find Quartiles (q1, q2 and q3) from the given example. Please solve correct and quickly. Example 2.3 Given below are the mean aunal death rates per 1,000 at ages 20-65 in each of 88occupational groups. Construct a grouped7.5826.29.910.910.87.449.711.612.610.29.97.37.38.41210.310.16.512.57.86.58.79.312.418hitps:#star9.3.10.49.19.36.210.36.67.48.67.79.4237.712.88.75.58.69.611.910.47.87.612.14.614.08.111.410.611.610.48.14.66.612.8-6.8716.68.88.810.710.86.07.97.39.39.38.910.13.96.06.99.08.89.411.410.9(B.I.S.E. Lahore, 1971)A scan of the data shows that the largest value is 14.0 and the smallest value is 3.9 so that the range140-3.9 = 10.1.As the data are recorded to one decimal place, we may therefore locate the lower limit of the firstgroup at 3.5. Let us choose a class interval of 1.0. Then the class limits are specified as 3.5 - 4.4,... amai.puTie ..https://stat9943.blogspot.com24INTRODUCTION TO STATISTICAL THEORY4.5-5.4, 5.5- 6.4, ... With this choice, the class-boundaries are 3.45 - 4.45, 4.45 - 5.45, 5.45-6.45, .,which do not coincide with the given values.The following table shows the required frequency distribution:FREQUENCY DISTRIBUTION OF MEAN DEATH RATESDeath RatesClass-boundariesTallyFrequency3.5-4.43.45-4.4514.5 - 5.44.45 - 5.454.5.5 - 6.45.45 - 6.456.5 - 7.4NA THA II6.45 - 7.45137.5 - 8.47.45-8.45128.5-9.48.45-9.45199.5 – 10.49.45 - 10.451310.5-11.410.45 - 11.451011.5 - 12.411.45 12.456.12.5 - 13.412.45-13.454.13.5- 14.413.45- 14.451Total88943.blogspot.com

Given data distribution of 88 points as shown below The total number of data points =88Its frequency…

Answered: 10.1 10.0ri.i Example 2.3 Given below…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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I need help with just part c. Someone on here already answered it for me but I'm really confused as there are steps missing and I'm not sure what to put in the calculator. " The first quartile separates the lowest 25% of the observations from the higher 75% of the observations. The first quartile is denoted by Q1. The second quartile separates the lower 50% of the observations from the higher 50% of the observations. The second quartile is denoted by Q2.The second quartile is same as median. The third quartile separates the lowest 75% of the observations from the higher 25% of the observations. The third quartile is denoted by Q3. The third quartile can obtained as follows...." Third Quartile is 1 somehow, and I don't understand how it is.

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