
Jon S. answered 03/18/21
Patient and Knowledgeable Math and English Tutor
unusual would be a value of pulse rate (x) that is 2 standard deviations above the mean.
z -score for x would be the number of standard deviations above or below the mean
z-score = (x - mean)/standard deviation
for x = 141.1, mean = 80.1 and standard deviation = 10.5
z = (141.1 - 80.1)/10.5 = 5.8
so value of 141.1 is 5.8 standard deviations above the mean which would be very unusual.
max usual value is 2 standard deviations above the mean, so value of x that corresponds to 2 standard deviations above the mean has z score = 2, so
2 = (x - 80.1)/10.5
x = 101.7 is the maximum usual value.