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Solution: Standard Normal Distribution. In Exercises, assume

Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola ISBN: 9780321836960 18

Solution for problem 20BSC Chapter 6.2

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola

Elementary Statistics | 12th Edition

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Problem 20BSC

Standard Normal Distribution. In Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph and find the probability of the given scores. If using technology instead of Table A-2, round answers to four decimal places.

Less than 1.96

Step-by-Step Solution:

Answer:

Step 1:

     Assume that a randomly selected subject is given a bone density. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. 

    Given the probability of score is less than 1.96.

     

   The type of test is determined by the Alternative Hypothesis ( H1 ):

   Left tailed test: 

        parameter < value.

       Decision rule:  Reject H0 if test statistic. < critical value.

      The graph of area under normal curve is given below:

       

     

         

    In finding the area under standard normal curve, you can use the statistical tables at the back of your textbook, or use your calculator.

     In using the tables, remember that areas are reported as P(Z < a)

Therefore, you need to use the following relationships

         P(Z > a) = 1 - P(Z < a)

        P(a < Z < b) = P(Z < b) - P(Z < a)

     Applying this principles, we get

  P(Z < 1.96) = 0.9750.

 This value from area under normal curve, here we have to take 1.9  row and 0.06 column. Then we get 0.9750. The shaded region is 0.9750.

Step 2 of 1

Chapter 6.2, Problem 20BSC is Solved
Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

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Solution: Standard Normal Distribution. In Exercises, assume