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 "Observations on the Faceted Initiation Site in the Dwell-Fatigue Tested Ti-6242 Alloy: Crystallographic Orientation and Size Effects"

Class: 0-<5 5-<10 10-<15 15-<20
Rel Fraq: .177 .166 .175 .136

Class: 20-<30 30-<40 40-<60 60-<90
Rel Fraq: .194 .078 .044 .030

A) Is it true that more that 50% of the sampled angles are smaller than 15 degrees, as asserted in the paper?

B) What proportion of the sampled angles are at least 30 degrees?

C) Roughly what proportion of angles are between 10 degrees and 25 degrees?

D) Construct a histogram and comment on any interesting features.

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

(B) The proportion of the sampled angles are at least 30 degrees is 15.2%.

(C) The probability of angles between 10° and 25° is less than 50%.

(D) The data is right skewed.

Step-by-step explanation:

Consider the frequency distribution table below.

(A)

Compute the probability of angles less than 15° as follows:

P (X < 15°) = P (0° ≤ X ≤ 5°) + P (5° ≤ X ≤ 10°) + P (10° ≤ X ≤ 15°)

                 = 0.177 + 0.166 + 0.175

                 = 0.518

                 ≈ 51.8%

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

(B)

Compute the probability of angles at least 30° as follows:

P (X ≥ 30°) = P (30° ≤ X ≤ 40°) + P (40° ≤ X ≤ 60°) + P (60° ≤ X ≤ 90°)

                 = 0.078 + 0.044 + 0.030

                 = 0.152

                 ≈ 15.2%

Thus, the proportion of the sampled angles are at least 30 degrees is 15.2%.

(C)

Compute the probability of angles between 10° and 20° as follows:

P (10° ≤ X ≤ 20°) = P (10° ≤ X ≤ 15°) + P (15° ≤ X ≤ 20°)

                          = 0.175 + 0.136

                          = 0.311

Compute the probability of angles between 10° and 30° as follows:

P (10° ≤ X ≤ 300°) = P (10° ≤ X ≤ 15°) + P (15° ≤ X ≤ 20°) + P (20° ≤ X ≤ 30°)

                            = 0.175 + 0.136 + 0.194

                            = 0.505

Then the probability of angles between 10° and 25° is less than 50%.

(D)

The histogram is shown below.

From the histogram it can be seen that the graph has a long tail towards the right or most of the observations are accumulated towards the left of the graph.

This implies that the data is positively skewed.

For a positively skewed data the Mean > Median > Mode.

step-by-step explanation: