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Pulse Rates of Women Women have pulse rates that are

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

Solution for problem 18BSC Chapter 6.5

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 18BSC

Pulse Rates of Women Women have pulse rates that are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute (based on Data Set 1).

a. Find the percentiles P1 and P99.

b. Dr. Puretz sees exactly 25 female patients each day. Find the probability that 25 randomly selected women have a mean pulse rate between 70 beats per minute and 85 beats per minute.

c. If Dr. Puretz wants to select pulse rates to be used as cutoff values for determining when further tests should be required, which pulse rates are better to use: the results from part (a) or the pulse rates of 70 beats per minute and 85 beats per minute from part (b)? Why?

Step-by-Step Solution:

Answer :

Step 1 of 3 :

Given, Women have pulse rates that are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute.

a)

The claim is to find the percentiles P1 and P99

The z-score corresponding to an area of 0.01 and 0.99 is 2.33

The corresponding value is the mean increased by the product of the z-score and the standard deviation

  x = - z 

    = 77.5 - 2.33(11.6)

   = 50.472

  x = + z 

    = 77.5 + 2.33(11.6)

   = 104.528

The z score for 1% is –2.33 which corresponds to a pulse rate of 50.472 beats per minute. The z score for 99% is 2.33 which corresponds to a pulse rate of 104.528 beats per minute.

Step 2 of 3 :

b) Dr. Puretz sees exactly 25 female patients each day.

The claim is to find the probability that 25 randomly selected women have a mean pulse rate between 70 beats per minute and 85 beats per minute.

  Z =

 

Where ,  = 70

                Z =

                    = -0.64

Then, P(z > -0.64)

= 1 - P(z < -0.64)

= 1 - 0.2611  (from area under normal curve table)

=  0.7389

 

Where ,  = 85

                Z =

                    = 0.65

Then, P(z > 0.65)

= 1 - P(z < 0.65)

= 1 - 0.7422  (from area under normal curve table)

= 0.2578

Step 3 of 3

Chapter 6.5, Problem 18BSC is Solved
Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

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