The body temperature of adults is normally distributed with a mean of 98.6°F and standard deviation...

The body temperature of adults is normally distributed with a mean of 98.6°F and standard deviation of 0.60°F. Suppose 36 adults are randomly selected, what is the probability that their mean body temperature is less than 98.0°F?

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Expert Solution

Solution :

Given that ,

mean = = 98.6

standard deviation = = 0.60

n = 36

= 98.6

= / n = 0.60 / 36=0.1

P( < 98.0) = P[( - ) / < (98.0-98.6) / 0.1]

= P(z <-6 )

Using z table

= 0.0000

probability= 0.0000

orchestra answered 1 year ago

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