Answer
a)$-15.3bpm.$
b)-1.49
c)-1.49
d)usual.
Work Step by Step
a) Difference between the student's pulse rate and the mean of the male pulse rates: $52bpm-67.3bpm=-15.3bpm.$
b)$z=\frac{x-\mu}{\sigma}=\frac{52bpm-67.3bpm}{10.3bpm}=-1.49.$
Hence the student's pulse rate is 1.49 standard deviations from the mean of the male pulse rates.
c)$z=\frac{x-\mu}{\sigma}=\frac{52bpm-67.3bpm}{10.3bpm}=-1.49.$
d) The z-score of the student's pulse rate is between -2 and 2, hence it is usual.
