Example 14-16. Fit a binomial distribution to the following data: 1 3. 28 62 46 10
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- * h6. In the frequency distribution of 100 families given below, the number of families corresponding to expenditure groups 20-40 and 60-80 are missing from the table. Find these missing frequencies Expenditure (in): 0-20 20-40 40-60 60-80 80-100 No. of families: 14 27 ? 15 However the median is known to be 50.The following graph illustrates 140 120 100 10 20 50 60 80 Multiple Choice О О a distribution for a single variable a positive or direct relationship a negative or inverse relationship no relationshipPLEASE PROVIDE ANSWER WITH EXPLANATION FOR BOTH
- 7 Confidence interval for a population proportion Twenty years ago, a very famous psychologist specializing in marriage counseling authored book detailing the way in which she believed spouses should communicate. She is now interested in the proportion of all couples who bought her book who stayed together. For a random sample of 200 couples who bought her book, she found that 162 of them stayed together. Based on this, compute a 95% confidence interval for the proportion of all couples who bought her book who stayed together. Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.) Lower limit: Upper limit: Explanation W Check 19 # 3 E $ 4 R S 5 T 6 A Y & 7 11 8 Español 9 CO ? As Ⓒ2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility MQ个4 3 of 6 200% 5.) If the number 12.3 were added to the dataset below, which statement best describes how its mean and median would change? 70.1, 72.3, 75.6, 78.9, 81.0, 83.8, 84.5, 91.2 A.) The mean and median both decrease, but the mean decreases more. B.) The mean and median both decrease, but the median decreases more. C.) The median doesn't change, but the mean decreases. D.) The mean doesn't change, but the median decreases. 021 Illuminate EducationTM, Inc. ? XP OCa 29Need help with this Intro to probability and statistics homework problem. Make sure your handwriting is neat and readable.
- arsSHar 32. Find the Geometric mean from the following frequency distribution of the marks: Marks No. of students Marks 1eem No. of students sem 4-8 6. 24-28 12 8-12 10 28-32 10 12-16 18 32-36 6. 16-20 30 36-40 20-24 15Lower Upper f cf rf crf 14 19 10 10 0.2 0.2 20 25 11 21 0.22 0.42 26 31 8 29 0.16 0.58 32 37 6. 35 0.12 0.7 38 43 15 50 0.3 1 The frequency table shown above shows ages from a random sample of 50 people. a. How many people are between 20 and 25 years of age? b. How many people are 25 years old or younger? c. How many people are older than 25? d. What percent of people are 25 years old or younger? Enter a whole number. cor e. What percent of people are between 20 and 25 years of age? Enter a whole number. Submit Question Jump to Answer MacBook Air O00 O00 F4 11 F8 F2 F3 F5 F6 F7 F9 %23 24 & * 3 4. 8. 9- E RI 6Chapter 7: 6. The mean room and board expense per year at a four-year college is $9200. Assume the room and board expenses are normally distributed with a standard deviation of $1200. (6 points each) You randomly select one four-year college. What is the probability that their average room and board is less than $8950? b. You randomly select 50 four-year colleges. What is the probability that the mean average room and board is less than $8950?
- 4. The frequency distribution below gives the cost of production of sugarcane in holdings. Obtain the arithmetic mean. Cost of Production No. of Holdings Cost of Production No. of Holdings 2 - 6 1 18 – 22 52 6 - 10 9. 22 – 26 36 10 – 14 21 26 – 30 19 14 – 18 47 30 - 34 379 The table below shows the weights of 8 different bears at a zoo. Type of Bear Asiatic Black Bear Black Bear Brown Bear Panda Bear Polar Bear Sloth Bear Spectacled Bear Sun Bear Weight (pounds) 225 300 550 200 1,000 300 280 100 If the weight of the polar bear is removed, which statement is true? A The mean decreases more than the median because the polar bear is a high outlier. The mean decreases less than the median because the polar bear is a high outlier.
