2.131 GPAs of students. At one university, the students aregiven z-scores at the end of each semester, rather than thetraditional GPAs. The mean and standard deviation ofall students' cumulative GPAs, on which the z-scores arebased, are 2.7 and .5, respectively.a. Translate each of the following z-scores to corre-sponding GPAGPA scores: z = 2.0, z=-1.0, z = .5,z = -2.5.b. Students with z-scores below -1.6 are put on probation.What is the corresponding probationary GPA?c. The president of the university wishes to graduate thetop 16% of the students with cum laude honors and thetop 2.5% with summa cum laude honors. Where (approx-imately) should the limits be set in terms of z-scores? Interms of GPAs? What assumption, if any, did you makeabout the distribution of the GPAs at the university?

Part (a): Translate Z-Scores to GPA ScoresThe formula to convert a z-score to a GPA is:…

Answered: 2.131 GPAs of students. At one…

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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A student graduated from a 4 -year college with an outstanding loan of $9563 where the average debt is $8558 with a standard deviation of $1860 . Another student graduated from a university with an outstanding loan of $12,370 where the average of the outstanding loans was $10,318 with a standard deviation of $2144 Find the corresponding z score for each student. Round z scores to two decimal places. College student: =z University student: =z
Q6
The table below shows the grades of the 20 students of a martial arts class. There are three grades: satisfactory(S), unsatisfactory(U), and incomplete(I). Question: Which measure of variability is appropriate for the variable? IQV Range IQR Standard deviation
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 78.6Mbps. The complete list of 50 data speeds has a mean of x=15.15 Mbps and a standard deviation of s=21.46Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between−2and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 34 and the age of the winner for best actress was 43. For all best actors, the mean age is 42.5 years and the standard deviation is 6.2 years. For all best actresses, the mean age is 31.9 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Question content area bottom Part 1 Since the z score for the actor is z=enter your response here and the z score for the actress is z=enter your response here, the ▼ actress actor had the more extreme age. (Round to two decimal places.)
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 33 and the age of the winner for best actress was 47. For all best actors, the mean age is 44.4 years and the standard deviation is 6.5 years. For all best actresses, the mean age is 32.8 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z= and the z score for the actress is z = the V had the more extreme age. (Round to two decimal places.)
Answer the letter E only. Thank you!
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 77.3 Mbps. The complete list of 50 data speeds has a mean of x = 18.24 Mbps and a standard deviation of s = 17.77 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between -2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? a. The difference is (Type an integer or a Mbps. decimal. Do not round.) b. The difference is (Round to two decimal standard deviations. places as needed.) c. The z score is z = (Round to two decimal places as needed.) d. The carrier's highest data speed is C
one z-score unit is equivalent to ______ of the distribution. a. the variance b. the range c. the standard deviation d. the mean e. the correlation coefficient
The following table gives the marks of a sample of students in Statistics. Marks No. of Students 40 to less than 50 50 to less than 60 12 60 to less than 70 22 70 to less than 80 34 80 to less than 90 39 90 to less than 100 19 The mean is: Select one: 71.75 74.25 75.98 76.27
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 72.6 Mbps. The complete list of 50 data speeds has a mean of x=18.29 Mbps and a standard deviation of s=19.75 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? Question content area bottom Part 1 a. The difference is 54.3154.31 Mbps. (Type an integer or a decimal. Do not round.) Part 2 b. The difference is enter your response here standard deviations. (Round to two decimal places as needed.)
Calculate the mean deviation (from mean) from the following data : Marks : 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of Students : 5 8 15 7 3

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