Answer to Question #270350 in Statistics and Probability for Lara

Question

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)?

Answer 1

a) $P(X>102)=P(Z>\frac{102-98}{10})=P(Z>0.4)=0.3446.$

b) $P(120<X<125)=P(\frac{120-98}{10}<Z<\frac{125-98}{10})=P(2.2<Z<2.7)=$

$=P(Z<2.7)-P(Z<2.2)=0.9965-0.9861=0.0104.$

c) $P(X>120)=P(Z>\frac{120-98}{10})=P(Z>2.2)=0.0139.$

$N=1500*0.0139\approx 21.$