Chapter 5: The Normal Distribution (Cont’d)site.iugaza.edu.ps/ymadhoun/files/2016/09/Lecture-15.pdf · 2017-10-23 · Chapter 5: The Normal Distribution (Cont’d) Section 5.3:

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun Page 1

Chapter 5: The Normal Distribution (Cont’d)

Section 5.3: Approximating Distributions with the

Normal Distribution

Problem (01):

There is a probability of 0.6 that an oyster produces a pearl with a diameter

of at least 4 mm, which is consequently of a commercial size.

(a) How many oysters does an oyster farmer need to farm in order to be

99% confident of having at least 1000 pearls of commercial value?

(b) Assume that pearl has an expected diameter of 5 mm and a variance

of 8.33. If a farmer collected 1050 pearls, compute the interval over

which a farmer can be 99.7% confident of having an average pearl

diameter lying within this interval?

(Example 30 in textbook)

Solution:

σ

P(X ≥ 1000) ≅ P(Y ≥ 999.5) ≥ 0.9

1 − Φ(999.5 − 0.6𝑛

√0.24𝑛) ≥ 0.9

Φ(999.5 − 0.6𝑛

√0.24𝑛) ≤ 0.1

999.5 − 0.6𝑛

√0.24𝑛≤ −2.33

𝑛 ≤ 1746.8

𝑛 = 1750



σ

μav = 5.0

σav2 =

σ2

n=8.33

1050= 0.00793

[ μ − c × σ , μ + c × σ ] = [ 5 − 3 × √0.00793 ,5 + 3 × √0.00793 ]

= [ 4.73 , 5.26 ]

Problem (02): Calculate the following probabilities both exactly and by using a normal

approximation:

(a) P(X ≥ 8) where X ∼ B(10, 0.7)

(b) P(2 ≤ X ≤ 7) where X ∼ B(15, 0.3)

(c) P(X ≤ 4) where X ∼ B(9, 0.4)

(d) P(8 ≤ X ≤ 11) where X ∼ B(14, 0.6)

(Problem 5.3.1 in textbook)

Solution:

≥ ∼

P(Z ≥8 − 10

√0.7) = 1.0 − P (Z ≤

8 − 10

√0.7)

= 1.0 − P(Z ≤ −2.3905)

= 1.0 − Φ(−2.3905)

= 1.0 − 0.6177

= 0.3823

×

σ × ×



P(Z ≥8 − 0.5 − 7

√2.1) = 1.0 − P (Z ≤

8 − 0.5 − 7

√2.1)

= 1.0 − P(Z ≤ 0.3450)

= 1.0 − Φ(0.3450)

= 1.0 − 0.9635

= 0.3650

≤ ≤ ∼

P(2 − 15

√0.3≤ Z ≤

7 − 15

√0.3) = P (Z ≤

7 − 15

√0.3) − 𝑃 (𝑍 ≤

2 − 15

√0.3)

= P(Z ≤ −14.6059) − P(Z ≤ −23.7346)

= Φ(−14.6059) − Φ(−23.7346)

= 0.9147

×

σ × ×

P(2 − 0.5 − 4.5

√3.15≤ Z ≤

7 + 0.5 − 4.5

√3.15)

= P (Z ≤7 + 0.5 − 4.5

√3.15) − P (Z ≤

2 − 0.5 − 4.5

√3.15)

= P(Z ≤ 1.7321) − 𝑃(𝑍 ≤ −1.1547)

= Φ(1.7321) − Φ(−1.1547)

= 0.9090

≤ ∼

P(Z ≤4 − 9

√0.4) = P(Z ≤ −7.9057)

= Φ(−7.9057)

= 0.7334

×



σ × ×

P(Z ≤4 + 0.5 − 3.6

√2.16) = P(Z ≤ 0.3450)

= Φ(0.3450)

= 0.6364

≤ ≤ ∼

P(8 − 14

√0.6≤ Z ≤

11 − 14

√0.6) = P (Z ≤

11 − 14

√0.6) − 𝑃 (𝑍 ≤

8 − 14

√0.6)

= P(Z ≤ −3.8730) − P(Z ≤ −7.7460)

= Φ(−3.8730) − Φ(−7.7460)

= 0.6527

×

σ × ×

P(8 − 0.5 − 8.4

√3.36≤ Z ≤

11 + 0.5 − 8.4

√3.36)

= P (Z ≤11 + 0.5 − 8.4

√3.36) − P (Z ≤

8 − 0.5 − 8.4

√3.36)

= P(Z ≤ 1.7898) − 𝑃(𝑍 ≤ 0.0577)

= Φ(1.7898) − Φ(0.0577)

= 0.6429

Problem (03): Suppose that a fair die is rolled 1000 times.

