Answer to Question #268304 in Statistics and Probability for ggggg

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,

what percent of the patients

a) has blood glucose level higher than 100 mg/dL?

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

c) has blood glucose level between 120 and 125 mg/dL?

Answer 1

a)

$$P(X>100)=1-P(Z\leq\dfrac{100-98}{10})$$

$$=1-P(Z\leq0.2)\approx0.420740$$

$42.0740\%$

b)

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

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

$1.3903\%$

c)

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

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

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

$$\approx0.9965330-0.9860966$$

$$\approx0.010436$$

$1.0436\%$