Calculate the Standard deviation of the grouped data shown in the frequency tablebelow:Classesfrequency50 – 59360 – 69570 – 79980 – 8912-90 – 998

Class Frequency (f) Mid value (x) f⋅x f⋅x2 50-59 3 54.5 163.5 8910.75 60-69 5 64.5…

Answered: Calculate the Standard deviation of the…

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.6: Summarizing Categorical Data

Problem 33PPS

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The inner diameters of washers produced by a company can be measured to the nearest thousandth of an inch. If the class marks of a frequency distribution of these diameters are given in inches by 0.587,0.602 0.617, 0.632, and 0.647, the class size is Blank 1
Give a brief analysis on the following Chi square for cross 1 (Wild type female × white eyes vestigial wings male) (1st image) and the chi square 2 (White eyes vestigial female × wild type Male)(2nd image). The chi-square value for cross1 is 7.259 and for cross 2 is 59.301.
Given the stem-and-leaf diagram. Answer the following questions: • The total frequency all stems Observations of the stem that has largest frequency Stem-and-leaf of C1 N - 82 Leaf Unit - 0.10 1 4 84 33 4 85 86 777 13 87 456789 24 88 89 23334556679 0233678899 90 0111344456789 0001112256688 22236777 023347 (13) 35 91 22 92 14 93 8 94 2247 95 4 96 97 15 2 2 98 8 1. 99 1 100 3 Q1 88.575 Median Q3 90.400 92.200 a. Total frequency = 18 Observations on the largest frequency: 90.0, 90.1, 90.1, 90.1, 90.3, 90.4, 90.4, 90.4, 90.5, 90.6, 90.7, 90.8, 90.9 b. Total frequency cannot be determined by the present diagram Observations on the largest frequency: 90.0, 90.1, 90.1, 90.1, 90.3, 90.4, 90.4, 90.4, 90.5, 90.6, 90.7, 90.8, 90.9 O.Total frequency = 82 Observations on the largest frequency: 90.0, 90.1, 90.1, 90.1, 90.3, 90.4, 90.4, 90.4, 90.5, 90.6, 90.7, 90.8, 90.9 O d. Total frequency = 82 Observations on the largest frequency: 87.4, 87.5, 87.6, 87.7, 87.8, 87.9 34 5 678 9012NM 435 2
A company collects data from 10,000 websites about how long it takes to load the site. The number of seconds it takes to fully load the website is summarized in the relative frequency table. seconds to load relative frequency 1.4–1.6 0.0003 1.6–1.8 0.0012 1.8–2.0 0.0053 2.0–2.2 0.0181 2.2–2.4 0.0442 2.4–2.6 0.0910 2.6–2.8 0.1555 2.8–3.0 0.1861 3.0–3.2 0.1938 3.2–3.4 0.1447 3.4–3.6 0.0923 3.6–3.8 0.0447 3.8–4.0 0.0166 4.0–4.2 0.0048 4.2–4.4 0.0012 4.4–4.6 0.0002 The relative frequency histogram summarizes the same data. Expand Image The mean time to load a website is 3 seconds and the standard deviation is 0.4 seconds. Would a normal distribution be a good model for this distribution? Explain your reasoning. What proportion of websites loaded within 1 standard deviation of the mean? What proportion of websites loaded within 2 standard deviations of the mean? What proportion of websites loaded within 1 standard deviation of 2.8 seconds?…
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Thirty AA batteries were tested to determine how long they would last. The results, to the nearest minute, were recorded as follows: Table 3: Life of AA batteries, in minutes Battery life, x (in minutes) Number of battery 360 - 369 2 370 - 379 3 380 - 389 390 - 399 7 400 - 409 410 - 419 4 420 - 429 430 - 439 Based on Table 3: (a) Draw a cumulative frequency curve (ogive) to represent the above data. [Use graph paper] (b) Find the semi inter-quartile range by using the cumulative frequency curve. (c) Estimate the percentage number of battery with the battery life between 400 and 430 minutes. (d) of battery life. Calculate the mean and standard deviation for the number LO 3.
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Calculate the Standard deviation of the grouped data shown in the frequency table below: Classes frequency 50 – 59 3 60 – 69 5 70 – 79 9 80 – 89 12 90 – 99 8