(a) Estimate the probability that the number of 6s is between 150 and

180.

(b) What is the smallest value of n for which there is a probability of at

least 99% of obtaining at least 50 6s in n rolls of a fair die?


Solution:



×

σ × ×

P(150 − 0.5 − 166.67

√138.89≤ Z ≤

180 + 0.5 − 166.67

√138.89)

= P (Z ≤180 + 0.5 − 166.67

√138.89)

− P(Z ≤150 − 0.5 − 166.67

√138.89)

= P(Z ≤ −1.4616) − 𝑃(𝑍 ≤ 1.0922)

= Φ(−1.4616) − Φ(1.0922)

= 0.8072

×

σ × ×

P

(

Z ≥50 − 0.5 −

n6

√5𝑛36 )

≥ 0.99

1.0 − P

(

Z ≤50 − 0.5 −

n6

√5𝑛36 )

≥ 0.99

1.0 − Φ(50 − 0.5 − n/6

√5𝑛/36) ≥ 0.99

Φ(50 − 0.5 − n/6

√5𝑛/36) ≤ 0.01

n ≥ 402



Problem (04): The number of cracks in a ceramic tile has a Poisson distribution with

parameter λ = 2.4.

(a) How would you approximate the distribution of the total number of

cracks in 500 ceramic tiles?

(b) Estimate the probability that there are more than 1250 cracks in 500

ceramic tiles.


Solution:

∼

σ λ

λ

λ

𝜆 =2.4 × 500.0

1.0= 1200.0

∼

σ λ

∼

P(X ≥ 1250) = 1.0 − P(X ≤ 1250)

= 1.0 − P (Z ≤1250 − 1200

√1200)

= 1.0 − P(Z ≤ 1.44)

= 1.0 − Φ(1.44)

= 1.0 − 0.9251

= 0.0749



Problem (05): A multiple-choice test consists of a series of questions, each with four

possible answers.

(a) If there are 60 questions, estimate the probability that a student who

guesses blindly at each question will get at least 30 questions right.

(b) How many questions are needed in order to be 99% confident that a

student who guesses blindly at each question scores no more than

35% on the test?


Solution:

×

σ × ×

P(Z ≥30 − 0.5 − 15

√11.25) = 1.0 − P (Z ≤

30 − 0.5 − 15

√11.25)

= 1.0 − P(Z ≤ 4.3719)

= 1.0 − Φ(4.3719)

= 1.0 − 1.0

= 0.0

≤ ≥

×

σ × ×

P

(

Z ≤0.35𝑛 + 0.5 − 𝑛/4

√3𝑛16 )

≥ 0.99

Φ

(

0.35𝑛 + 0.5 − 𝑛/4

√3𝑛16 )

≥ 0.99

n ≥ 92



Exam Problems (applications on chapter 5 as a whole):

Problem (01): Adults salmon fish have lengths that are normally distributed with a mean

of µ = 70 cm and a standard deviation of σ = 5.4 cm.

(a) (5 points) What is the probability that an adult salmon fish is longer

than 80 cm?

(b) (5 points) What is the value of "c" for which there is 90% probability

that an adult salmon has length within [70 – c , 70 + c]?

(c) (5 points) If you go fishing with a friend, what is the probability that

the first adult salmon you catch is longer than the first adult salmon

your friend catches?

(d) (5 points) What is the probability that the average length of the first

two adult salmon fish you catch is at least 10 cm longer than the first

salmon fish your friend catches?

(e) (5 points) What is the probability that 15 fishes selected at random

have an average length less than 72 cm?

(Question 2: (25 points) in Final Exam 2010)

Solution:

P (X ≥ 80) = 1.0 − P (X ≤ 80)

= 1.0 − P (Z ≤80 − 70

5.4)

= 1.0 − P(Z ≤ 1.85)

= 1.0 − Φ(185)

= 1.0 − 0.9678

= 0.0322

P(70 − c ≤ X ≤ 70 + c) = 0.9

P ((70 − c) − 70

5.4≤ X ≤

(70 + c) − 70

5.4) = 0.9

P(−0.185c ≤ X ≤ 0.185c) = 0.9

Φ(0.185c) − Φ(−0.185c) = 0.9 ………𝑒𝑞𝑛. 1

Φ(x) + Φ(−x) = 1.0



Φ(0.185c) + Φ(−0.185c) = 1.0

Φ(−0.185c) = 1.0 − Φ(0.185c) ………𝑒𝑞𝑛. 2

Φ(0.185c) − [1.0 − Φ(0.185c)] = 0.9

Φ(0.185c) − 1.0 + Φ(0.185c) = 0.9

2Φ(0.185c) = 1.9

Φ(0.185c) =1.9

2

Φ(0.185c) = 0.95

0.185c = 1.645

c = 8.883

P(70 − c ≤ X ≤ 70 + c) = 0.9

μY = a1μX1 + a2μX2 = 1.0 × 70 − 1.0 × 70 = 0.0

σY2 = a1

2 × σY2 + a2

2 × σY2 = (1.0)2 × (5.4)2 + (−1.0)2 × (5.4)2

= 58.32

P(Y ≥ 0) = 1.0 − P(Y ≤ 0)

