Question #269898

2. A monthly record of glucose level of patients was recorded and found to have a mean of


98 mg/dL and a standard deviation of 10 mg/dL. If the data is normally distributed,


a) what percent of the patients has blood glucose level higher than 102 mg/dL?


b) what percent of the patients has blood glucose level between 120 and 125 mg/dL?


c) Assuming there are 1,500 patients tested, how many of the patients would have


glucose level above normal (glucose level ≥ 120 mg/dL)?

1
Expert's answer
2021-11-22T19:09:39-0500

a)


P(X>102)=1P(Z1029810)P(X>102)=1-P(Z\leq\dfrac{102-98}{10})

=1P(Z0.4)0.344578=1-P(Z\leq0.4)\approx0.344578


34.4578%34.4578\%


b)


P(120<X<125)=P(X<125)P(X120)P(120<X<125)=P(X<125)-P(X\leq120)

=P(Z<1259810)P(Z1209810)=P(Z<\dfrac{125-98}{10})-P(Z\leq\dfrac{120-98}{10})

=P(Z<2.7)P(Z2.2)=P(Z<2.7)-P(Z\leq2.2)

0.99653300.9860966\approx0.9965330-0.9860966

0.010436\approx0.010436

1.0436%1.0436\%


c)


P(X120)=1P(Z<1209810)P(X\geq120)=1-P(Z<\dfrac{120-98}{10})

=1P(Z<2.2)0.013903=1-P(Z<2.2)\approx0.013903

0.013903(1500)=210.013903(1500)=21

21 patients.


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