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Chapter 1: Q26E (page 28)

Automated electron backscattered diffraction is now being used in the study of fracture phenomena. The following information on misorientation angle (degrees) was extracted from the article 鈥淥bservations on the Faceted Initiation Site in the Dwell-Fatigue Tested Ti-6242 Alloy: Crystallographic Orientation and Size Effects鈥� (Metallurgical and Materials Trans., 2006: 1507鈥�1518).

Class: 0-<5 5-<10 10-<15 15-<20

Relfreq: .177 .166 .175 .136

Class: 20-<30 30-<40 40-<60 60-<90

Relfreq: .194 .078 .044 .030

a. Is it true that more than 50% of the sampled angles are smaller than 15掳, as asserted in the paper?

b. What proportion of the sampled angles are at least 30掳?

Short Answer

Expert verified

a.

Yes, it is true that more than 50% of the sampled angles are smaller than 15掳, as asserted in the paper.

b.

The percentage of the sampled angles that are at least 30 degrees is 0.152.

Step by step solution

01

Given information

The data for the misorientation angle (degrees) is provided.

02

Check the proportion

The relative frequency table is given as,

Class

Relative frequency

0-5

0.177

5-10

0.166

10-15

0.175

15-20

0.136

20-30

0.194

30-40

0.078

40-60

0.044

60-90

0.030

a.

Let x represents the sampled angles.

The proportion of the sampled angles that are smaller than 15 degrees is computed as,

\(\begin{aligned}P\left( {x < 15} \right) &= P\left( {0 - 5} \right) + P\left( {5 - 10} \right) + P\left( {10 - 15} \right)\\ &= 0.177 + 0.166 + 0.175\\ &= 0.518\end{aligned}\)

Therefore, the percentage of the sampled angles that are smaller than 15 degrees is 51.8% which is more than 50%.

Yes, it is true that more than 50% of the sampled angles are smaller than 15掳, as asserted in the paper.

03

Given information

The data for the misorientation angle (degrees) is provided.

04

Compute the proportion

The relative frequency table is given as,

Class

Relative frequency

0-5

0.177

5-10

0.166

10-15

0.175

15-20

0.136

20-30

0.194

30-40

0.078

40-60

0.044

60-90

0.030

b.

Let x represents the sampled angles.

The proportion of the sampled angles that are atleast 30 degrees is computed as,

\(\begin{aligned}P\left( {x \ge 30} \right) &= P\left( {30 - 40} \right) + P\left( {40 - 60} \right) + P\left( {60 - 90} \right)\\ &= 0.078 + 0.044 + 0.030\\ &= 0.152\end{aligned}\)

Therefore, the percentage of the sampled angles that are at least 30 degrees is 0.152.

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Most popular questions from this chapter

If the amount of soft drink that I consume on any given day is independent of consumption on any other day and is normally distributed with\(\mu = 13\)oz and\(\sigma = 2\)and if I currently have two six-packs of\({\bf{16}}\)-oz bottles, what is the probability that I still have some soft drink left at the end of\({\bf{2}}\)weeks?

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Obs

InitSe

InitCont

FinalSe

FinalCont

1

11.4

9.1

138.3

9.3

2

9.6

8.7

104.0

8.8

3

10.1

9.7

96.4

8.8

4

8.5

10.8

89.0

10.1

5

10.3

10.9

88.0

9.6

6

10.6

10.6

103.8

8.6

7

11.8

10.1

147.3

10.4

8

9.8

12.3

97.1

12.4

9

10.9

8.8

172.6

9.3

10

10.3

10.4

146.3

9.5

11

10.2

10.9

99.0

8.4

12

11.4

10.4

122.3

8.7

13

9.2

11.6

103.0

12.5

14

10.6

10.9

117.8

9.1

15

10.8

121.5

16

8.2

93.0

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b. Obtain an expression for the \({\rm{(100p)th}}\) percentile. What is the value of \({\rm{\tilde \mu }}\)?

c. Compute \({\rm{E(X)}}\) and \({\rm{V(X)}}\).

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