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Given the frequency distribution, compute for each quantile and match it with the letter code of its corresponding value to find the hidden word below. P60 Q1₁ D4 Q2 D7 Q₂ D4 P75 I. 48.5 L. 65 N. 66 O. 45.17 P. 38.83 S. 55.5 T. 63.5 Daily Allowance of 50 Grade 10 Students Amount in Number of Peso Students 11-20 2 21-30 8 31-40 3 41-50 15 51-60 4 61-70 10 71-80 1 81-90 2 91-100 5

Given the frequency distribution, compute for each quantile and match it with the letter code of its corresponding value to find the hidden word below. P60 Q1₁ D4 Q2 D7 Q₂ D4 P75 I. 48.5 L. 65 N. 66 O. 45.17 P. 38.83 S. 55.5 T. 63.5 Daily Allowance of 50 Grade 10 Students Amount in Number of Peso Students 11-20 2 21-30 8 31-40 3 41-50 15 51-60 4 61-70 10 71-80 1 81-90 2 91-100 5

MATLAB: An Introduction with Applications
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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PLEASE ANSWER Q2 D7 P75 in the format same with the 2nd photo

Given the frequency distribution, compute for each quantile and match it with the letter code of its corresponding value to find the hidden word below. Q1 D4 P60 Q2 D7 Q2 D4 P75 I. 48.5 L. 65 N. 66 O. 45.17 P. 38.83 S. 55.5 T. 63.5 Daily Allowance of 50 Grade 10 Students Amount in Number of Peso Students 11-20 2 21-30 8 31-40 3 41 - 50 15 51-60 4 61-70 10 71-80 1 81-90 2 91-100 5
Transcribed Image Text:Given the frequency distribution, compute for each quantile and match it with the letter code of its corresponding value to find the hidden word below. Q1 D4 P60 Q2 D7 Q2 D4 P75 I. 48.5 L. 65 N. 66 O. 45.17 P. 38.83 S. 55.5 T. 63.5 Daily Allowance of 50 Grade 10 Students Amount in Number of Peso Students 11-20 2 21-30 8 31-40 3 41 - 50 15 51-60 4 61-70 10 71-80 1 81-90 2 91-100 5
2₁ class: Class with (1) value of the observation in cf column 50th value of the observation in of column = (12.5) value of the observation in of column and it lies in the class 31 - 40. : Q₁ class : 30.5 - 40.5 The lower boundary point of 30.5 - 40.5 is 30.5. :. L = 30.5 PL 1-2-3-1 10 7-of 2₁=1+ 12.5-10 = 30.5 + 3 2.5 = 30.5+ 10 3 = 30.5 +8.3333 = = 38.8333 || . 10
Transcribed Image Text:2₁ class: Class with (1) value of the observation in cf column 50th value of the observation in of column = (12.5) value of the observation in of column and it lies in the class 31 - 40. : Q₁ class : 30.5 - 40.5 The lower boundary point of 30.5 - 40.5 is 30.5. :. L = 30.5 PL 1-2-3-1 10 7-of 2₁=1+ 12.5-10 = 30.5 + 3 2.5 = 30.5+ 10 3 = 30.5 +8.3333 = = 38.8333 || . 10
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