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Standard Normal Distribution. In Exercises, | Ch 6.2 - 27BSC

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

Solution for problem 27BSC 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 27BSC

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.

Between and − 2.75 and − 2.00

Step-by-Step Solution:
Step 1 of 3

Problem 27BSC

Answer:

Step1 of 3:

We have In Exercises, 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.

Between and −2.75 and −2.00.

Step2 of 3:

A graph of area under normal curve is given below

Step3 of 3:

From the above graph we have P(p < z <q ).

P(p < z <q ) = P(-2.75 < z < -2.00)                    

                    = P(z < q) - P(z < p)

                    = P(z < -2.75) - P(z < -2.00)

Area under normal curve can be calculated by using standard normal table or statistical table that is P(z < -2.00) = 0.0228 (i.e, row -2.0 under column 0.00)

   P(z < -2.75) = 0.0030(i.e, row -2.75 under column 0.05)                    

consider,

P(-2.75 < z < -2.00) = P(z < -2.00) - P(z < -2.75)

                               = 0.0228 - 0.0030

                               = 0.0198.

Therefore, the area of the shaded region is 0.0198.

Step 2 of 3

Chapter 6.2, Problem 27BSC is Solved
Step 3 of 3

Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

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Standard Normal Distribution. In Exercises, | Ch 6.2 - 27BSC