×
Log in to StudySoup
Get Full Access to Elementary Statistics - 12 Edition - Chapter 6.2 - Problem 26bsc
Join StudySoup for FREE
Get Full Access to Elementary Statistics - 12 Edition - Chapter 6.2 - Problem 26bsc

Already have an account? Login here
×
Reset your password

Standard Normal Distribution. In Exercises, | Ch 6.2 - 26BSC

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

Solution for problem 26BSC Chapter 6.2

Elementary Statistics | 12th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola

Elementary Statistics | 12th Edition

4 5 1 305 Reviews
22
2
Problem 26BSC

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 1.23 and 2.37

Step-by-Step Solution:
Step 1 of 3

Problem 26BSC

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 1.23 and 2.37.

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(1.23 < z < 2.37)                    

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

                    = P(z < 2.37) - P(z < 1.23)

area under normal curve can be calculated by using standard normal table or statistical table that is P(z < 2.37) = 0.9911 (i.e, row 2.3 under column 0.07)

   P(z < 1.23) = 0.8907 (i.e, row 1.2 under column 0.03)                    

consider,

P(1.23 < z < 2.37) = P(z < 2.37) - P(z < 1.23)

                               = 0.9911 - 0.8907

                               = 0.1004.

Therefore, the area of the shaded region is 0.1004.

Step 2 of 3

Chapter 6.2, Problem 26BSC is Solved
Step 3 of 3

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

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Standard Normal Distribution. In Exercises, | Ch 6.2 - 26BSC