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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

Respuesta :

Given data :

[tex]\mu=98.25[/tex][tex]\text{Standard deviation , }\sigma=0.73[/tex]

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

Finding probability ,

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

The correct option is (A)

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