Answer to Question #269898 in Statistics and Probability for ggggg

Question

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)?

Answer 1

a)

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

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

$34.4578\%$

b)

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

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

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

$$\approx0.9965330-0.9860966$$

$$\approx0.010436$$

$1.0436\%$

c)

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

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

$$0.013903(1500)=21$$

21 patients.