Suppose X and Y have a bivariate normal distribution with ox = 0.04, y = 0.08, x = 3.00, y = 7.70, and p = 0.Determine the following. Round your answers to three decimal places (e.g. 98.765).(a) P (2.98 < X < 3.05) = i(b) P (7.60

Answered: Image /qna-images/answer/9832ae84-5d73-4c7f-919d-a4a5730df59b.jpg

Answered: Suppose X and Y have a bivariate normal…

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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Suppose the scores on a standardized test are normally distributed with mean μ=340μ=340 and the standard deviation σ=32σ=32. Use the normal distribution to find: The lowest test score in the top 15% bracket.Ans: The highest score in the bottom 20% bracket.Ans:
Consider the production data shown in this example. (Assume a normal distribution. Due to the nature of this problem, (a) What is the mean (in minutes)? x = 9.22 minutes (b) What is the standard deviation (in minutes)? 515.16 S = Z = X-X S 99 (c) Determine the z value corresponding to a 5-minute-assembly time. 5- 9.22 = -1.85 2.2811 = 2.2811 minutes (d) Referring to this table, determine the A value. A = (e) Determine the probability that it will take a person longer than 5 minutes to assemble the computer parts. 0.5 +
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T3.6
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The undergraduate grade point averages (UGPA) of students taking an admissions test in a recent year can be approximated by a normal distribution, as shown in the figure. (a) What is the minimum UGPA that would still place a student in the top 15% of UGPAs? (b) Between what two values does the middle 50% of the UGPAs lie? 3.3042.6Grade point average μ=3.30 σ=0.16 x A normal curve labeled mu = 3.30 and sigma = 0.16 is over a horizontal x-axis labeled Grade point average from 2.6 to 4 in increments of 0.35 and is centered on 3.30. (a) The minimum UGPA that would still place a student in the top 15% of UGPAs is enter your response here. (Round to two decimal places as needed.)
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Suppose X and Y have a bivariate normal distribution with ox = 0.04, y = 0.08, x = 3.00, y = 7.70, and p = 0. Determine the following. Round your answers to three decimal places (e.g. 98.765). (a) P (2.98 < X < 3.05) = i (b) P (7.60 < Y < 7.80) = (c) P (2.98 < X < 3.05, 7.60 < Y < 7.80) = IN