SOLUTION: Totally lost on this. 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
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-> SOLUTION: Totally lost on this. 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
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Question 263045: Totally lost on this. 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.
Any clarification for these would be great. Thanks for all your help Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 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.
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