Find the median of the following data :FrequencyClass intervalClass intervalFrequency7:5-10:51222'5-25'51410 5-13 51025 5-28'5313'5-16'51528:5-31'516'5-19 519 5-22 51931'5-34'541234'5-37'51

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Answered: Find the median of the following data :…

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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Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the frequency distribution below.   Lengths (mm)      Frequency    180 - 184 1 185 - 189 16 190 - 194 71 195 - 199 108 200 - 204 83 205 - 209 18 210 - 214 3 What is the lower class boundary for the first class?class boundary =
Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.   Lengths (mm)      Frequency    160 - 169 1 170 - 179 16 180 - 189 71 190 - 199 108 200 - 209 83 210 - 219 18 220 - 229 3 What is the class midpoint for the sixth class?class midpoint =
Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.   Lengths (mm)      Frequency    220 - 229 1 230 - 239 16 240 - 249 71 250 - 259 108 260 - 269 83 270 - 279 18 280 - 289 3 What is the upper class limit for the seventh class?
435 507 448 435 463 440 448 413 432 458 473 465 428 472 439 a. Summarize the data on femur lengths in a frequency table. Use class intervals that start at 400 and have width 20. b. Add a column to your table from part (a) for the relative frequencies. c. Draw a histogram that represents your frequency table. (Use either frequency or relative frequency for the vertical axis.)
Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below. Lengths (mm) Frequency 140 - 159 1 160 - 179 16 180 - 199 71 200 - 219 108 220 - 239 83 240 - 259 18 260 - 279 3 What is the lower class boundary for the first class?class boundary =
The boxplot shown below results from the heights (cm) of males listed in a data set. What do the numbers in that boxplot tell us? 151 172.5 192 168.5 182.6 A boxplot consists of a box that extends from 168.5 to 182.6, a vertical line segment drawn through the box at 172.5, and a horizontal line segment extending from a vertical line segment at 151 to a vertical line segment at 192 that bisects the box. Question content area bottom Part 1 The minimum height is enter your response here cm, the first quartile Q1 is enter your response here cm, the second quartile Q2 (or the median) is enter your response here cm, the third quartile Q3 is enter your response here cm, and the maximum height is enter your response here cm. (Type integers or decimals. Do not round.)
The following is a list of P/E ratios (current stock price divided by company's earnings per share) for 15 companies. 57, 53, 50, 46, 42, 35, 31, 56, 52, 37, 37, 34, 34, 30, 30 Send data to calculator Draw the histogram for these data using an initial class boundary of 29.5 and a class width of 9. Note that you can add or remove classes from the figure. Label each class with its endpoints. Frequency 9 8 7- 6+ S43N- 5 3+ 2- 0 I 0 I P/E ratio 0 " 0 0:00 00:0 × Ś
According to the table below How do you construct of frequency distribution using 4 classes, a width of 2 hours, and a lower limit of 3 for class 1?
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter. Less than Cumulative frequency of the data obtained is represented in the following table. Length (mm) |Less than cum freq 118-126 3 127-135 136-144 17 145-153 29 154-162 34 163-171 38 172-180 40 If you have to find the median of the distribution, what will be value of cf in the formula? O a 17 O b.29 Oc 12 Od. 20

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Find the median of the following data : Frequency Class interval Class interval Frequency 7:5-10:5 12 22'5-25'5 14 10 5-13 5 10 25 5-28'5 3 13'5-16'5 15 28:5-31'5 16'5-19 5 19 5-22 5 19 31'5-34'5 4 12 34'5-37 5 1