Question 263045
Assume that the mean systolic blood pressure of normal 
adults is 120 mmHg and the standard deviation is 5.6. 
Assume a normal distribution. Thats one part. 
Then they are asking if an individual is selected find the probability that the indivdual's pressure will be between 120 and 121.8 mmhg.
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Find the z-value of 120 and of 121.8
z(120) = (120-120)/5.6 = 0
z(121.8) = (121.8-120)/5.6 = 1.8/5.6 = 0.3214..
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P(120<= x <=121.8) = P(0<= z <=0.3214) = 0.1260,,
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Cheers,
Stan H.