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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

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Guidelines to problems chapter 6

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  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

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