Automated electron backscattered diffraction is now being used in the study of fracture phenomena. The following information is on misorientation angle (degrees). Class: 0−<5 5−<10 10−<15 15−<20 20−<30 30−<40 40−<60 60−<90 Rel freq: 0.176 0.169 0.179 0.139 0.193 0.075 0.040 0.029
A)What proportion of the sampled angles are at least 30°? (Round your answer to three decimal places.) B)Roughly what proportion of angles are between 10° and 25°? (Round your answer to three decimal places.) C)Construct a histogram. Comment on any interesting features. Select all that apply
1.There are two obvious outliers.
2.Without more precise information, we cannot tell if the data contain outliers.
3.The distribution of misorientation angles is approximately symmetric.
4.The distribution of misorientation angles is heavily negatively skewed.
5.The distribution of misorientation angles is heavily positively skewed.
6.The distribution of misorientation angles is bimodal

a) Proportion of the sampled angles are at least 30° = 0.144

b) Proportion of sampled angles between 10 and 25 = 0.415

Step-by-step explanation:

ClassRelative Frequency

0-<5                 0.176

5-<10                0.169

10-<15          0.179

15-<20          0.139

20-<30          0.193

30-<40          0.075

40-<60          0.04

60-<90          0.029

A) Proportion of the sampled angles which are at least 30° can be calculated by adding the relative frequencies which are greater than 30. This will include the classes 30-<40, 40-<60 and 60-<90, So,

proportion of the sampled angles are at least 30° = 0.075 + 0.04 + 0.029

proportion of the sampled angles are at least 30° = 0.144

B) For the proportion of angles between 10 and 25, we need to consider the classes 10-<15, 15-<20 and 20-<30. Since we are interested in the angles up till 25 degrees, we need to divide the relative frequency of the class 20-<30 by 2 because 25 lies in the middle of this class. So,

Proportion of sampled angles between 10 and 25 = 0.179 + 0.139 + (0.193/2)

Proportion of sampled angles between 10 and 25 = 0.415

C) For constructing the histogram, you need to plot the relative frequencies on the y axis and the classes on the x-axis by choosing a suitable scale for both the axes. I have plotted the graph on paint but its a bit untidy. I am attaching the graph as a snippet here so that you can have an idea of the heights.  

By looking at the heights of the classes, we can say that The distribution of misorientation angles is heavily positively skewed because there is a tail at the positive side of the graph. The graph is also unimodal since there is only one highest peak present.

the first one

step-by-step explanation:

just id the test.