Math 11 Elementary Statistics In Class 3.3 Measures of Position (or location) Name Date Example 1: Computing and interpreting z- Scores A National Center for Health Statistics study states that the mean height for adult men in the United States is u = 69.4 inches, with a standard deviation of o = 3.1 inches. The mean height for adult women is u = 63.8 inches, with a standard deviation of o = 2.8 inches. Who is taller relative to their gender, a man 73 inches tall, or a woman 68 inches tall? Work through this problem in steps: The man's height is X = The men's mean is = The men's standard deviation is The man's z-score is z = (x - p)/o = ( _- So standard deviations from the mean Zman = The woman's height is x = The women's mean is The women's standard deviation is The woman's z-score is z = (x - p)/o = (___- standard deviations from the mean So Zwoman %3D Which one is larger? Example 2: 0.00, 0.08, 0.13, 0.14, 0.16, 0.17, 0.20, 0.29, .056, 0.70, 0.79 L = .25(11) = 2.75 which rounds to 3, so the 3rd number is the 25th percentile and its value is L = .50(11) = 5.5 which rounds to 6, so the 6th number is the 50th percentile, or the median and its value is L = .75(11) = 8.25 which rounds to 9, so the 9th number is the 75th

Math 11 Elementary Statistics In Class 3.3 Measures of Position (or location) Name Date Example 1: Computing and interpreting z- Scores A National Center for Health Statistics study states that the mean height for adult men in the United States is u = 69.4 inches, with a standard deviation of o = 3.1 inches. The mean height for adult women is u = 63.8 inches, with a standard deviation of o = 2.8 inches. Who is taller relative to their gender, a man 73 inches tall, or a woman 68 inches tall? Work through this problem in steps: The man's height is X = The men's mean is = The men's standard deviation is The man's z-score is z = (x - p)/o = ( _- So standard deviations from the mean Zman = The woman's height is x = The women's mean is The women's standard deviation is The woman's z-score is z = (x - p)/o = (___- standard deviations from the mean So Zwoman %3D Which one is larger? Example 2: 0.00, 0.08, 0.13, 0.14, 0.16, 0.17, 0.20, 0.29, .056, 0.70, 0.79 L = .25(11) = 2.75 which rounds to 3, so the 3rd number is the 25th percentile and its value is L = .50(11) = 5.5 which rounds to 6, so the 6th number is the 50th percentile, or the median and its value is L = .75(11) = 8.25 which rounds to 9, so the 9th number is the 75th

Name: Math 11 Elementary StatisticsIn Class 3.3Measures of Position (or loc
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Math 11 Elementary StatisticsIn Class 3.3Measures of Position (or location)NameDateExample 1: Computing and interpreting z-ScoresA National Center for Health Statistics study states that the mean height foradult men in the United States is u = 69.4

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

estion Help In a psychology journal, researchers reported that a chief executive officer's facial structure can be used to predict a firm's financial performance. The study involved measuring the facial width-to-height ratio (WHR) for each in a sample of 51 CEOS at publicly traded Fortune 500 firms. These WHR values (determined by computer analyzing a photo of the CEO's face) had a mean of x = 1.81 and a standard deviation of s = 0.12. Use this information to complete parts a and b below. a. Find and interpret a 90% confidence interval for u, the mean facial WHR for all CEOS at publicly traded Fortune 500 firms. (Round to three decimal places as needed.) O A. The 90% confidence interval for mean facial WHR indicates that the true mean facial WHR for the CEOS in the sample is between and with 90% confidence. O B. The 90% confidence interval for mean facial WHR indicates that 90% of all CEOS at publicly traded Fortune 500 firms have a WHR between and O C. The 90% confidence interval for…
Consider the following sample data. Sample A: 5, 8, 11 Sample B: 64, 67, 70 Sample C: 1,015; 1,018; 1,021 (a) Find the mean and standard deviation for each sample. (b) What does this exercise show about the standard deviation? multiple choice The idea is to illustrate that the standard deviation is not a function of the value of the mean. The idea is to illustrate that the standard deviation is a function of the value of the mean.
open file attached for question
hi -looking for the answer to this item
A successful basketball player has a height of 6 feet 11 inches, or 211 cm. Based on statistics from a data set, his height converts to the z score of 5.17. How many standard deviations is his height above the mean?
Answer and provide solution
Recently, a random sample of 25-34 year olds was asked, "How much do you currently have in savings, not including retirement savings?" The data in the table represent the responses to the survey. Approximate the mean and standard deviation amount of savings. EEB Click the icon to view the frequency distribution for the amount of savings Frequency distribution of amount of savings The sample mean amount of savings is $ (Round to the nearest dollar as needed ) Savings $0-$199 Frequency 344 The sample standard deviation is $ $200-$399 93 (Round to the nearest dollar as needed.) 65 $400-$599 $600-$799 24 $800-$999 10 $1000-$1199 $1200-$1399 1 Print Done %23 e here to search lyi
Adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the height of a man with a z-score of 3.2 (to 2 decimal places) Question Help: Submit Question Video
z Scores LeBron James, one of the most successful basketball players of all time, has a height of 6 feet 8 inches, or 203 cm. Based on statistics from Data Set 1 “Body Data” in Appendix B, his height converts to the z score of 4.07. How many standard deviations is his height above the mean?
Detecting gypsy moths The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. Each month, an SRS of 50 traps is inspected, the number of moths in each trap is recorded, and the mean number of moths is calculated. Based on years of data, the distribution of moth counts is discrete and strongly skewed with a mean of 0.5 and a standard deviation of 0.7. a. Explain why it is reasonable to use a Normal distribution to approximate the sampling distribution of for SRSS of size 50. b. Estimate the probability that the mean number of moths in a sample of size 50 is greater than or equal to 0.6. c. In a recent month, the mean number of moths in an SRS of size 50 was = 0.6. Based on this result, is there convincing evidence that the moth population is getting larger in this state? Explain your reasoning.
The histogram below shows the ages (in years) of Honda Civic automobiles listed for sale on the Seattle Times Web site on November 26, 2003. 20 15 10 2 4 6 8 10 12 14 16 18 Age (years) Which of the following terms apply to this histogram? (Select all that apply.) O negatively skewed(to the left) O uniform O symmetric O bimodal O unimodal O positively skewed(to the right) Which measure of center is most appropriate for this data? O median O mean Which measure of spread is most appropriate for this data? O standard deviation M # of cars