The body temperatures of adults are normally distributed with a mean of 98.6° F and a...

The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.50° F. If 25 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° F.

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Expert Solution

Solution :

Given that,

mean = = 98.60 F

standard deviation = = 0.500 F

n = 25

= = 98.60 F

= / n = 0.50 / 25 = 0.10

P( > 98.40 F) = 1 - P( < 98.40 F)

= 1 - P[( - ) / < (98.4 - 98.6) / 0.10]

= 1 - P(z < -2.00)

Using z table,

= 1 - 0.0228

= 0.9772

orchestra answered 1 year ago

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