The body temperatures of adults are normally distributed with a mean of 98.6 degrees Fahrenheit and a standard deviation of 1.2 degrees Fahrenheit. If 31 adults are randomly selected, find the probability that their mean body temperature is greater than 99.3 degrees Fahrenheit.
Solution :
Given that ,
mean = = 98.6
standard deviation = = 1.2
_{} = / n = 1.2 / 31 = 0.2155
P( > 99.3) = 1 - P( < 99.3)
= 1 - P[( - _{} ) / _{} < (99.3 - 98.6) / 0.2155]
= 1 - P(z < 3.25)
= 1 - 0.9994
= 0.0006
Probability = 0.0006
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