Answer:
[tex]z=-1.49[/tex]
Step-by-step explanation:
[tex]\text{Standard Score, z} =\dfrac{X-\mu}{\sigma} $ where:\\\\Mean Pulse rate, \mu =67.3$ beats per minute\\Standard Deviation, \sigma = 10.3$ beats per minute.\\[/tex]
For a male student who has a measured pulse rate of 52 beats per minute.
Raw Score, X =52 beats per minute.
Therefore:
[tex]\text{Standard Score, z} =\dfrac{52-67.3}{10.3}\\z=-1.49[/tex]
Since the usual pulse rates are within 2 standard deviations of the mean, a z-score of -1.49 tells us that the selected student's pulse rate is within the usual pulse rates.