The data on the major and home state for a randomly chosen group ofstudents at a private college are: a11 =80, a12 =204, a13 -68, a₁4 =68, q21 =67, a22 =213, α23=66, α24 =104, α31 =330,a34 =283, a41 =127, a42=183,=67,=76,a32a33a4366, α44 =134The following contingency table was created based on the above data,where you will need to calculate the grand total, the total for each row, andthe total for each column on your own before answering the question thatappears after the table.BiologyPennsylvania 80New Jersey67New York330127MarylandTotalAccounting20421367183History68667666Education Total68104283134Calculate the probability of a student majoring in education, given that thestudent's home state is Maryland (For answers up to four decimal places,enter the number in the box below without rounding it. For answers that havemore than four decimal values, round your answer to four decimal places).

The table shows the data of student majoring and their state.

Answered: The data on the major and home state…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A nutritionist working for the United States Department of Agriculture (USDA) randomly selected three cartons of eggs from all of the available cartons of standard large eggs at a neighborhood grocery store. Each egg in the randomly selected cartons had their nutritional content analyzed.The data provided here are the amounts of milligrams of cholesterol in each of the sampled eggs. 186, 188, 179, 180, 192, 186, 183, 177, 184, 178, 191, 174,189, 176, 190, 188, 196, 187, 184, 184, 192, 194, 198, 183,183, 181, 187, 190, 186, 176, 183, 185, 191, 180, 184, 182 Test the validity of the claim, at the 0.01 level of significance, that the mean amount of cholesterol in a standard large egg is under 190 mg.
The World Bank collects information on the life expectancy of a person in each country ("Life expectancy at," 2013) and the fertility rate per woman in the country ("Fertility rate," 2013). The data for 12 randomly selected countries for the year 2011 are shown in the table below. Table: Data of Fertility Rates versus Life Expectancy Fertility Rate Life Expectancy 1.7 77 5.8 55 2.2 70 2.1 76 1.8 75 2.0 78 2.6 73 1.5 81 6.9 54 2.9 72 4.7 63 6.8 57 The most commonly used type of correlation is Pearson correlation, named after Karl Pearson, introduced around the turn of the 20th century. Pearson's r measures the linear relationship between two variables, say X and Y. Find the Pearson’s correlation coefficient for the above data In reference to the calculated correlation coefficient above, describe the strength of the correlation. Briefly give an explanation to meaning of the above results. Obtain the…
About 2% of the population has a particular genetic mutation. 900 people are randomly selected. Find the mean for the number of people with the genetic mutation in such groups of 900.
A college student is interested in whether presoaking time has an effect on cleanliness of clothing. He collects the dirty clothes from all of the people in his dorm and randomly assigns clothing to one of four presoak time conditions: 5 minutes, 10 minutes, 15 minutes, 20 minutes. He then asks members of another dorm to rate the cleanliness of the items (they don’t know which clothing got which soaking time). The data are presented below. C) Perform an LSD test and answer the questions that follow (SHOW ALL WORK!!! This can be on the same page as your other ANOVA hand work. Make sure it is inserted and/or uploaded): df error = tcrit (df error) MS error = s = LSD = _____________ Mean Comparisons: Mean5 min vs. Mean10 min = | ___ __ | = ___ ___ Greater than (>) LSD value? __ ___ Significant?__ ___ Mean10 min vs. Mean15 min = |__________________| = ______ Greater…
A healthcare provider monitors the number of CAT scans perfomed each month in each of its clinics. The most recent year of data for a particular clinics follows (the reported variable is the number of CAT scans each month expressed as the number of CAT scans per thousand members of the health plan): 2.31, 2.09, 2.36, 1.95, 1.98, 2.25, 2.16, 2.07, 1.88, 1.94, 1.97, 2.02. Find a two-sided 95% confidence interval for the standard deviation. What should you do to address any reservations about this confidence interval? Round your answers to two decimal places (e.g. 98.76).
We are interested in estimating the proportion of people that can identify Nebraska but not Vermont. In a class of 55 students, each student was requested to select a sample of 6 people, show each person a map of the US and ask the person to identify the states of Vermont and Nebraska. The following numbers are the counts that each student got, that is, each number is the count of people in the student sample of 6, that could identify Nebraska but not Vermont.3344010110020030002020001100002010000001000200000001010Note: You can cut and past these numbers into an excel spreadsheet to help you calculate the mean and the standard deviation and compute the confidence interval. If you have trouble pasting them, paste them to a text file first and then copy from there.When we collect all the students' data, we have a sample size of 6x55=330.Of the sample of 330 people, what proportion were able to locate Nebraska but not Vermont? (round answer to 3 decimals) Find a 95% confidence interval for…
About 5% of the population has a particular genetic mutation. 600 people are randomly selected.Find the mean for the number of people with the genetic mutation in such groups of 600.
The data gathered from a survey of a random group of people about their smoking habits and their bodily weights are: a11 =159, a12 =203, a13 =10, a21 =285, a22 =500, a23 =38, a31 214, a32 -300, a33 =28, a41 =113, a42 =189, a43 =30, a51 =467, a52 1,969, a53 71 The following contingency table was created based on the above data, where you will need to calculate the grand total, the total for each row, and the total for each column on your own before answering the question that appears after the table. Very heavy Heavy Moderate Light SMOKING STATUS Non-Smoker Total BODILY Normal 159 285 214 113 467 WEIGHT Overweight 203 500 300 189 1,969 - STATUS Underweight Total 10 38 28 30 71 Calculate the probability of someone who has normal weight, given that the individual is non- smoker (For answers up to four decimal places, enter the number in the box below without rounding it. For answers that have more than four decimal values, round your answer to four decimal places).
A manufacturing plant makes computer chips 24 hours a day. Employees work one of three, 8 hour shifts during the morning (2am-10am), day (10am-6pm), and night (6pm-2am). Below are data on the number of defective chips on a randomly chosen 24 hour period. Shift Morning Day Night Number of Defective Chips 23 26 47 Total Chips Manufactured 330 330 330 The plant manager thinks there might be more defective computer chips manufactured during the night shift compared to the day shift. Give appropriate statistical evidence to test the managers' claim.
IBL Company manager asked his statistic team to give to the production department the data on the number of phones manufactured over the 17 months.. The data collected are as follows: Months 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 REQUIRED # of phones produced 2045 1566 1207 4563 5052 3465 2542 4134 3567 5163 4563 7453 2621 4772 4563 5052 2846 1. Calculate the following using the data provided on the table: a. the mean b. median c. mode 2. Find the Quartile 1, Quartile 3 and inter-quartile of the phones produced 3. Why is the quartile important? What purpose does it serve? 4. Why is median preferred to mean in certain circumstances? 5. How do we call data when Mode >Median >Mean and show with a diagram how such distribution looks? 6. Explain and compare parameters and statistics.

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