For a data set of the pulse rates for a sample of adult females, the lowest pulse rate is 31 31 beats per minute, the mean of the listed pulse rates is x overbar x equals = 71.0 71.0 beats per minute, and their standard deviation is s equals = 12.6 12.6 beats per minute.

Complete question is;

For a data set of the pulse rates for a sample of adult females, the lowest pulse rate is 31 beats per minute, the mean of the listed pulse rates is x over bar equals 71.0 beats per minute, and their standard deviation is s equals 12.6 beats per minute.

a. What is the difference between the pulse rate of 31 beats per minute and the mean pulse rate of the females?

b. How many standard deviations is that [the difference found in part (a)]?

c. Convert the pulse rate of 31 beats per minutes to a z score.

d. If we consider pulse rates that convert to z scores between minus 2 and 2 to be neither significantly low nor significantly high, is the pulse rate of 31 beats per minute significant?

A) 40 beats per minute

B) 3.1746

C) z = -3.17

D) the pulse rate of 31 beats per minute is significantly low.

Step-by-step explanation:

A) The mean pulse rate is given as x bar = 71

Thus difference between this and pulse rate of 31 beats per minute is;

Difference = 71 - 31 = 40 beats per minute

B) number of standard deviations of the difference found in part a is given as;

number = difference/standard deviation(s)

number = 40/12.6

number = 3.1746

C) The z-score is calculated from;

z = (x - xbar)/s

z = (31 - 71)/12.6

z = -3.17

D) from the question we are told that the z scores between minus2 and 2 to be neither significantly low nor significantly high. However, we have a Z-score of -3.17 which doesn't fall into that range. Thus, the pulse rate of 31 beats per minute is significantly low.

y > 1/3x

y < = -x + 4

hope this : )