= 1.0 − P (𝑍 ≤0 − 0

√58.32) = 1 − P(Z ≤ 0)

= 1.0 − Φ(0)

= 1.0 − 0.5

= 0.5



X̅ =X1 − X2

2

Y = X̅ − X3

μX̅ =70 + 70

2= 70

σX̅2 =

σ2

n=(5.4)2

2= 14.58

μY = a1μX1 + a2μX2 = 1.0 × 70 − 1.0 × 70 = 0.0

σY2 = a1

2 × σY2 + a2

2 × σY2 = (1.0)2 × (14.58) + (−1.0)2 × (5.4)2

= 43.74

P(Y ≥ 10) = 1.0 − P(Y ≤ 10)

= 1.0 − P (𝑍 ≤10 − 0.0

√43.74)

= 1.0 − P(Z ≤ 1.51)

= 1.0 − Φ(1.51)

= 1.0 − 0.9345

= 0.0655

Y = X̅

μY =15 × 70

15= 70

σY2 =

σ2

n=(5.4)2

15= 1.944

P(Y ≤ 72) = P (𝑍 ≤72 − 70

√1.944)



= P(Z ≤ 1.434)

= Φ(1.434)

= 0.9236

Problem (02): Adults salmon fish have lengths that are normally distributed with a mean

of µ = 70 cm and a standard deviation of σ = 5.4 cm.

(a) (5 points) Suppose that a restaurant accepts only fish with length

from 60 cm to 83 cm. What is the expectation of the proportion of

fish that will not be accepted by the restaurant?

(b) (5 points) If a fisherman brings 150 fish to the restaurant, what are

the expected number of fish that will not be accepted by the

restaurant.

(c) (5 points) For "b", what is the probability that the number of fish

rejected by the restaurant is less than four fish?


Solution:

σ

P(60 ≤ X ≤ 83) = P(60 − 70

5.4≤ Z ≤

83 − 70

5.4)

= P(−1.85 ≤ Z ≤ 2.4)

= Φ(2.4) − Φ(−1.85)

= 0.9920 − 0.0322

= 0.9598



×

×

σ ×

P(×≤ 4) ≃ P(×≤ 4.5)

= P (Z ≤4.5 − 6.03

√5.784)

= P(Z ≤ −0.6361)

= Φ(−0.6842)

= 0.2466

Problem (03): A multiple-choice exam has 200 questions, each with four possible

answers, of which only one is the correct answer. Suppose that a student

guesses on each question.

(a) (5 points) What is the probability that a student will manage to obtain

38 to 40 correct answers?



(b) (5 points) What is the probability that a student will manage to obtain

at least 42 correct answers?

(c) (additional) In how many ways can a student check off one answer

to each question?

(d) (5 points) In how many ways can a student check off one answer to

each question and get all the answers wrong?


Solution:

×

×

×

σ ×

P(38 ≤×≤ 40) = P (37.5 − 50

√37.5≤ Z ≤

40.5 − 50

√37.5)

= P(−2.04 ≤ Z ≤ −1.55)

= Φ(−1.55) − Φ(−2.04)

= 0.06615 − 0.0207

= 0.04545

P(×≥ 42) = 1.0 − P(×≤ 42)

= 1.0 − P (Z ≤42.5 − 50

√37.5)



= 1.0 − P(Z ≤ −1.22)

= 1.0 − Φ(−1.22)

= 1.0 − 0.1112

= 0.8888

Problem (04): The probability that an electrical equipment will fail in less than 1000 hours

of continuous use is 0.25. Estimate that the probability that among 200 such

components:

(a) fewer than 45 will fail in less than 1000 hours of continuous use?

(b) between 20 and 100 will fail in less than 1000 hours of continuous

use?

(Question 2: in Midterm Exam 2005)

Solution:

×

×

×

×



σ ×

P(×≤ 45) = P (Z ≤45.5 − 50

√37.5)

= P(Z ≤ −0.73)

= Φ(−0.73)

= 0.2327

P(20 ≤×≤ 100) = P (100.5 − 50

√37.5≤ Z ≤

19.5 − 50

√37.5)

= P(−4.48 ≤ Z ≤ 8.24)

= Φ(8.24) − Φ(−4.48)

Problem (05): The thickness of glass sheets produced by certain process are normally

distributed with a mean of µ = 3 mm and a standard deviation of σ = 0.12

mm.

