1 / 7

Guidelines to problems chapter 6

Guidelines to problems chapter 6. Nutan S. Mishra Department of Mathematics and Statistics University of South Alabama. Exercise 6.26. Draw a picture of the z-curve before computing each probability. p(z > -1.06)= area to the right side of z= -1.06

nemo
Download Presentation

Guidelines to problems chapter 6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Guidelines to problemschapter 6 Nutan S. Mishra Department of Mathematics and Statistics University of South Alabama

  2. Exercise 6.26 Draw a picture of the z-curve before computing each probability. • p(z > -1.06)= area to the right side of z= -1.06 = p(z < 1.06)= which is same as area to the left of z= 1.06 • p(-.68  z  1.84) = area between -.68 and 1.84. Since -.64 is on the left of 0 and 1.84 is to the right of 0, we add the two areas to compute the total area = p(0< z < 1.84) + p( 0< z <.68) • p(0  z  3.85) = area between 0 and 3.85.Since area covered between 0 and 3.09 is .4990 which is close to .5 We conclude that area between 0 and 3.85 is almost .5 and hence p(0  z  3.85)  .5 • p(-4.34 z  0) = area between 0 and -4.34 =p (0  z  4.34)= which is same as area between 0 and 4.34  .5 ( by the same argument as in part c) • p(z > 4.82 ) = 1-p(z < 4.82)  0 (write the explanations and draw curve) • p(z < -6.12) = p( z > 6.12)  0 (same argument as in part c)

  3. Exercise 6.36 Given that x~ N(65, 15) • p(x<43) = p( z < ) = p(z < -1.47)=p(z > 1.47) • p(x>74) = p(z > ) = p(z > .6) • p(x> 56) = p(z > ) = p( z > -.6) = p(z < .6) • p(x < 71) = p(z < .4) Draw the picture of z-curve for each of above and then find the probabilities from normal table.

  4. Exercise 6.48 X = stress score of a dental patient X ~ N(7.59, .73) • Percentage of the patients with stress score less than 6.00 = 100* p(x < 6.0) =100* p( z < ) • = 100* p( 7.0 < x < 8.0) = 100* p( < z < ) • A patient needs sedative if her stress score is more than 9.0 Percentage of the patients that would need sedative 100* p(x > 9.0) = 100*p( z > )

  5. Exercise 6.52 X= weight of a hockey puck in ounces X ~ N(5.75, .11) Specifications given by NHL for weight of the puck is 5.5<x<6.0 Percentage of pucks can not be used by NHL= percentage of the pucks falling outside the specification limits = 100* p(x<5.5 or x>6.0) =100* {p(x<5.5) + p(x>6.0) =100* {p(z<____) +p(z> ____) Complete this problem

  6. Exercise 6.58 It is given that X~N(550, 75) • p(X< x )= .0250 = p(Z<z) = .0250 First find the value of -z Using the normal table Then use the transformation formula to find the corresponding value of x x = µ+ zσ

  7. Exercise 6.58 part b p(X>x) = .9345 p(Z>z) = .9345 First find the value of -z Using the normal table Then use the transformation formula to find the corresponding value of x x = µ+ zσ

More Related