Guidelines to problemschapter 6

Nutan S. MishraDepartment of Mathematics and Statistics

University of South Alabama

Exercise 6.26Draw a picture of the z-curve before computing each probability.a. 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 b. 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)c. 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) .5d. 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)e. p(z > 4.82 ) = 1-p(z < 4.82) 0 (write the explanations and draw

curve)f. p(z < -6.12) = p( z > 6.12) 0 (same argument as in part c)

Exercise 6.36

Given that x~ N(65, 15)a. p(x<43) = p( z < ) = p(z < -1.47)=p(z > 1.47)b. p(x>74) = p(z > ) = p(z > .6) c. p(x> 56) = p(z > ) = p( z > -.6) = p(z < .6)d. p(x < 71) = p(z < .4)

Draw the picture of z-curve for each of above and then find the probabilities from normal table.

156543

156574

156556

Exercise 6.48X = stress score of a dental patientX ~ N(7.59, .73)a. Percentage of the patients with stress score less than

6.00 = 100* p(x < 6.0) =100* p( z < ) b. = 100* p( 7.0 < x < 8.0) = 100* p( < z < )

c. 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 > )

73.59.70.6

73.59.70.8

73.59.70.7

73.59.70.6

Exercise 6.52X= weight of a hockey puck in ouncesX ~ N(5.75, .11) Specifications given by NHL for weight of the puck

is 5.5<x<6.0Percentage 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

Exercise 6.58

It is given that X~N(550, 75)a. p(X< x )= .0250 = p(Z<z) = .0250First find the value of -zUsing the normal tableThen use the transformation formula to find the corresponding value of xx = µ+ zσ

Exercise 6.58 part bp(X>x) = .9345p(Z>z) = .9345First find the value of -zUsing the normal tableThen use the

transformation formula to find the corresponding value of xx = µ+ zσ