(a) What is the probability that a glass sheet is thinner than 2.3 mm?

(b) What is the value of "c" for which there is a 99% probability that a

glass sheet has a thickness within the interval [3.0 - c , 3.0 + c]?

(c) What is the probability that three glass sheets placed one on top of

another have a total thickness greater than 9.5 mm?

(d) What is the probability that seven glass sheets have an average

thickness less than 3.1 mm?

(e) What is the smallest number of glass sheets required in order for

their average thickness to be between 2.95 and 3.05 mm with a

probability of at least 99.7%?

(Question 2: in Midterm Exam 2005 / Final Exam 2012)

Solution:

≤



𝑃(𝑋 ≤ 2.3) = 1.0 − 𝑃 (𝑍 ≤2.3 − 3.0

0.12)

= 1.0 − 𝑃 (𝑍 ≤2.3 − 3.0

0.12)

= 1.0 − 𝛷(−5.8333)

− ≤ ≤

σ × ×

× × ×

σ σ σ σ

P(Y ≥ 9.5) = 1.0 − P(Y ≤ 9.5)

= 1.0 − P (𝑍 ≤9.5 − 9

√0.0432)

= 1.0 − P(Z ≤ 2.4) = 1.0 − Φ(2.4) = 1.0 − 0.9918

= 0.0082

μ = 3.0

σ2 =(0.12)2

7= 0.00205

P(X ≤ 3.1) = P (Z ≤3.1 − 3

√0.00205)

= Φ(2.2)

= 0.9861



𝑉𝑎𝑟(𝑋) = σ2 =(0.12)2

𝑛=0.0144

𝑛

σ = √𝑉𝑎𝑟(𝑋) = √(0.12)2

𝑛=0.12

√𝑛

P(2.95 ≤ Z ≤ 3.05) ≥ 0.997

P(2.95 − 3.0

0.12

√𝑛

≤ Z ≤3.05 − 3.0

0.12

√𝑛

) ≥ 0.997

ϕ(3.05 − 3.0

0.12

√𝑛

) − ϕ(2.95 − 3.0

0.12

√𝑛

) ≥ 0.997

ϕ(0.4167√𝑛) − ϕ(−0.4167√𝑛) ≥ 0.997

ϕ(0.4167√𝑛) − [1 − ϕ(0.4167√𝑛) ≥ 0.997

ϕ(0.4167√𝑛) − 1 + ϕ(0.4167√𝑛) ≥ 0.997

2ϕ(0.4167√𝑛) ≥ 1.997

ϕ(0.4167√𝑛) ≥1.997

2ϕ(0.4167√𝑛) ≥ 0.9985

0.4167√𝑛 ≥ 2.965

𝑛 ≥ 51

Problem (06): The probability that cylindrical concrete specimen strength exceeds 2000

N/cm2 is 15%. The coefficient of variation of these strengths is 20%.

Assuming a normal distribution, calculate:

(a) The mean of the strength.

(b) The probability that the strength will be between 1300 and 1900

N/cm2.




Solution:

𝐶𝑉 =𝜎

𝜇× 100

20 =𝜎

𝜇× 100

𝜎 = 0.2 𝜇………𝐸𝑞𝑛. (1)

𝑃(𝑋 > 2000) = 1.0 − 𝑃(𝑋 ≤ 2000)0.15 = 1.0 − 𝑃(𝑋 ≤ 2000)𝑃(𝑋 ≤ 2000) = 1 − 0.15

𝑃(𝑍 ≤ 𝑧) = 0.85𝑧 = 1.037𝑥 = 𝜇 + 𝜎𝑧2000 = 𝜇 + 𝜎 × 1.037………𝐸𝑞𝑛. (2)

𝜇 = 1656.4519 𝑁/𝑐𝑚2

𝜎 = 331 𝑁/𝑐𝑚2

𝑃(1300 ≤ 𝑋 ≤ 1900) = 𝑃 (1300 − 1656.5

331≤ 𝑍 ≤

1900 − 1656.5

331)

= 𝑃(−1.0769 ≤ 𝑍 ≤ 0.7358)

= 𝑃(𝑍 ≤ 0.7358) − 𝑃(𝑍 ≤ −1.0769)

= ϕ(0.7358) − ϕ(−1.0769)

= 0.7689 − 0.1412 = 0.6277

Documents

Chapter 5: The Normal Distribution (Cont’d)site.iugaza.edu.ps/ymadhoun/files/2016/09/Lecture-15.pdf · 2017-10-23 · Chapter 5: The Normal Distribution (Cont’d) Section 5.3: