Question

Resting body temperatures are normally distributed with an average of 98.25 degreesFahrenheit and a standard deviation of 0.73 degrees Fahrenheit. What is the probability a random patient examined by a nurse will have a body temperature greater than 99 degrees Fahrenheit? (Round to the hundredths)A)0.15B)0.85C)96.55D)99.94

Answer 1

Given data :

$\mu=98.25$$\text{Standard deviation , }\sigma=0.73$

To find : Probability a random patient examined by a nurse will be greater that 99 degree .

Finding probability ,

$P(X>99)=P(\frac{x-\mu}{\sigma}>\frac{99-98.25}{0.73})$$P(X>99)=P(z>1.027)$$P(X>99)=0.15\text{ (from z-table)}$

The correct option is (A)