Question

It is known that the average body temperature of healthy individuals is is normally distributed with a mean, μ = 37 ^oC and standard deviation, σ = 0.6 ^oC.  

We found the average temperature of a random sample of 24 SIBT students in a Statistics tutorial.

1. What is the probability that the average body temperature of this sample of students will be less than 37.4 ^oC? (4 dp) Answer   
2. What is the probability that average body temperature of these students will be more than 37.1 ^oC?  (4 dp)

Answer 1

Solution :

Given that,

mean = = 37

standard deviation = = 0.6

n = 24

= 37

=  / n = 0.6 / 24=0.1225

P( < 37.4) = P[( - ) / < (37.4-37) / 0.1225]

= P(z < 3.27)

Using z table  

= 0.9995   

probability=0.9995

(B)

P( >37.1 ) = 1 - P( <37.1 )

= 1 - P[( - ) / < (37.1-37) / 0.1225]

= 1 - P(z <0.82 )

Using z table

= 1 - 0.7939

= 0.2061

probability= 0.2061