Question

The body temperatures of adults have a mean of 98.6° F and a standard deviation of 0.60° F. If 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.8° F. Hint: You will need to use the sampling distribution of the sample mean. 0.8188 0.9772 0.9360 0.0228

Answer 1

Solution :

Given that ,

mean = = 98.6

standard deviation = = 0.60

n = 36

= 98.6

= / n = 0.60 / 36 = 0.1

P( >98.8 ) = 1 - P( <98.8 )

= 1 - P[( - ) / < (98.8-98.6) /0.1 ]

= 1 - P(z <2 )

Using z table

= 1 - 0.9772

= 0.0228

probability= 0.0228