Question

For a data set of the pulse rates for a sample of adult​ females, the lowest pulse rate is 32 beats per​ minute, the mean of the listed pulse rates is x̄ =71.0 beats per​ minute, and their standard deviation is s =23.2 beats per minute.

a. What is the difference between the pulse rate of 32 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 32 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 32 beats per minute​ significant?

Answer 1

Answer:

Given,

pulse rate = 32

xbar = 71

standard deviation = 23.2

a)

Here the difference between the pulse rate of 32 beats per minute and the mean pulse rate of the​ females = 71 - 32

= 39

b)

Number of standard deviations = 39 / 23.2

= 1.68

c)

Now to give the z score

consider,

z = (x - xbar) / s

substitute values

= (32 - 71) / 23.2

= -1.68

z score = - 1.68

d)

Here the z score is - 1.68 , so we can say that the pulse rate is low & it is not significant.