The figure below represents the data from Table 3.6 in the book. It summarizes the lengths of hospital stays (days) for a sample 25 patients.Duration of hospitalization (days)1099.8.4.3.2.06-1011-1516-20>2001-05What is the probability that someone,stayed in the hospital 6 days or more?64%36%72%None of these choices are correct.O O 0 O

Answered: The figure below represents the 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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Car speeds, the picture For the car speed data of Exercise18, here is the histogram, boxplot, and Normal probabilityplot of the 100 readings. Do you think it is appropriate toapply a Normal model here? Explain.
The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 88 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 68.74 79.38 70.94 84.71 79.74 85.74 100.91 98.02 LOADING... Click the icon to view the table of critical t-values. (a) Determine a point estimate for the population mean travel tax. A point estimate for the population mean travel tax is $11.51 (Round to two decimal places as needed.) (b) Construct and interpret a 95% confidence interval for the mean tax paid for a three-day business trip. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to two decimal places as needed.) A. One can be nothing% confident that the all cities have a travel tax between $nothing and $nothing.…
Question #5 Part C
You get a random sample of 30 students, and you give them an Anxiety inventory called Form A. Below are the scores for 30 students. What is the Standard Deviation for the 30 Form A scores? 2 3 5 B UD 7 8 9 10 Form A 6 6 9 8 5 6 8 4 6
A random variable follows the normal probability distribution with a mean of 100 and a standard deviation of 10. Determine the probability for a randomly selected value from this population in parts a through d below. Click here to view page 1 of the standard normal probability table. Click here to view page 2 of the standard normal probability table. Question content area bottom Part 1 a. What is the probability that the value is less than 90? The probability that the value is less than 90 is enter your response here. Part 2 b. What is the probability that the value is less than 85? The probability that the value is less than 85 is (Round to four decimal places as needed.) Part 3 c. What is the probability that the value is more than 130? The probability that the value is more than 130 is enter your response here. Part 4 d. What is the probability that the value is more than 75? The probability that the value is more than 75 is (Round to four decimal…
The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 68.92 79.66 69.56 84.65 80.91 87.03 101.01 98.34 O E Click the icon to view the table of critical t-values. (a) Determine a point estimate for the population mean travel tax. A point estimate for the population mean travel tax is $ (Round to two decimal places as needed.) (b) Construct and interpret a 95% confidence interval for the mean tax paid for a three-day business trip. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to two decimal places as needed.) O A. One can be % confident that the all cities have a travel tax between $ and $ O B. One can be % confident that the mean travel tax for all cities is between $ and $ O C. There is a %…
Workers at a private business was given a grade (A, B, C) based on their yearly performance. The data was recorded along with the department they belong. The results obtained using MINITAB are shown below: Expected counts are printed below observed counts A B C Total 12 8 7 Accounting (12.960) (7.830) (6.210) * 17 10 8 Sales (16.800) (10.150) (8.050) 35 19 11 8 Marketing (18.240) (**) (8.740) 38 Total 48 29 23 100 Chi-sq = 0.0711 + 0.0037 + 0.1005 + 0.0024 + 0.0022 + 0.003 + 0.0317 + 0.0000 +…
You want to assess the differences in frequency counts for four different groups. You know there are 150 people total. Without any additional information about how the data is distributed, what would you use for the "expected frequency" in each group? a.1/4 b.37.5 c.25 d..375
An achievement test in Accounting contained 40 questions. The distribution below summarizes the results of the test. Number of Answers 0-3 4-6 7-9 10-12 13-15 16-18 19-21 22-24 25-27 28-30 Find the mean, median and the mode. f 1 3 12 18 21 19 13 8 4 1 N = 100
The second slide is the first part of the question and the first slide is the second part of the question
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* The table shows a frequency distribution of Mathematic session marks for 120 students in the second stage at engineering faculty. Find the Quartiles of the distribution, then 30 students have marks equal to 65 or lower? Choose either True or : False Mark: 30-39 40-49 50-59 60-69 70-79 80-89 90-100 Number of Students 1 3 11 21 43 32 9 False O True O

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