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Solution: Normal Distributions. Answer the given questions

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

Solution for problem 14BSC Chapter 2.2

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 14BSC

Normal Distributions. In Exercises 11–14, answer the given questions which are related to normal distributions.

Normal Distribution Refer to the frequency distribution given in Exercise 10 and ignore the given frequencies. Assume that the first two frequencies are 2 and 6, respectively. Assuming that the distribution of the 40 sample values is a normal distribution, identify the remaining four frequencies.

Step-by-Step Solution:

Answer :

Step 1 of 2 :

Given, the first two frequencies are 2 and 6, respectively and the distribution of the 40 sample values is a normal distribution,

The claim is to identify the remaining four frequencies.

We know that the lower two frequencies must mirror the first two frequencies.

The two frequencies in the middle will be peak frequencies. To find the peak frequencies add the known frequencies 2+6+6+2 = 16.

We have to subtract 16 by total number of frequencies i.e, 40 - 16 = 24.

To find the two middle frequencies divide 24 by 2

Therefore 12 and 12 are the middle two frequencies.

The remaining four frequencies are 12 , 12 , 6 , 2.

Step 1 of 2 :

class width : the difference between the upper and lower class limit.

Example : 4.9 - 3

                  = 1.9

class midpoints : (lower class limit + upper class limit)/ 2 

Example : (3+4.9)/2

              = 3.95

 class boundaries :

Upper class boundary = Upper limit + ()  unit of measurement

Ex: 4.9 is the upper class limit and unit of measurement is 0.1

Therefore ,

Upper class boundary = 4.9 + ()  0.1

                                    = 4.95

Lower class boundary = Lower limit + ()  unit of measurement

Ex: 4.9 is the upper class limit and unit of measurement is 0.1

Therefore ,

Upper class boundary = 3 + ()  0.1

                                    = 3.05

White blood cell count of female

Class width

Class midpoints

Class boundaries

3 - 4.9

1.9

3.95

(3.05 , 4.95)

5 - 6.9

1.9

5.95

(5.05 , 6.95)

7 - 8.9

1.9

7.95

(7.05 , 8.95)

9 - 10.9

1.9

9.95

(9.05 , 10.95)

11 - 12.9

1.9

11.95

(11.05 , 12.95)

13 - 14.9

1.9

13.95

(13.05 , 15.95)

Step 2 of 1

Chapter 2.2, Problem 14BSC is Solved
Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

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Solution: Normal Distributions. Answer the given questions