝐵𝑒−𝑥 = 0.0

1 + 𝐵 = 0.0

𝐵 = −1.0

𝐹(𝑥) = 1 − 𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 9

𝒙

𝑭(𝒙)

≤ ≤ = = ≤ ≤

𝑃(1 ≤ 𝑋 ≤ 7) = 𝐹(𝑋 = 7) − 𝐹(𝑋 = 1)

= (1 − 𝑒−7) − (1 − 𝑒−1)

= 0.9991 − 0.6321 = 0.3670

𝑓(𝑥) =𝑑𝐹(𝑥)

𝑑𝑥=

𝑑(1 − 𝑒−𝑥)

𝑑𝑥= 𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

𝑓(𝑥) = 𝑒−𝑥 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ ∞

𝒙

𝒇(𝒙)

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 10

Problem (6): Suppose that the random variable X is the time taken by a garage to

service a car. These times are distributed between 0 and 10 hours with a

cumulative distribution function:

𝐹(𝑥) = 𝐴 + 𝐵 𝑙𝑛(3𝑥 + 2) 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 10.

(a) Find the values of A and B and sketch the cumulative distribution

function.

(b) What is the probability that a repair job takes longer than two

hours?

(c) Construct and sketch the probability density function.

(problem 2.2.7 in textbook)

Solution:

𝐹(10) = 1.0

𝐴 + 𝐵 𝑙𝑛(3 × 10 + 2) = 1.0

𝐴 + 𝐵 ln (32) = 1.0

𝐹(0) = 0.0

𝐴 + 𝐵 𝑙𝑛(3 × 0 + 2) = 0.0

𝐴 + 𝐵 ln (2) = 0.0

𝐴 = −0.25

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 11

𝐵 = 0.3607

𝐹(𝑥) = −0.25 + 0.3607 𝑙𝑛(3𝑥 + 2) 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 10

𝒙

𝑭(𝒙)

≤

𝑃(𝑋 > 2) = 1 − 𝐹(𝑋 = 2)

= 1 − [−0.25 + 0.3607 𝑙𝑛(3 × 2 + 2)]

= 1 − 0.5007 = 0.4993

𝑓(𝑥) =𝑑𝐹(𝑥)

𝑑𝑥=

𝑑(−0.25 + 0.3607 𝑙𝑛(3𝑥 + 2))

𝑑𝑥=

0.3607 × 3

3𝑥 + 2

𝑓(𝑥) =1.0821

3𝑥 + 2 𝑓𝑜𝑟 0 ≤ 𝑥 ≤ 10

𝒙

𝒇(𝒙)

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 12

Problem (7): The resistance X of an electrical component has a probability density

function:

𝑓 (𝑥) = 𝐴𝑥(130 − 𝑥2)

for resistance values in the range 10 ≤ x ≤ 11.

(a) Calculate the value of the constant A.

(b) Calculate the cumulative distribution function.

(c) What is the probability that the electrical component has a

resistance between 10.25 and 10.5?

(problem 2.2.11 in textbook)

Solution:

𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑃𝐷𝐹 = ∫ 𝑓 (𝑥)11

10

= ∫ 𝐴𝑥(130 − 𝑥2)11

10

= 1.0

∫ (130𝐴𝑥 − 𝑥2𝐴𝑥)11

10

= 1.0

[130𝐴𝑥2

2−

𝐴𝑥4

4]

10

11

= 1.0

[130𝐴(11)2

2−

𝐴(11)4

4] − [

130𝐴(10)2

2−

𝐴(10)4

4] = 1.0

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 13

𝐴 =4

819

𝑓(𝑥) =4𝑥

819(130 − 𝑥2) 𝑓𝑜𝑟 10 ≤ 𝑥 ≤ 11

𝐹(𝑥) = ∫ 𝑓(𝑥)𝑥

10

= ∫4𝑦

819(130 − 𝑦2)

𝑥

10

𝐹(𝑥) = [130 (

4819) 𝑦2

2−

(4

819) 𝑦4

4]

10

𝑥

𝐹(𝑥) =130 (

4819) 𝑥2

2−

(4

819) 𝑥4

4−

16000

819 𝑓𝑜𝑟 10 ≤ 𝑥 ≤ 11

≤ ≤ ≤ ≤

𝑃(10.25 ≤ 𝑋 ≤ 10.5) = 𝐹(𝑋 = 10.5) − 𝐹(𝑋 = 10.25)

= 𝑓(𝑋 ≤ 10.5) − 𝑓(𝑋 ≤ 10.25)

= (130 (

4819) 10.52

2−

(4

819) 10.54

4−

16000

819)

− (130 (

4819) 10.252

2−

(4

819) 10.254

4−

16000

819)

= 0.6226 − 0.3396

= 0.2830

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 14

Problem (8): The resistance X of an electrical component has a probability density

function:

𝑓 (𝑥) = 𝐴(2𝑥 + 1) 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 2

(a) Find the value of A?

(b) Find the cumulative ditribution function?

(Question 1: in Midterm Exam 2005)

Solution:

∫ 𝑓 (𝑥)2

1

= ∫ 𝐴(2𝑥 + 1)2

1

= 1.0

[𝐴(𝑥2 + 𝑥)]12 = 1.0

[𝐴(22 + 2) − 𝐴(12 + 1)] = 1.0

𝐴 =1

4

𝑓(𝑥) =1

4(2𝑥 + 1) 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 2

𝐹(𝑥) = ∫1

4(2𝑦 + 1)

𝑥

1

=1

4(𝑥2 + 𝑥) −

1

2

𝒙

𝑭(𝒙)