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((x-\mu)/(\sigma)>(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)