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Bluman 5th_Chapter 6 HW Soln

# Bluman 5th_Chapter 6 HW Soln - Ch 6.1#7-49 odd The area is...

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Ch. 6.1 #7-49 odd The area is found by looking up z= 0.75 in Table E and subtracting 0.5. Area = 0.7734- 0.5= 0.2734 The area is found by looking up z= 2.07 in Table E and subtracting from 0.5. Area = 0.5- 0.0192 = 0.4808 The area is found by looking up z=0.23 in Table E and subtracting it from 1 Area = 1-0.5910= 0.4090 The area is found by looking up z= 1.43 in Table E. Area = 0.0764

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15) Between z = 1.05 and z= 1.78 The area is found by looking up the values 1.05 and 1.78 in Table E and subtracting the areas. Area= 0.9625– 0.8531 = 0.1094. The area is found by looking up the values z=1.56 and z=1.83 in Table E and subtracting the areas. Area = 0.0594 - 0.0336 = 0.0258 19) Between z = -1.53 and z= - 2.08 The area is found by looking up the values z=1.53 and z= 2.08 in Table E and subtracting the areas. Area = 0.0630- 0.0188 = 0.0442 . 0.1094 Z=1.78 Z=1.05
The area is found by looking up z= 2.11 in Table E. Area = 0.9826 23) To the right of z=-0.25 The area is found by looking up z= 0.25 in Table E and subtracting it from 1. Area= 1- 0.4013 = 0.5987 For z = - 0.44, the area is 0.3300. For z = 1.92, the area is 1 - 0.9726 = 0.0274 Area = 0.3300 + 0.0274 = 0.3574 Area = 0.7486 - 0.5 = 0.2486

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Area = 0.5 - 0.0582 = 0.4418 Area = 1 - 0.9977 = 0.0023 Area = 0.1131
Area= 0.9591 - 0.0069 = 0.9522 Area = 0.9985 - 0.9279 = 0.0706 Area = 0.9222 For exercises 40 through 45, find the z value that corresponds to the given area. 41)

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Since the z score is on the left side of 0, use the negative z table. Areas in the negative z table are in the tail, so we will use 0.5 - 0.4175 = 0.0825 as the area. The closest z score corresponding to an area of 0.0825 is z=- 1.39. 43) Z=- 2.08, found by using the negative z table. 45) Use the negative z table and 1 - 0.8962 = 0.1038 for the area. The z score is z= -1.26. 47) Find the z values to the left of the mean so that a) 98.87% of the area under the distribution curve lies to the right of it. Using the negative z table, area = 1 - 0.9887 = 0.0113. Hence Z=- 2.28. b) 82.12% of the area under the distribution curve lies to the right of it. Using the negative z table, area = 1 0.8212 = 0.1788. Hence Z=- 0.92.
c) 60.64% of the area under the distribution curve lies to the right of it. Using the negative z table, area = 1 0.6064 = 0.3936. Hence Z=- 0.27. 49) Find two z values, one positive and one negative, so that the areas in the two tails total the following values. a) 5% For total area = 0.05, there will be area = 0.025 in each tail. The z scores are 1.96. b) 10% For total area = 0.10, there will be area = 0.05 in each tail. The z scores are 1.645.

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c) 1% For total area = 0.01, there will be area = 0.005 in each tail. The z scores are z= 2.58. Section 6-2 # 1, 2, 9, 11, 15, 21, 22, 23, 26, 28, 30 1. Admission Charges for Movies The average admission charge for a movie is \$5.81. If the distribution of movie admission charges is approximately normal with a standard deviation of \$0.81, what is the probability that a randomly selected admission charge is less than \$3.50? \$3.50 \$5.81 2.85 \$0.81 X z    then look up z = -2.85 in Table E and you get that the area = 0.0022 or 0.22% ( 2.85) 0.0022 Pz   or 0.22% -2.85 0 2. Teachers’ Salaries The average annual salary for all U.S. teachers is \$47,750. Assume that the distribution is normal and the standard deviation is \$5680. Find the probability that a randomly selected teacher earns a. Between \$35,000 and \$45,000 a year 35,000 47,750 2.24 5680 45,000 47,750 0.48 5680 X z z     ( 2.24 0.62) 0.3156 0.0125 0.3031 or 30.31%  
b. More than \$40,000 a year 40,000 47,750 1.36 5680 X z    ( 1.36) 1 0.0869 0.9131 or 91.31% Pz     c. If you were applying for a teaching position and were offered \$31,000 a year, how would you feel

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Bluman 5th_Chapter 6 HW Soln - Ch 6.1#7-49 odd The area is...

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