SOLUTION: assuming that the mean systolic blood pressure of a normal adult is 120 millimeters of mercury(mmHg)and the standard deviation is 5.6. assume the variable is normally distributed,
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Question 465156: assuming that the mean systolic blood pressure of a normal adult is 120 millimeters of mercury(mmHg)and the standard deviation is 5.6. assume the variable is normally distributed,
i). if an individual is selected, find the probability the individual's pressure will be between 120 and 121.8 mmHg
ii). if a sample of 30 adults is randomly selected, find the probability that the sample mean will be between will be between 120 and 121.8 mmHg
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
i)z=(121.8-120)/5.6=.321
The area under the normal curve between 0 and .321 is .126 the probability the individual's pressure will be between 120 and 121.8 mm Hg.
.
ii) (121.8-120)/(5.6/sqrt(30))
=(1.8/1.022)
=1.761
The area under the normal curve between 0 and 1.761 is .461, the probability that the sample mean will be between will be between 120 and 121.8 mm Hg.
.
Ed
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